NCERT Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions

Important Points

1. Sequence : An arrangement of numbers in a definite order according to some rule is called a sequence. e.g., 1, 2, 3, 4, 5, 6, ... 2, 4, 6, 8, 10, ... 2, 4, 8, 16, 32, ... Sequence is said to be finite or infinite accordingly it has a finite or infinite number of terms. Consider a sequence a₁, a₂, a₃, a₄,...... Then, the first term is a1, second term is a₂, third term is a₃ and so on. Its nth term can be written as an which is known as the general term of the sequence.

2. Arithmetic Progression : A succession of number is said to be an Arithmetic Progression (AP), if each term of the sequence can be obtained by adding a fixed numbered to the preceding term except the first term a. The fixed numbered may be positive, negative or zero and is called the common difference of AP.
The
general term of an AP is a, a + d, a + 2d, a + 3d,...
Examples :
(i) 3, 7, 11, 15, 19, ...
(ii) 15, 12, 9, 6, 3, ...
(iii) 2, 2, 2, 2, ....
(iv) – 2.5, – 3, – 3.5, – 4, – 4.5, ....
Note : A given list of numbers a₁, a₂, a₃ ,..... is an AP, if the difference a₂ – a₁, a₃, give the same value.

3. nth term of an AP : The nth term an (or the general term) of an AP is a_n = a + (n – 1)d, where a is the first term and d is the common difference.

4. nth term of an AP from the End : The nth term a_n (or the general term) of an AP from the end an = l – (n – 1)d, where l is the last term and d is the common difference.

5. Sum of n terms of an AP : The sum Sn of the first n terms of an AP is given by
$$s_n=\frac{n}{2}[2a+(n-1)d]$$
Note : If l is the last term of the finite AP say the nth term, then the sum of all terms of the AP is given by
$$s_n=\frac{n}{2}[a+l]$$

Exercise 5.1

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why ?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.

Sol. (i) According to question, the fare for journey of first km, i.e., 1 km is ₹ 15 and next 2 km, 3 km, 4 km ....... are respectively ₹ (15 + 8),
₹ (15 + 2 × 8), ₹ (15 + 3 × 8), .... so on.
i.e., 15, 23, 31, 39, .....
Here, each term is obtained by adding 8 to the preceding term except first term. So, it is an AP.

(ii) Let the amount of air present in the cylinder be y units.
So, according to question, the terms giving the air present in the cylinder is given by
$$y,y-\frac{y}{4}=\frac{3y}{4},\frac{3y}{4}-\frac{1}{4}×\frac{3y}{4}\\ory,\frac{3y}{4},\frac{9y}{4}....\\\text{Here,}\frac{3y}{4}-y=-\frac{y}{4} (T2 – T1)\\\text{and}\frac{9y}{16}-\frac{3y}{4}=\frac{9y-12y}{16}\\=-\frac{3y}{16} (T3 – T2)\\⇒\frac{3y}{4}-y≠\frac{9y}{16}-\frac{3y}{4}\\⇒ \text{It does not form an AP, because common difference is not same.}$$

(iii) According to question , the cost of digging for the first metre, second metre, third metre and so on are respectively ₹ 150, ₹ (150 + 50), ₹ (200 + 50), .... and so on.
i.e., 150, 200, 250 ,.....
Here, each term is obtained by adding 60 to the preceding term except first term. So, it is an AP.

(iv) According to question, the amount of money in the account in the first year, second year, third year and so on are respectively.
$$10000,10000(1+\frac{8}{100}),10000(1+\frac{8}{100})^2,...\\i.e., 10000,10000×\frac{108}{100},10000×\frac{108}{100}×\frac{108}{100},...\\i.e., 10000, 10800, 11664.\\Now, d = 10800 – 10000 = 800\\and\space d = 11664 – 10800 = 864\\800 ≠ 864\\\text{Since, the common difference are not same. So, it does not form an AP.}$$

2. Write first four terms of the AP, when the first term a and the common difference d are given as follows :
(i) a = 10, d = 10 (ii) a = – 2, d = 0
(iii) a = 4, d = – 3 (iv) a = – 1, d =1/2
(v) a = – 1.25, d = – 0.25

Sol. (i) First term a = 10
The common difference d = 10
a₁ = a = 10
a₂ = a₁ + d = 10 + 10 = 20
a₃ = a₂ + d = 20 + 10 = 30
a₄ = a₃ + d = 30 + 10 = 40
Let us assume that, arithmetic progression series be
a₁, a₂, a₃, a₄, .....
Thus, the first four terms of AP will be
= 10, 20, 30, 40

(ii) First term a = – 2
The common difference d = 0
For the first four terms of an AP
a₁ = a = – 2
a₂ = a₁ + d = – 2 + 0 = – 2
a₃ + a₂ + d = – 2 + 0 = – 2
a₄ = a₃ + d = – 2 + 0 = – 2
Let us assume that, the first four arithemtic progression series will be a₁, a₂, a₃, a₄.
Thus, the first four terms of AP = – 2, – 2, – 2 and –2.

(iii) Given,
First term a = 4
The common difference d = –3
Let us assume that, the first four arithmetic progression series will be a₁, a₂, a₃, a₄.
Now,
a₁ = a = 4
a₂ = a₁ + d = 4 – 3 = 1
a₃ = a₂ + d = 1 – 3 = – 2
a₄ = a₃ + d = – 2 – 3 = – 5
Thus, the first four terms of AP = 4, 1, – 2, – 5.

(iv) Given, a = –1, d = 12
Let us assume that, the first four arithmetic progression series will be a₁, a₂, a₃, a₄.
Now, a₁ = a = –1
$$a_2 = a_1 + d = −1+\frac{1}{2}=-\frac{1}{2}\\a_3 = a_2 + d =-\frac{1}{2}+\frac{1}{2}=0\\a_4 = a_3 + d = 0+\frac{1}{2}+\frac{1}{2}\\\text{Thus, the first four terms of AP =}\\-1,-\frac{1}{2},0,\frac{1}{2}$$

(v) Given,
First term a = – 1.25
d = – 0.25
Let us assume that, the first four arithmetic progression series will be a₁, a₂, a₃, a₄.
Now, a₁ = a = – 1.25
a₂ = a₁ + d = – 1.25 + (– 0.25) = – 1.5
a₃ = a₂ + d = – 1.5 + (– 0.25) = – 1.75
a₄ = a₃ + d = – 1.75 + (– 0.25) = – 2.0
Thus, the first four terms of AP = – 1.25,
– 1.5, – 1.75, – 2.0.

3. For the followng APs, write the first term and the common difference
(i) 3, 1, – 1, – 3, ........
(ii) – 5, – 1, 3, 7, ........
$$\textbf{(iii)}-\frac{1}{3}.\frac{5}{3}.\frac{9}{3}.\frac{13}{3},......$$
(iv) 0.6, 1.7, 2.8, 3.9, ..........

Sol. (i) 3, 1, – 1, – 3, .......
First term a1 = 3 = a
Common difference d = a2 – a1
= 1 – 3
= – 2
Thus, the first term ‘a’ is 3 and the common difference ‘d’ is – 2.

(ii) – 5, – 1, 3, 7, .......
First term a1 = – 5 = a
The common difference d = a2 – a1
= – 1 – (– 5)
= – 1 + 5
= 4
Thus, the first term ‘a’ is – 5 and the common difference ‘d’ is 4.

$$\textbf{(iii)}-\frac{1}{3}.\frac{5}{3}.\frac{9}{3}.\frac{13}{3},......$$
$$\text{First term a}=-\frac{1}{3}$$
The common difference d = Second term
– First term = a₂ – a₁

$$=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}\\\text{Thus, the first term ‘a’ is}\frac{1}{3}\text{and the common differrence ‘d’ is}\frac{4}{3}$$

(iv) 0.6, 1.7, 2.8, 3.9, ..........
First term a1 = 0.6 = a
The common difference d
= second term – first term
= a₂ – a₁
= 1.7 – 0.6
d = 1.1
Thus, the first term ‘a’ is 0.6 and the common difference ‘d’ is 1.1.

4. Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
$$(i) 2, 4, 8, 16, .......\\(ii) 2,\frac{5}{3},3,\frac{7}{2},.....\\(iii) – 1.2, – 3.2, – 5.2, – 7.2, .....\\(iv) – 10, – 6, – 2, 2, ......\\(v) 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},.....\\(vi) 0.2, 0.22, 0.222, 0.2222, .........\\(vii) 0, – 4, – 8, – 12, .......\\(viii)-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},......\\(ix) 1, 3, 9, 27, ......\\(x) a, 2a, 3a, 4a, ......\\(xi) a, a^2, a^3, a^4 , ......\\(xii)\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},......\\(xiii)\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12}......\\(xiv) 1^2, 3^2, 5^2, 7^2, .......\\(xv) 1^2, 5^2, 7^2, 7^3, .......$$

Sol. (i) Here, a₁ = 2, a₂ = 4, a₃ = 8, a₄ = 16
The common difference is
a₂ – a₁= 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
Clearly a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP, because the common difference is not same.

$$\textbf{(ii)} Here, a_1 = 2,a_2 = \frac{5}{2},a_3=3,a_4 =\frac{7}{2},....\\\text{The common difference is}\\a_2 – a_1 =\frac{5}{2}-2=2\frac{5-4}{2}=\frac{1}{2}\\a_3 – a_2 = 3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}\\a_3 – a_3 =\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\\\text{Clearly, the difference of successive terms is constant, therefore list of numbers form an}\\\text{AP. So, common difference, d =}\frac{1}{2}\\\text{Next three terms of AP are}\\a_5 = a_4+d=\frac{7}{2}+\frac{1}{2}=\frac{7+1}{2}=\frac{8}{2}=4\\a_6 = a_5+d=4+\frac{1}{2}=\frac{8+1}{2}=\frac{9}{2}=4.5\\a_7 = a_6+d=\frac{9}{2}+\frac{1}{2}=\frac{9+1}{2}=\frac{10}{2}=5$$

(iii) Here, a₁ = – 1.2, a₂ = – 3.2, a₃ = – 5.2, a₄ – 7.2
The common difference is
a₂ – a₁ = – 3.2 – (– 1.2)
= – 3.2 + 1.2
= – 2
a₃ – a₂ = – 5.2 – (– 3.2)
= – 5.2 + 3.2 = – 2
a₄ – a₃ = – 7.2 – (– 5.2)
= – 7.2 + 5.2
= – 2
Clearly, the difference of successive terms is constant, therefore list of numbers form an AP. So, common difference, d = – 2.
Next three terms of AP are
a₅ = a₄ + d = – 7.2 + (– 2) = – 9.2
a₆ = a₅ + d = – 9.2 + (– 2) = – 11.2
a₇ = a₆ + d = – 11.2 + (– 2) = – 13.2

(iv) – 10, – 6, – 2, 2, ....
Here, a₁ = – 10, a₂ = – 6, a = – 2, a₄ = 2
The common difference is
a₂ – a₁ = – 6 – (– 10) = – 6 + 10 = 4
a₃ – a₂ = – 2 – (– 6) = – 2 + 6 = 4
a₄ – a₃ = 2 – (– 2) = 2 + 2 = 4
Clearly, the difference of successive terms is constant, therefore list of numbers form an AP. So, common difference, d = 4.
Next three terms of AP are
a₅ = a₄ + d = 2 + 4 = 6
a₆ = a₅ + d = 6 + 4 = 10
a₇ = a₆ + d = 10 + 4 = 14

$$\textbf{(v)}3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},....\\\text{Here,}a_1 = 3, a_2 = 3+\sqrt{2},a_3=3+2\sqrt{2},\\a_4 = 3 + 3\sqrt{2}\\\text{The common difference is}\\a_2 – a_1 = 3+\sqrt{2}-\sqrt{3}=\sqrt{2}\\a_3 – a_2 = 3+2\sqrt{2}-(3+\sqrt{2})=\sqrt{2}\\a_4 – a_3 =3+3\sqrt{2}-(3+2\sqrt{2})=\sqrt{2}\\\text{Clearly, the difference of successive terms is constant, therefore list of numbers form an AP. So, common difference,}d=\sqrt{2}\\\text{Next three terms of AP are}\\a_5 – a_4+d = 3+3\sqrt{2}+\sqrt{2}=3+4\sqrt{2}\\a_6 – a_5+d = 3+4\sqrt{2}+\sqrt{2}=3+5\sqrt{2}\\a_7 – a_6+d = 3+5\sqrt{2}+\sqrt{2}=3+6\sqrt{2}$$

(vi) 0.2, 0.22, 0.222, 0.2222, ....
Here, a₁ = 0.2, a₂ = 0.22, a₃ = 0.222, a₄ = 0.2222
The common difference is
a₂ – a₁ = 0.22 – 0.2
= 0.02
a₃ – a₂ = 0.22 – 0.22
= 0.002
a₄ – a₃ = 0.2222 – 0.222
= 0.0002
Clearly, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP, because the common difference is not same.

(vii) 0, – 4, – 8, – 12, ....
Here, a₁ = 0, a₂ = – 4, a₃ = – 8, a₄ = – 12
The common difference is
a₂ – a₁ = – 4 – 0 = – 4
a₃ – a₂ = – 8 – (– 4)
= – 8 + 4 = – 4
a₄ – a₃ = – 12 – (– 8)
= – 12 + 8 = – 4
Clearly, the difference of successive terms is constant, thererfore list of numbers form an AP. So, common difference, d = – 4.
a₅ = a₄ + d
= – 12 + (– 4)
= – 16
a₆ = a₅ + d
= – 16 + (– 4)
= – 20
a₇ = a₆ + d
= – 20 – 4
= – 24

$$\textbf{(viii)}-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},.....\\\text{Here},a1=-\frac{1}{2},a2=-\frac{1}{2},a3=-\frac{1}{2},a4=-\frac{1}{2}\\\text{The common difference is}\\a_2 – a_1 =-\frac{1}{2}-(-\frac{1}{2})\\=-\frac{1}{2}+\frac{1}{2}=0\\a_3 – a_2 =-\frac{1}{2}-(-\frac{1}{2})\\=-\frac{1}{2}+\frac{1}{2}=0\\a_4 – a_3 =-\frac{1}{2}-(-\frac{1}{2})\\=-\frac{1}{2}+\frac{1}{2}=0$$
Clearly, difference of successive terms is constant, therefore list of numbers form an AP. So, common difference, d = 0.
$$a_5 = a_4 + d\\=-\frac{1}{2}+0=-\frac{1}{2}\\a_6 = a_5 + d\\=-\frac{1}{2}+0=-\frac{1}{2}\\a_7 = a_6 + d\\=-\frac{1}{2}+0=-\frac{1}{2}$$

(ix) 1, 3, 9, 27, .......
Here, a₁ = 1, a₂ = 3, a₃ = 9, a₄ = 27
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
a₄ – a₃ = 27 – 9 = 18
Clearly, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not for an AP, because the common difference is not same.

(x) a, 2a, 3a, 4a, ....
Here, a₁ = a, a₂ = 2a, a₃ = 3a, a₄ = 4a
a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃ = 4a – 3a = a
Clearly, difference of successive terms is contant, therefore list of numbrs form an AP. So, common difference, d = a.
Next three terms of AP are
a₅ = a₄ + d = 4a + a = 5a
a₆ = a₅ + d = 5a + a = 6a
a₇ = a₆ + d = 6a + a = 7a

(xi) a, a², a³, a⁴ ,.....
Here, a₁ = a, a₂ = a₂, a₃ = a₃, a₄ = a₄
a₂ – a₁ = a₂ – a = a(a – 1)
a₃ – a₂ = a₃ – a₂ = a₂(a – 1)
a₄ – a₃ = a₄ – a₃ = a₃(a – 1)
Clearly, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP, becaue the common difference is not same.

$$\textbf{(xii)}\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},.....$$
$$\text{Here},a_1=\sqrt{2},a_2=\sqrt{8}=2\sqrt{2},a_3=\sqrt{18}\\=3\sqrt{2},a4=\sqrt{32}=4\sqrt{2}\\\text{The common difference is}\\a_2 – a_1 =2\sqrt{2}-\sqrt{2}=\sqrt{2}\\a_3 – a_2 =3\sqrt{2}-\sqrt{2}=\sqrt{2}\\a_4 – a_3 =2\sqrt{2}-\sqrt{2}=\sqrt{2}\\\text{Clearly, difference of successive terms is constant,}\\\text{ therefore list of numbers form an AP. So, common difference,}\\\space d=\sqrt{2}\text{Next three terms of AP are}\\a_5 = a_4+d=4\sqrt{2}+\sqrt{2}=5\sqrt{2}\\a_6 = a_5+d=5\sqrt{2}+\sqrt{2}=6\sqrt{2}\\a_7 = a_6+d=6\sqrt{2}+\sqrt{2}=7\sqrt{2}$$

$$\textbf{(xiii)}\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},.....$$
$$\text{Here},a_1=\sqrt{3},a_2=\sqrt{6},a_3=\sqrt{9},a_4=\sqrt{12}\\\text{The common difference is}\\a_2 – a_1=\sqrt{6}-\sqrt{3}=\sqrt{3×2}-\sqrt{3}\\=\sqrt{3}×\sqrt{2}-\sqrt{3}=\sqrt{3}(\sqrt{2}-1)\\a_3 – a_2=\sqrt{9}-\sqrt{6}=\sqrt{3}×\sqrt{3}-\sqrt{3×2}\\=\sqrt{3}(\sqrt{3}-\sqrt{2})\\a_4 – a_3=\sqrt{12}-\sqrt{9}\\=\sqrt{4×3}-\sqrt{3}×\sqrt{3}\\=2\sqrt{3}-3\\2×\sqrt{3}-\sqrt{3}×\sqrt{3}\\=\sqrt{3}(2-\sqrt{3})\\\text{Clearly}, a_2 – a_1 ≠ a_3 – a_2\\\text{Hence, the given list of numbers does not form an AP, because the common difference is not same.}$$

(xiv) 1², 3², 5², 7², .....
Here, a₁ = 1², a₂ = 3² = 9, a3 = 5² = 25,
a₄ = 7² = 49
The common difference is
a₂ – a₁ = 9 – 1 = 8
a₃ – a₂ = 25 – 9 = 16
a₄ – a₃ = 49 – 25 = 24
Clearly, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP, because the common difference is not same.

(xv) 1², 5², 7², 73
Here, a₁ = 1² = 1, a² = 5² = 25,
a³ = 7² = 49, a₄ = 73
a₂ – a₁ = 25 – 1 = 24
a₃ – a₂ = 49 – 25 = 24
a₄ – a₃ = 73 – 49 = 24
Clearly, difference of successive terms is constant, therefore list of numbers form an AP. So, common difference, d = 24.
Next three terms of AP are
a₅ = a₄ + d = 73 + 24 = 97
a₆ = a₅ + d = 97 + 24 = 121
a₇ = a₆ + d = 121 + 24 = 145

Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP.

	a	d	n	a _n
(1)	7	3	8	...
(2)	-18	...	10	0
(3)	...	-3	18	-5
(4)	-18.9	2.5	...	3.6
(5)	3.5	0	105	...

Sol. (i) The nth term of an AP is a_n = a + (n – 1)d
= 7 + (8 – 1) 3 = 7 + 7 × 3 = 7 + 21 = 28
a_n = 28

(ii) The nth term of an AP is a_n = a + (n – 1)d
⇒ 0 = – 18 + (10 – 1)d
⇒ 18 = 9d ⇒d=18/2=2
d = 2

(iii) The nth term of an AP is a_n = a + (n – 1)d
⇒ – 5 = a + (18 – 1)(– 3)
⇒ – 5 = a + 17 (– 3)
⇒ – 5 = a – 51
⇒ a = – 5 + 51 = 46
a = 46

(iv) The nth term of an AP is a_n = a + (n – 1)d
⇒ 3.6 = – 18.9 + (n – 1) 2.5
3.6 + 18.9 = (n – 1) 2.5
⇒ 22.5 = (n – 1) 2.5
⇒ n – 1 =(22.5/2.5)
⇒ n – 1 = 9
⇒ n = 9 + 1 = 10
n = 10

(v) The nth term of an AP is a_n = a + (n – 1)d
= 3.5 + (105 – 1) × 0
= 3.5 + 0 = 3.5
an = 3.5

2. Choose the correct choice in the following and justify.

(i) 30th term of the AP : 10, 7, 4, ....., is
(a) 97
(b) 77
(c) – 77
(d) – 87

(ii) 11th term of the AP :-3,\frac{1}{2},2,..., is
(a) 28
(b) 22
(c) – 38
(d) −48(1/2)

Sol. (i) (c) – 77
Explanation: Here, first term a = 10, common difference d = 7 – 10 = – 3, number of terms n = 30
Since, the nth term of an AP is a_n = a + (n – 1)d
⇒ a30 = 10 + (30 – 1)(– 3)
= 10 + 29(– 3)
= 10 – 87 = – 77

Sol. (ii) (b) 22
$$\textbf{Explanation:}\text{Here}\space a = – 3, d = –\frac{1}{2}-(-3)\\=\frac{-1+6}{2}=\frac{5}{2}\\\text{Since, the nth term of an AP is}\\=a_n = a + (a – 1)d\\a_{11} = 3+(11-1)\frac{5}{2}\\= -3+10×\frac{5}{2}\\= – 3 + 25 = 22$$

3. In the following APs, find the missing terms in the boxes.
(i) 2,▢ , 26
(ii) ▢, 13, ▢, 3
(iii) 5,▢ ,▢ ,9(1/2)
(iv) – 4,▢ ,▢ ,▢ ,▢ , 6
(v)▢ , 38, ▢,▢ , ▢, –22

Sol. (i) Let 2, , 26 be a, a + d and a + 2d respectively are in AP.
⇒ a = 2 and a3 = a + 2d = 26
⇒ 2 + 2d = 26 (.. a = 2)
⇒ 2d = 26 – 2 = 24
⇒ d =24/2=12
Hence, the missing term a₂ = a + d = 2 + 12 = 14

(ii) Let , 13, , 3 be a , a + d, a + 2d and a + 3d respectively are in AP
∴ a₂ = a + d = 13 ...(i)
and a₄ = a + 3d = 3 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
2d = – 10 ⇒ d = – 5
Put, d = – 5 in Eq. (ii), we get
a + 3(– 5) = 3
a – 15 = 3
⇒ a₁ = a = 3 + 15 = 18
Hence, the missing term are a = 18 and
a₃ = a + 2d = 18 + 2(– 5)
= 18 – 10 = 8

$$\textbf{(iii)} Let 5,\square,\square,9\frac{1}{2}\text{be}\space a, a + d, a + 2d\space and\space a + 3d\space \text{respectively are in AP}.\\∴ a = 5 ...(i)\\\text{and}\space a + 3d =9\frac{1}{2}=\frac{19}{2}...(ii)\\\text{On subtracting Eq.(i) from Eq. (ii), we get}\\ 3d =\frac{19}{2}-5=\frac{19-10}{2}=\frac{9}{2}\\⇒ 3d =\frac{9}{2}\\⇒d=\frac{1}{3}×\frac{9}{2}=\frac{3}{2}\\\text{Hence, the missing terms are} a2 = a + d = 5 +\\\frac{3}{2}=\frac{10+2}{2}=\frac{13}{2}=6\frac{1}{2}\\\text{and} a_3 = a + 2d = 5 + 2×\frac{3}{2}=5+3=8$$

(iv) Let – 4, , , , , 6 be a, a + d, a + 2d, a + 3d, a + 4d and a + 5d respectively are in AP.
∴ a = – 4 ...(i)
and a6 = a + 5d = 6 ...(ii)
On putting a = – 4 in Eq. (ii), we get
– 4 + 5d = 6
⇒ 5d = 6 + 4 = 10
⇒d=10/2=2
Hence, the missing terms are
a + d = – 4 + 2 = – 2
a + 2d = – 4 + 2 × 2
= – 4 + 4 = 0
a + 3d = – 4 + 3 × 2
= – 4 + 6 = 2
a + 4d = – 4 + 4 × 2
= – 4 + 8 = 4

(v) Let , 38, , , , –22 be a, a + d, a + 2d,
a + 3d, a + 4d and a + 5d respectively are in AP.
∴ a₂ = a + d = 38 ...(i)
and a₆ = a + 5d = – 22 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
4d = – 60
⇒ d =-60/4=-15
On putting d = – 15 in Eq. (i), we get
a + (– 15) = 38
⇒ a – 15 = 38
⇒ a = 38 + 15 = 53
Hence, the missing terms are a = 53
a₃ = a + 2d = 53 + 2 × (– 15)
= 53 – 30 = 23
a₄ = a + 3d = 53 + 3 × (– 15)
= 53 – 45 = 8
and a₅ = a + 4d = 53 + 4 × (– 15)
= 53 – 60 = – 7
Therefore, the missing terms are 53, 23, 8 and – 7 respctively.

4. Which term of the AP : 3, 8, 13, 18, ...., is 78 ?

Sol. Let nth term be 78.
Given, 3, 8, 13, 18, ...., are in AP.
First term, a = 3, common difference d = 8 – 3 = 5
∴ nth term a_n = 78
∴ a + (n – 1)d = an
⇒ 3 + (n – 1)d = 78
⇒ (n – 1)5 = 78 – 3
⇒ (n – 1)5 = 75
⇒ n – 1 = 15
⇒ n = 15 + 1
⇒ n = 16
Hence, 16th term of the AP be 78.

5. Find the number of terms in each of the following APs
(i) 7, 13, 19, ....., 205
(ii) 18,15(1/2),13,...,-47

Sol. (i) Suppose, there are n terms in the given AP. Then, nth term an = 205, first term a = 7, common difference d = 13 – 7 = 6
a + (n – 1)d = an
⇒ 7 + (n – 1)6 = 205
⇒ 6(n – 1) = 205 – 7
⇒ 6(n – 1) = 198
⇒ n – 1 =198/6=33
⇒ n = 33 + 1 = 34
Hence, the given AP contains 34 terms.

(ii) Suppose, there are n terms in the given AP.
Given, first term a = 18, common difference
$$d=15\frac{1}{2}-18=\frac{31}{2}-18\\=\frac{31-36}{2}=\frac{-5}{2}\\\text{Then, nth term an = – 47}\\⇒ 18+(n-1)(\frac{-5}{2})=-47 [.. a + (n – 1)d = a_n]\\⇒ 18+(n-1)(\frac{-5}{2})=-47\\⇒(\frac{-5}{2})(n-1)=– 47 – 18 = – 65\\⇒ (n – 1) =-65×\frac{-2}{5}\\= 13 × 2 = 26\\⇒ n = 26 + 1 = 27\\\text{Hence, the given AP contains 27 terms.}$$

6. Check whether – 150 is a term of the AP : 11, 8, 5, 2, .....

Sol. Here, a₁ = 11, a_2v = 8, a₃ = 5, a₄ = 2
a₂ – a₁ = 8 – 11 = – 3
a₃ – a₂ = 5 – 8 = – 3
a₄ – a₃ = 2 – 5 = – 3
Clearly, the successive difference of terms is constant. So, the list of numbers are in AP. Hence, common difference d = – 3.
Let, – 150 be the nth term of the given AP.
We know that, the nth term of a_n AP is
a_n = a₁ + (n – 1)d
⇒ – 150 = 11 + (n – 1)(– 3)
⇒ – 3(n – 1) = – 150 – 11 = – 161
$$⇒ n – 1 =\frac{161}{3}\\⇒ n =\frac{161}{3}+1=\frac{164}{3}$$
But n should be positive integer. So, our assumption was wrong and so, – 150 is not a term of the given AP.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Sol. Let a be the first term and d be the common difference of an AP.
Now, the nth term of an AP
a_n = a + ( n – 1)d
a₁₁ = a + 10d = 38 [ a₁₁ = 38 given)] ...(i)
and a₁₆ = a + 15 d = 73[ a₁₆ = 73 given)] ...(ii)
On substituting Eq. (i) from Eq. (ii), we get
5d = 35 ⇒ d=35/5=7
From equation (i),
a + 10 × 7 = 38
⇒ a = 38 – 70 = – 32
∴ The 31st term of an AP
a₃₁ = a + 30d
= – 32 + 30 × 7
= – 32 + 210 = 178
Hence, the 31st term of the AP is 178.

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Sol. Let a be the first term and d the common difference
Now, the nth term of AP
a_n = a + (n – 1)d
Given, a₃ = 12 and a₅₀ = 106
a₃ = a + 2d = 12 ...(i)
and a₅₀ = a + 49d = 106 ...(ii)
On subtracting Eq. (i) From Eq. (ii), we get
47d = 94
⇒ d =94/47=2
From equation (i),
a + 2 × 2 = 12
⇒ a = 12 – 4 = 8
∴ The 29th term of an AP
a29 = a + (29 – 1)d
= 8 + 28 × 2
= 8 + 56 = 64

9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

Sol. Let a be the first term and d be the common difference of an AP.
∴ The nth term of an AP is
a_n = a + (n – 1)d
∴ a₃ = a + 2d
= 4 [... a₃ =4 (given)] ...(i)
and a₉ = a + 8d
= – 8 [.. a₉ = –8 (given)] ...(ii)
On substituting Eq. (i) from Eq. (ii), we get
6d = – 12
⇒ d =\frac{-12}{6}=-2
From equation (i),
a + 2 × (– 2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let the nth term of an AP is zero.
Then a_n = 0
⇒ a + (n – 1)d = 0
⇒ 8 + (n – 1)(– 2) = 0
⇒ (n – 1)(– 2) = – 8
$$⇒ n – 1 =\frac{-8}{-2}=4$$
⇒ n = 4 + 1 = 5
Hence, 5th term of an AP is zero.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Sol. Let a be the first term and d the common difference of an AP.
Given, a₁₇ – a₁₀ = 7
(a + 16d) – (a + 9d) = 7 [...a_n = a (n – 1(d)]
⇒ 7d = 7 ⇒ d = 1
Hence, the common difference of an AP is 1.

11. Which term of the AP : 3, 15, 27, 39, ...., will be 132 more than its 54th term?

Sol. Here, first term a = 3, common difference d = 15 – 3 = 12
Then, a₅₄ = a + 53d = 3 + 53 × 12
[...a_n = a + (n – 1)d]
a₅₄ = 3 + 636 = 639
Let an be 132 more than its 54th term.
Then a_n = a₅₄ + 132
(By given condition)
a_n = 639 + 132
a_n = 771
⇒ a + (n – 1)d = 771 [a_n = a + (n – 1)d]
⇒ 3 + (n – 1) 12 = 771
⇒ 12 (n – 1) = 771 – 3
⇒ 12 (n – 1) = 768
⇒ n – 1 = 768/12=64
⇒ n = 65
Hence, 65th term is 132 more than its 54th term of an AP.

12. Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?

Sol. Let the two APs be a₁, a₂, a₃, ..... a_n and b₁, b₂, b₃, ..... b_n
Also, let d be the same common difference of two APs, then the nth term of first AP
an = a₁ + (n – 1)d
and the nth term of second AP
b_n = b₁ + (n – 1)d
Now, a_n – b_n = [a₁ + (n – 1)d] – [b₁ + (n – 1)d]
⇒ a_n – b_n = a₁ – b₁ for all n ∈ N
⇒ a₁₀₀ – b₁₀₀ = a₁ – b₁ = 100 (Given)
∴ a₁₀₀₀ – b₁₀₀₀ = a₁ – b₁
⇒ a₁₀₀₀ – b₁₀₀₀ = 100 [ a₁ – b₁ = 100]
Hence, the difference between their 1000th terms is also 100 for all n ∈ N.

13. How many three digit numbers are divisible by 7?

Sol. We know that, 105 is the first and 994 is the last 3 digit number divisible by 7. Thus, we have to determine the number of terms in the list 105, 112, 119, ......, 994.
Clearly, the successive difference of terms is constant with common difference
d = 112 – 105 = 7
So, it forms an AP.
Let there be n terms in the AP, then nth term
= 994
... a_n = a + (n – 1)d
⇒ 105 + (n – 1)7 = 994
⇒ 7(n – 1) = 994 – 105
⇒ 7(n – 1) = 889
⇒ n – 1 =889/7
= 127
⇒ n = 127 + 1 = 128
So, there are 128 numbers of three digit which are divisible by 7.

14. How many multiplies of 4 lie between 10 and 250?

Sol. Here, we know that 12 is the first integer between 10 and 250, which is a multiple of 4. Also, when we divide 250 by 4, the remainder is 2. Therefore, 250 – 2 = 248 is the greatest integer divisible by 4 and lying between 10 and 250. Thus, we are to find the number of terms in an AP whose first term = 12, last term = 248 and common difference = 4.
a_n = 248
⇒ 12 + (n – 1)4 = 248 [.. a_n = a + (n – 1)d]
⇒ 4(n – 1) = 248 – 12
⇒ 4(n – 1) = 236
⇒ n – 1 =236/4=59
⇒ n = 59 + 1 = 60
Hence, there are 60 multiples of 4 lie betwen 10 and 250.

15. For what value of n, are the nth terms of two APs 63, 65, 67, ... and 3, 10, 17...... equal?

Sol. If nth terms of the APs : 63,65,67, .... and 3, 10, 17, .... are equal.
Here, first term of first AP (a₁) = 63
Common difference of first AP (d₁) = 65 – 63 = 2
and first term of second AP (b₁) = 3
Common difference of second AP (d₂)
= 10 – 3 = 7
Then by condition nth terms of both APs are equal.
∴ 63 + (n – 1)2 = 3 + (n – 1)7
⇒ 7(n – 1) –2(n – 1) = 63 – 3
⇒ (n – 1)(7 – 2) = 63 – 3
[.. a_n = a + (n – 1)d]
⇒ (n – 1)(7 – 2) = 60
⇒ 5(n – 1) = 60
⇒ (n – 1) =60/5=12
⇒ n = 12 + 1 = 13
Hence, the 13th terms of the two given APs are same.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Sol. nth term of an AP
⇒ a_n = a + (n – 1)d
According to the question,
a₃ = a + (3 – 1)d = 16 ...(i)
where a is first term and d be the common difference of an AP.
Now,
7th term of an AP = 12 + 5th term of an AP
⇒ a₇ = 12 + a₅
⇒ a₇ – a₅ = 12
⇒ (a + 6d) – (a + 4d) = 12
⇒ a + 6d – a – 4d = 12
2d = 12
d = 6
On putting the value of d in Eqn. (i), we get
a + (3 – 1)d = 16
a + 2 × 6 = 16
a = 16 – 12
a = 4
Since, the terms of an AP in the form a, a + d,
a + 2d, a + 3d, .....
Therefore, A.P. will be 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6
Hence, the series of an AP will be 4, 10, 16, 22.

17. Find the 20th term from the last term of the AP 3, 8, 13, ....., 253.

Sol. Given, l = last term = 253
d = common difference = 8 – 3 = 5
∴ 20th term from the end
= l – (20 – 1)d
= l – 19d
= 253 – 19 × 5
(.. n = 20)
= 253 – 95 = 158
Hence, the 20th term from the last term of the AP be 158.

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Sol. Let a be the first term and d the common difference of an AP.
Given, a₄ + a₈ = 24 (By definition)
⇒ (a + 3d) + (a + 7d) = 24 [a_n = a + (n – 1)d]
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ...(i)
and a₆ + a₁₀ = 44 (By definition)
⇒ (a + 5d) + (a + 9d) = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
2d = 10 ⇒ d = 5
∴ From equation (i), we get
a + 25 = 12
⇒ a = – 13
Hence, the first three terms are a, (a + d), (a + 2d)
i.e., – 13, (– 13 + 5) and (– 13 + 2 × 5)
i.e., – 13, – 8 and – 3.

19. Subha Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000 ?

Sol. The annual salary received by Subha Rao in the years 1995, 1996, 1997 etc. is ₹ 5000, ₹ 5200, ₹ 5400 ..... ₹ 7000.
Hence, the list of numbers 5000, 5200, 5400, ...., 7000 terms forms an AP.
The common difference of an AP
... d = a₂ – a₁ = a₃ – a₂ = 200
Let nth term of an AP.
an = 7000
⇒ 7000 = a + (n – 1)d
[... a_n = a + (n– 1)d]
⇒ 7000 = 5000 + (n – 1) (200)
⇒ 200 (n – 1) = 7000 – 5000 = 2000
⇒ n – 1 =2000/200=10
⇒ n = 10 + 1 =11
Thus, 11th year of his service or in 2005 Subha Rao received an annual salary ₹ 7000.

20. Ramkali save ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

Sol. Ramkali’s savings in the subsequent weeks are respectively ₹ 5, ₹ 5 + ₹ 1.75, ₹ 5 + 2 × ₹ 1.75, ₹ 5 + 3 × ₹ 1.75 ,...
In nth week her saving will be ₹ 5 + (n – 1) × ₹ 1.75.
⇒ 5 + (n – 1) × 1.75 = ₹ 20.75 (Given)
⇒ (n – 1) × 1.75 = 20.75 – 5 = 15.75
⇒ (n – 1) =15.75/1.75=9
⇒ n = 9 + 1 = 10
Hence, in the 10th week, Ramkali’s saving will be ₹ 20.75.

Exercise 5.3

1. Find the sum of the following APs :
(i) 2, 7, 12, ......., to 10 terms.
(ii) – 37, – 33, – 29, ...., to 12 terms.
(iii) 0.6, 1.7, 2.8, ....., to 100 terms.
$$\textbf{(iv)}\frac{1}{15},\frac{1}{12},\frac{1}{10},........,\text{to 11 terms.}$$

Sol. (i) 2, 7, 12, ..........., to 10 terms.
According to the given question :
First term a = 2
The common difference d = 7 – 2 = 5
Number of terms n = 10
We know that, sum of n terms of an AP
$$S_n =\frac{n}{2}[2a+(n-1)d]\\S_{10} =\frac{n}{2}[2×2+(10-1)5]$$
= 5[4 + 9 × 5]
= 5[4 + 45]
= 5 × 49
= 245
The sum of 10 terms is 245.

(ii) – 37, – 33, – 29, ......., to 12 terms.
According to the given question :
First term a = – 37
The common difference d = a₂ – a₁
= (– 33) – (– 37)
= – 33 + 37 = 4
Number of terms, n = 12
We know that, sum of n terms of an AP
$$S_n =\frac{n}{2}[2a+(n-1)d]\\S_{12} =\frac{12}{2}[2(-37)+(12-1)4]$$
= 6[– 74 + 44]
= 6[– 30]
= – 180
The sum of 12 terms is – 180.

(iii) 0.6, 1.7, 2.8, ......., to 100 terms.
According to the given question :
First term a = 0.6
The common difference d = 1.7 – 0.6 = 1.1
Number of terms n = 100
Sum of n terms of an AP
$$S_n =\frac{n}{2}[2a+(n-1)d]\\S_{100} =\frac{100}{2}[2×0.6+(100-1)1.1]$$
= 50[1.2 + 108.9]
= 50[110.1]
= 5505
The sum of 100 terms is 5505.

$$\textbf{(iv)}\frac{1}{15},\frac{1}{12},\frac{1}{10},......,\text{to 11 terms. Given,}\\\text{First term a}=\frac{1}{15}\\\text{The common difference} d = a_2 – a_1\\=\frac{1}{12}-\frac{1}{15}=\frac{1}{60}\\\text{Number of terms n = 11}\\\text{Sum of n terms of an AP}\\S_n =\frac{n}{2}[2a+(n-1)d]\\S_{100} =\frac{11}{2}[2×\frac{1}{15}+(11-1)\frac{1}{60}]\\=\frac{11}{2}[\frac{2}{5}+\frac{10}{60}]\\=\frac{11}{2}[\frac{2}{15}+\frac{1}{6}]\\=\frac{11}{2}×\frac{3}{10}\\=\frac{33}{20}$$

2. Find the sums given below
$$\textbf{(i)}7+10\frac{1}{2}+14+...+84$$
(ii) 34 + 32 + 30 + .... + 10
(iii) – 5 + (– 8) + (– 11) + ..... + (– 230)

Sol. (i)$$\textbf{(i)}7+10\frac{1}{2}+14+...+84$$
Here, first term a = 7, common difference
$$d=10\frac{1}{2}-7=3\frac{1}{2}=\frac{7}{2}\\\text{and last term}\space l = a_n = 84\\84 = a + (n – 1)d [ a_n = a + (n – 1)d]\\⇒ 84 = 7+(n-1)\frac{7}{2}\\⇒\frac{7}{2}(n-1)= 84 – 7\\⇒\frac{7}{2}(n-1)= 77\\⇒ n – 1 = 77×\frac{2}{7}\\⇒ n – 1 = 22\\⇒ n = 23\\\text{We know that sum of n terms of an AP,}\\S_n=\frac{n}{2}(a+l)\\S_23=\frac{23}{2}(7+84)\\=\frac{23}{2}×(91)=\frac{2093}{2}\\1046\frac{1}{2}$$

(ii) 34 + 32 + 30 + .... + 10
Here, first term a = 34,
Common difference d = 32 – 34 = – 2,
Last term l = an + 10
∴ We know that a_n = a + (n – 1)d
⇒ 10 = 34 + (n – 1)(– 2)
⇒ (– 2)(n – 1) = 10 – 34
⇒ (– 2)(n – 1) = – 24
⇒ n – 1 = 12
⇒ n = 12 + 1 = 13
$$\text{By sum of n terms of an AP,}S_n=\frac{n}{2}(a+l)\\\text{we get}\\S_{13}\\S_{13}=\frac{13}{2}(34+10)\\=\frac{13}{2}×44\\= 13 × 22 = 286$$

(iii) – 5 + (– 8) + (– 11) + .... + (– 230)
Here, first term a = – 5, common difference d = – 8 – (– 5 )= – 8 + 5 = – 3, last term l = an = – 230
∴ an = a + (n – 1)d
⇒ – 230 = – 5 + (n – 1)(– 3)
⇒ (– 3)(n – 1) = – 230 + 5
⇒ (– 3)(n – 1) = – 225
⇒ n – 1 =\frac{-225}{-3}
⇒ n – 1 = 75
⇒ n = 75 + 1 = 76
$$\text{By sum of n terms of an AP},S_n=\frac{n}{2}(a+l)\\\text{we get}\\S_{76}=\frac{76}{2}(-5-230)\\= 38 × (– 235) = – 8930$$

3. In an AP
(i) given a = 5, d = 3, a_n = 50, find n and S_n.
(ii) given a = 7, a₁₃ = 35, find d and S13.
(iii) given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) given a₃ = 15, S₁₀ = 125, find d and a10.
(v) given d = 5, S₉ = 75, find a and a₉.
(vi) given a = 2, d = 8, S_n = 90, find n and an.
(vii) given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) given a_n = 4, d = 2, S_n = –14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms, find a.

Sol. (i) Given, a = 5, d = 3 and an = 50
[..a_n = a + (n – 1)d]
⇒ 5 + (n – 1)3 = 50
⇒ 3(n – 1) = 50 – 5
$$⇒(n-1)=\frac{45}{3}=15$$
⇒ n = 15 + 1 = 16
Putting, n = 16, a = 5 and l = a_n = 50 in
$$S_n=\frac{n}{2}(a+l),\text{we get}\\S_16=\frac{16}{2}(5+50)\\= 8 × 55 = 440\\So, n = 16\text{ and} S_{16} = 440$$
= 8 × 55 = 440
So, n = 16 and S₁₆ = 440

(ii) Given, a = 7 and a₁₃ = 35
Let d be the common difference of the
given AP,
Then, a₁₃ = 35
[... a_n = a + (n – 1)d]
⇒ 7 + 12d = 35 (.. a = 7)
⇒ 12d = 35 – 7 = 28
$$⇒ d =\frac{28}{12}=\frac{7}{3}\\\text{Putting}, n = 13, a = 7 \text{and} l = a_{13} = 35 in\\S_n=\frac{n}{2}(a+l),\text{we get}\\S_n =\frac{13}{2}(7+35)\\=\frac{13}{2}×42\\= 13 × 21 = 273\\\text{Hence,}\\d=\frac{7}{3}\text{and}S_{13}=273$$

(iii) Given, a₁₂ = 37, d = 3
a be the first term of given AP, then
a₁₂ = 37
⇒ a + 11d = 37 [... an = a + (n – 1)d]
⇒ a + 11(3) = 37
⇒ a = 37 – 33 = 4
Putting, n = 12, a = 4 and l = a₁₂ = 37 in
$$\\S_n=\frac{n}{2}(a+l),\text{we get}\\S_{12} =\frac{12}{2}(4+47)\\= 6 × 41 = 246\\Hence, a = 4\\\text{and} S_{12} = 246$$

(iv) Given, a₃ = 15, S₁₀ = 126
Let a be the first term and d be the common difference of the given AP,
Then, a₃ = 15
and S₁₀ = 125
⇒ a + 2d = 15 ...(i)
[.. a_n = a + (n – 1)d]
$$\frac{10}{2}[2a+(10-1)d]=125\\S_n=\frac{n}{2}[2a+(n-1)d]\\⇒ 5(2a + 9d) = 125\\⇒ 2a + 9d = 25 ...(ii)\\\text{On multilying Eq. (i) by 2 and subtracting Eq. (ii), we get}\\2(a + 2d) – (2a + 9d) = 2 × 15 – 25\\⇒ 4d – 9d = 30 – 25\\⇒ – 5d = 5\\⇒ d =-\frac{5}{5}=-1\\\text{From Equation (i), we get}\\15 = a + 2d\\15 = a + 2(– 1)\\a = 17\\a_{10} = a + 9d\\= 17 + 9(– 1)\\= 17 – 9 = 8\\a_{10} = 8\\\text{Hence}, d = – 1, a_{10} = 8$$

(v) Given, d = 5, S₉ = 75
Then, S₉ = 75
$$⇒\frac{9}{2}[2a+(9-1)5]=75\\{[..S_n=\frac{n}{2}[2a+(n-1)d]]}\\⇒\frac{9}{2}(2a+40)=75\\⇒ 9a + 180 = 75\\⇒ 9a = 75 – 180\\⇒ 9a = – 105\\⇒\text{First term a} =\frac{-105}{9}=\frac{-35}{3}\\So, a_9 = a + 8d\\=\frac{-35}{3}+8×5\\{[..an = a + (n – 1)d]}\\\frac{-35+120}{3}=\frac{85}{3}\\\text{Hence}, a=\frac{-35}{3}\text{and}\space a_9=\frac{85}{3}$$

(vi) Given, first term a = 2, common difference d = 8, sum of n terms Sn = 90
Then, Sn = 90
$$⇒\frac{n}{2}[2×2+(n-1)8]=90\\{[..S_n=\frac{n}{2}[2a+(n-1)d]]}\\⇒\frac{n}{2}(4+8n-8)=90\\⇒\frac{n}{2}(8n-4)=90\\n(8n – 4) = 180\\8n^2 – 4n – 180 = 0\\2n^2 – n – 45 = 0\\2n^2 – 10n + 9n – 45 = 0\\2n(n – 5) + 9(n – 5) = 0\\(n – 5)(2n + 9) = 0\\n = 5\text{and}-\frac{9}{2}\\\text{n always be a positive integer.}\\n=5\\\text{Then,}\\a_n = a + (n – 1)d\\a_5 = 2 + (5 – 1)8\\= 2 + 4 × 8\\= 2 + 32\\=4\\\text{Hence}, n = 2 and a_5 = 34.$$

(vii) Given, a = 8, l = a_n = 62, S_n = 210
Since, S_n = 210
$$⇒S_n=\frac{n}{2}(a+l)=210\space\space\space{[..S_n=\frac{n}{2}(a+l)}]\\⇒\frac{n}{2}(8+62)=210\space\space\space( ..a = 8, l = an = 62)\\⇒\frac{n}{2}×70=210\\⇒ n = 210×\frac{n}{2}×70\\= 3 × 2 = 6\\n = 6\\\text{and} a_n = 62 ⇒ a_6 = 62\\⇒ a + 5d = 62 [ a_n = a + (n – 1)d]\\⇒ 8 + 5d = 62 ( a = 8)\\⇒ 5d = 62 – 8 = 54\\⇒d=\frac{54}{5}\\\text{Hence,}d=\frac{54}{5}\text{and}n=6$$

(viii) Given, l = an = 4, d = 2, S_n = – 14
Then, a_n = 4
⇒ a + (n – 1)2 = 4
(.. d = 2) [.. a_n = a + (n – 1)d]
⇒ a = 4 – 2 (n – 1) ...(i)
and S_n = – 14
$$⇒\frac{n}{2}(a+l)=-14\space\space\space{[..S_n=\frac{n}{2}(a+l)}]$$
⇒ n(a + 4) = – 28 (.. l = a_n)
⇒ n{4 – 2(n – 1) + 4} = – 28 [From Eq. (i)]
⇒ n(4 – 2n + 2 + 4) = – 28
⇒ n (– 2n + 10) = – 28
⇒ n (– n + 5) = – 14
⇒ – n² + 5n = – 14
⇒ n² – 5n – 14 = 0
⇒ (n – 7)(n + 2) = 0
(By factorisation method)
⇒ n = 7 or n = – 2
Since, n cannot be negative.
∴ n = 7
Putting n = 7 in Eq. (i), we get
a = 4 – 2(7 – 1)
= 4 – 2 × 6
= 4 – 12 = – 8
Hence, n = 7 and a = – 8

(ix) Given, a = 3, n = 8, S_n = 192
We know that,
$$S_n=\frac{n}{2}[2a+(n-l)d]\\192=\frac{8}{2}[2×3+(8-1)d]\space\space\space\\⇒ 192 = 4(6 + 7d)\\⇒ 48 = 6 + 7d\\⇒ 7d = 48 – 6 = 42\\⇒d=\frac{42}{7}=6\\\text{Hence, the common difference d = 6}$$

(x) Given, l = 28, S_n = 144, n = 9
Now, S_n = 144
$$⇒\frac{n}{2}(a+l)=144\\{[S_n=\frac{n}{2}[\text{(First term + Last term)}]]}\\⇒\frac{9}{2}(a+28)=144\\⇒a+28=144×\frac{2}{9}\\⇒ a + 28 = 32\\⇒ a = 32 – 28 = 4\\\text{Hence, first term a} = 4$$

4. How many terms of the AP : 9, 17, 25, .... must be taken to give a sum of 636?

Sol. Given, a = 9, d = 17 – 9 = 8, S_n = 636
$$⇒\frac{n}{2}[2a+(n-1)d]=636\\{[S_n=\frac{n}{2}[2a+(n-1)d]]}\\⇒\frac{n}{2}(18+8n-8)=636\\⇒ n(4n + 5) = 636\\⇒ 4n^2 + 5n – 636 = 0\\n=\frac{-5±\sqrt{(5)^2-4×4×(-636)}}{2(4)}\\\Bigg(\text{By quadratic formula,}x=\frac{-b±\sqrt{(b)^2-4ac}}{2a}\Bigg)\\=\frac{-5±\sqrt{25+10176}}{8}\\=\frac{-5±\sqrt{10201}}{8}\\=\frac{-5±101}{8}\\=\frac{96}{8}\\=\frac{-106}{8}\\12,\frac{-53}{4}\\\text{Since, n cannot be negative integer}\\n = 12\\\text{Hence, the sum of 12 terms is 636.}$$

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Sol. Given, first term a = 5, last term l = 45 and sum of n terms S_n = 400
We know that,
$$\\{S_n=\frac{n}{2}(a+l)}\\⇒400=\frac{n}{2}(5+45)\\⇒ 400 × 2 = 50n\\⇒n=\frac{400×2}{50}\\= 8 × 2 = 16\\n = 16\\\text{and}\space \text{l} = 45\\⇒ a + (n – 1)d = 45\space\space\space [ l = an = a + (n – 1)d]\\⇒ 5 + (16 – 1)d = 45\\⇒ 15d = 45 – 5 = 40\\⇒d=\frac{40}{15}=\frac{8}{3}\\\text{The number of terms is 16 and the common difference is}\frac{8}{3}$$

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Sol. Given, first term a = 17, last term l = an = 350, common difference d = 9
We know that,
a + (n – 1)d = 350
[.. l = a_n = a + (n – 1)d]
⇒ 17 + (n – 1)9 = 350
⇒ 9(n – 1) = 350 – 17 = 333
⇒ n – 1=333/9=37
⇒ n = 37 + 1 = 38
Now,
$$\text{Sum of n terms}\space{S_n=\frac{n}{2}(a+l)},\text{we get}\\{S_{38}=\frac{38}{2}(17+350)}\\= 19(367) = 6973\\\text{So, there are 38 terms in the AP having their sum as 6973.}$$

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Sol. Given, d = 7 and a₂₂ = 149
⇒ a + (22 – 1)d = 149 [.. a_n = a + (n – 1)d]
⇒ a + 21 × 7 = 149
⇒ First term, a = 149 – 147 = 2
Put, a = 2, n =22 and d = 7 in
$$\text{Sum of n terms}\space{S_n=\frac{n}{2}[2a+(n-1)d]},\text{we get}\\{S_{22}=\frac{22}{2}[2×2+(22-1)7]}\\= 19(367) = 6973\\\text{So, there are 38 terms in the AP having their sum as 6973.}\\= 11(4 + 2 × 7)\\= 11(4 + 147)\\= 11(151) = 1661\\\text{Hence, the sum of first 22 terms is 1661.}$$

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Sol. Given that, a₂ = 14 and a₃ = 18
⇒ a₂ = a + d = 14
[.. a_n = a + (n – 1)d] ...(i)
and a₃ = a + 2d = 18 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
d = 4
Put d = 4 in Eq. (i), we get
a + 4 = 14
⇒ a = 14 – 4 = 10
a = 10
$$\text{Sum of n terms}\space{S_n=\frac{n}{2}[2a+(n-1)d]},\text{we get}\\{S_{51}=\frac{51}{2}[20+50×4]}\\=\frac{51}{2}(20+200)\\=\frac{51}{2}×220\\= 51 × 110 = 5610\\S_{51} = 5610$$

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. (Given)

Sol. Given, sum of 7 terms S₇ = 49 and sum of 17 terms S₁₇ = 289
$$=\frac{7}{2}[2a+(7-1)d]=49\\{[S_n=\frac{n}{2}[2a+(n-1)d]]},\\=\frac{7}{2}(2a+6d)=49\\⇒ a+ 3d = 7 ...(i)\\\text{and}\space \frac{17}{2}[2a+(17-1)d]=289\\⇒\frac{17}{2}[2a+16d]=289\\⇒ a + 8d = 17 ...(ii)\\\text{On subtracting Eq. (i) from Eq. (ii), we get}\\5d = 10 ⇒ d = 2\\\text{Put, d = 2 in Eq. (i), we get}\\a + 3(2) = 7\\⇒ a + 6 = 7\\⇒ \text{First term}, a = 1\\\text{Sum of n terms,}{[S_n=\frac{n}{2}[2a+(n-1)d]]}\\=\frac{n}{2}[2×1+(n-1)2]\\=\frac{n}{2}[2+2n-2]\\=\frac{n}{2}×2+2n=n^2\\S_n = n^2$$

10. Show that a₁, a₂, ..... an, .... form an AP where an is defined as below
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5_n. Also find the sum of the first 15 terms in each case.

Sol. (i) Given, a_n = 3 + 4_n
Put, n = 1, 2, 3, 4, .........
We get,
a₁ = 3 + 4 × 1 = 7
a₂ = 3 + 4 × 2 = 11
a₃ = 3 + 4 × 3 = 15
a₄ = 3 + 4 × 4 = 19
The AP will be : 7, 11, 15, 19, ......, (3 + 4n)
Common difference d = a₂ – a₁ = 11 – 7 = 4
d = 4
Here, first term a = a₁ = 7, common difference d = 4 and number of terms n = 15
We know that,
$${[S_n=\frac{n}{2}[2a+(n-1)d]]},\\=\frac{15}{2}[2×7+(15-1)4]\\=\frac{15}{2}[14+14×4]\\=\frac{15}{2}[14+56]\\=\frac{15}{2}[70]\\= 15 × 35\\S_n = 525$$

(ii) Given,
a_n = 9 – 5_n
Put, n = 1, 2, 3, 4, .............
We get,
a₁ = 9 – 5 × 1= 4
a₂ = 9 – 5 × 2 = – 1
a₃ = 9 – 5 × 3 = – 6
a₄ = 9 – 5 × 4 = – 11
The AP will be : 4, – 1, – 6, – 11, ......, (9 – 5n)
Common difference d = a₂ – a₁
= – 1 – 4
= – 5
Here, first term a = a₁ = 4 common difference d = 5 and number of terms n = 15
We know that,
$${[S_n=\frac{n}{2}[2a+(n-1)d]]},\\=\frac{15}{2}[2×4+(15-1)(-5)]\\=\frac{15}{2}[8+14×(-5)]\\=\frac{15}{2}[8+(-70)]\\= 15 × (– 31)\\= – 465$$

11. If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Sol. Given that,
S_n = 4n – n²
Put n = 1
S₁ = 4 × 1 – (1)²
S₁ = 4 – 1
a = S₁ = 3
Here, a is first term
Put n = 2
S₂ = 4 × 2 – 2²
= 8 – 4
= 4
S2 = 4
Second term, a₂ = S₂ – S₁
= 4 – 3 = 1
Common difference, d = a₂ – a₁
= 1 – 3
= – 2
Now, nth term of an AP
a_n = a + (n – 1)^d
a_n = 3 + (n – 1)(– 2)
= 3 – 2n + 2
= 5 – 2n
Therefore, a₃ = 5 – 2 × 3
= 5 – 6
= – 1
a₁₀ = 5 – 2(10)
= 5 – 20
= – 15
Hence, the sum of first two terms S₂ = 4. The second term a₂ = 1, the 3rd, the 10^th and the n^th terms are – 1, – 15 and 5 – 2n respectively.

12. Find the sum of the first 40 positive integers divisible by 6.

Sol. The first 40 positive integers divisible by 6 are 6, 12, 18, ....
Clearly, it is an AP with first term a = 6 and common difference d = 6
$${[S_n=\frac{n}{2}[2a+(n-1)d]]},\\S_{40}=\frac{40}{2}[2×6+(40-1)5]\\= 20(12 + 39 × 6)\\= 20(12 + 234)\\= 20 × 246\\= 4920$$

13. Find the sum of the first 15 multiples of 8.

Sol.The first 15 multiples of 8 are 8 × 1, 8 × 2, 8 × 3, ...., 8 × 15
i.e., 8, 16, 24, ...., 120 are in AP
Here, first term a = 8, last term, l = 120 and number of terms, n = 15
$${S_n=\frac{n}{2}(a+l)},\\S_{15}=\frac{15}{2}(8+120)\\=\frac{15}{2}×128\\= 15 × 64\\= 960$$

14. Find the sum of the odd numbers between 0 and 50.

Sol. The odd numbers between 0 and 50 are 1, 3, 5, ...., 49.
Here, first term a = 1 and last term, l = 49
Let n be the number of terms and the common difference d = 3 – 1 = 2
nth term an = a + (n – 1)d = l
∴ 1 + (n – 1)(2) = 49
⇒ 2(n – 1) = 48
⇒ n – 1 = 24
⇒ n = 25
Now, sum of n terms,
$${S_n=\frac{n}{2}(a+l)},\\S_{25}=\frac{25}{2}(1+49)\\=\frac{25}{2}×50\\= 25 × 25 = 625$$

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows : ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Sol. According to the given question, the penalty are form an A.P.
Let us denote the penalty for the nth day by an, then
a1 = ₹ 200, a2 = ₹ 250, a3 = ₹ 300
Here, a = 200, d = ₹ 250 – ₹ 200 = ₹ 50 and
n = 30 days.
$${S_n=\frac{n}{2}[2a+(n-1)d]}\\{=\frac{30}{2}[2×2+(30-1)50]}\\= 15 (400 + 29 × 50)\\= 15(400 + 1450)\\= 15 × 1850\\= 27750$$
So, a delay of 30 days, the contractor has to pay ₹ 27750 as penalty.

16. A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.

Sol. Let us consider that, the respective prizes be x, x – 20, x – 40, ........, which are form an A.P.
Here, first term a = x and the common difference d = x – 20 – x = – 20
Given, S_n = 700
We know that,
$${S_n=\frac{n}{2}[2a+(n-1)d]}\\{700=\frac{7}{2}[2x+(7-1)(-20)]}\\{700=\frac{7}{2}[2x+(-120)]}\\100 = x – 60\\x = 160$$
Hence, the values of the seven prizes are ₹ 160,
₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60 and ₹ 40.

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Sol. According to the given question, there are three sections of each class, then the number of trees that each section planted by students are
1 × 3, 2 × 3, 3 × 3, .......... 12 × 3, which will be form an AP.
⇒ 3, 6, 9, .......... 36
⇒ First term a₁ = 3
The common difference d = a₂ – a₁ = 6 – 3 = 3
We know that the n^th term of a_n AP,
a_n = a + (n – 1)d
36 = 3 + (n – 1)3
3(n – 1) + 3 = 36
3(n – 1) = 33
n – 1 = 11
n = 12
The number of trees planted by the students
$${S_n=\frac{n}{2}(a+l)}\\\text{Here,}l = a_n = 36\\{S_n=\frac{12}{2}(3+6)}\\= 6 × (39)\\S_n = 234$$

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ..... as shown in figure. What is the total length of such as spiral made up of thirteen consecutive semicircles?(Take π=22/7)

[Hint : Length of successive semicircles is l₁, l₂, l₃, l₄, .... with centres at A, B, A, B,...... respectively.]

Sol. Length of spiral made up of thirteen consecutive semi circles
= (π × 0.5 + π × 1.0 + π × 1.5 + π × 2.0 + ... + π × 6.5)
= π × 0.5 (1 + 2 + 3 + .... + 13)
which form an AP with first term, a = 1, common difference, d = 2 – 1 = 1 and number of terms, n = 13
$$\text{Sum of n terms}{=\frac{n}{2}[2a+(n-1)d]}=S_n\\\text{Therefore, sum of the length of 13 consecutive}\\\text{circle is :}\\{=\frac{13}{2}[1.0π+6π]}\\=\frac{13}{2}×7π\\=\frac{13}{2}×7×\frac{22}{7}\\= 143 cm$$

19. 200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row ?

Sol. Since, logs are stacked in each row form a series 20 + 19 + 18 + 17 + .... Clearly, it is an AP with first term, a = 20 and common difference,
d = 19 – 20 = – 1.
Suppose, Sn = 200
$$S_n=200\\{S_n=\frac{n}{2}[2a+(n-1)d]}=S_n\\⇒{200=\frac{n}{2}[2×20+(n-1)(-1)]}\\⇒ 400 = n(40 – n + 1)\\⇒ n^2 – 41n + 400 = 0\\\text{(By factorisation method)}\\⇒ n(n – 25) – 16(n – 25) = 0\\⇒ (n – 25)(n – 16) = 0\\⇒ n = 16\\or n = 25\\\text{Hence, the number of rows is either 25 or 16.}\\\text{When n} = 16,\\t_n = a + (n – 1)d\\= 20 + (16 – 1)(– 1)\\= 20 – 15\\t_{16} = 5\\\text{When n = 25,}\space t_n = a + (n – 1)d\\= 20 + (25 – 1)(– 1)\\= 20 – 24\\t_{25} = – 4 \text{(Not possible)}$$
Hence, the number of row is 16 and number of logs in the top row = 5.

20. In a potato race, a bucket is palced at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the lines (see figure).
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)].

Sol. According to question, a competitor pick up the 1st potato, second poato, third potato, fourth potato .....
The distances sum by competitor are
2 × 5, 2 × (5 + 3), 2 × (5 + 3 + 3), 2 × (5 + 3 + 3 + 3)
i.e., 10, 16, 22, 28, .....
Clearly, it is an AP with first term, a = 10 and common difference, d = 16 – 10 = 6
The sum of n terms,
$${S_n=\frac{n}{2}[2a+(n-1)d]}=S_n\\\text{The sum of 10 terms,}\\{S_n=\frac{10}{2}[2×10+(9-1)×6]}\space[n = 10 (given)]\\= 5 (20 + 54)\\= 5 × 74\\= 370$$
Hence, the total distance the competitor has to run = 370 m

Exercise 5.4

1. Which term of the AP : 121, 117, 113, ...., is its first negative term ?
[Hint : Find n for an < 0]

Sol. Given, first term, a = 121, common difference, d = 117 – 121 = – 4
nth term of an AP,
a_n = a + (n – 1)d
= 121 + (n – 1) × (– 4)
= 121 – 4n + 4
= 125 – 4n
For first negative term,
an < 0
⇒ 125 – 4n < 0
⇒ 125 < 4n
⇒ 4n > 125
$$⇒n>\frac{125}{4}\\⇒n>31\frac{1}{4}$$
Least integral value of n = 32,
Hence, 32nd term of the given AP is the first negative term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Sol. Let the first term and the common difference of the AP be a and d, respectively.
According to the question,
Third term + Seventh term = 6
a₃ + a₇ = 6
⇒ [a + (3 – 1)d] + [a + (7 – 1)d] = 6
[..a_n = a + (n – 1)d]
⇒ (a + 2d) + (a + 6d) = 6
⇒ 2a + 8d = 6
⇒ a + 4d = 3 ...(i)
and (third term) × (seventh term) = 8
a₃ × a₇ = 8
⇒ (a + 2d)(a + 6d) = 8
⇒ {(a + 4d) – 2d}{(a + 4d) + 2d} = 8
⇒ (3 – 2d)(3 + 2d) = 8
[Using Eq. (i)]
⇒ 9 – 4d² = 8
[ (a – b)(a + b) = a² – b²]
⇒ 4d² = 9 – 8
$$⇒d^2=\frac{1}{4}\\⇒d=±\frac{1}{2}\\\text{Taking}d=\frac{1}{2}\text{from Eq. (i), we get}\\a+4(\frac{1}{2})=3\\⇒ a + 2 = 3\\⇒ a = 3 – 2 ⇒ a = 1\\\text{Sum of first sixteen terms of the AP} = S_{16}\\S_n=\frac{n}{2}[2a+(n-1)d]\\S_{16}=\frac{16}{2}[2a+(16-1)d]\\= 8[2a + 15d]\\=8[2(1)+15(\frac{1}{2})]\\=8(2+\frac{15}{2})=8(\frac{19}{2})=76\\\text{Taking d}=-\frac{1}{2}\text{from Eq. (i), we get}\\a+4(-\frac{1}{2})=3\\⇒ a – 2 = 3\\⇒ a = 5\\\text{Sum of first sixteen terms of the AP} = S_{16}\\S_{16}=\frac{16}{2}[2a+(16-1)d]\\{[S_n=\frac{n}{2}[2a+(n-1)d]]}\\= 8(2a + 15d]\\=8[2(5)+15(-\frac{1}{2})]\\=8(10-\frac{15}{2})\\=8(\frac{5}{2})=20\\So,\space S_{16} = 20, 76$$

3. A ladder has rungs 25 cm apart (see figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the botton rungs are
$$2\frac{1}{2}\textbf{m}$$
apart, what is the length of the wood required for the rungs?

Sol. According to the question,
$$\text{Number of rungs }=\frac{2\frac{1}{2}m}{25cm}\\=\frac{250cm}{25cm}=10\\\text{Hence, there are 10 rungs.}\\\text{The length of the wood required for the rungs = Sum of 10 rungs}\\\frac{10}{2}(25+40)\space {[..S_n=\frac{n}{2}(a+l)]}\\= 5 × 70 = 350 cm$$

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint: S_{x – 1} = S₄₉ – S_x]

Sol. The consecutive numbers on the houses of a row are 1, 2, 3, ......... 49. Clearly, this list of numbers forming an AP.
Here, a = 1, d = 2 – 1 = 1.
$$S_x=\frac{x}{2}[2a+(x-1)d]\\S_{x-1}=\frac{x-1}{2}[2×1+(x-1)×1]\\\frac{x-1}{2}(2+x-2)\\=\frac{(x-1)}{2}x=\frac{x^2-x}{2}\\S_x=\frac{x}{2}[2×1+(x-1)×1]\\=\frac{x}{2}(x+1)=\frac{x^2+x}{2}\\\text{and}\\S_{49}=\frac{49}{2}[2×1+(49-1)×1]\\\frac{49}{2}[2+48]\\=\frac{49}{2}×50=49×25\\\text{Given condition,}\\S_{x – 1} = S_{49} – S_x\\\frac{x^2-x}{2}= 49×25-\frac{x^2-x}{2}\\⇒\frac{x^2-x}{2}+\frac{x^2+x}{2}=49×25\\⇒ x^2 = 49 × 25\\⇒ x = ± 7 × 5 = ± 35\\\text{Since, x is a counting number.}\\\text{Taking positive sign, we get}\\x = 35$$

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of
$$\frac{1}{4}m\text{and a tread of}\frac{1}{2}m$$
(see figure). Calculate the total volume of concrete required to build the terrace.

Sol. Since, volume of concrete required to build the Ist step, IInd step, IIIrd step ........ are
$$\frac{1}{4}m×\frac{1}{2}m×50\\(2×\frac{1}{4})×\frac{1}{2}×50,(3×\frac{1}{4})×\frac{1}{2}×50,..\\\text{i.e.,}\frac{50}{8},2×\frac{50}{2},3×\frac{50}{8},...\\\text{So, total volume of concrete required}\\=\frac{50}{8},2×\frac{50}{2},3×\frac{50}{8},...\\\text{Here,}n = 15\\S_n=\frac{15}{2}[2×\frac{50}{8}+(15-1)\frac{50}{8}]\\=\frac{15}{8}[\frac{25}{2}+\frac{175}{2}]\\=\frac{15}{2}[\frac{200}{2}]\\= 15 × 50\\= 750 m^3$$

Selected NCERT Exemplar Problems

Exercise 5.1

• Choose the correct answer from the given four options.

1. In an AP, if d = – 4, n = 7, a_n = 4, then a is
(a) 6
(b) 7
(c) 20
(d) 28

Sol. (a) 6

Explanation:

a_n = a + (n – 1)d
⇒ 4 = a + (7 – 1) (– 4)
⇒ 4 = a + 6(– 4)
⇒ 4 + 24 = a
⇒ a = 28

2. In an AP, if a = 3.5, d = 0, n = 101, then a_n will be
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5

Sol. (b) 3.5

Explanation:

a_n = a + (n – 1)d = 3.5 + (101 – 1) × 0
= 3.5

3. The list of numbers – 10, – 6, – 2, 2, .... is
(a) an AP with d = – 16
(b) an AP with d = 4
(c) an AP with d = – 4
(d) not an AP

Sol. (b) an AP with d = 4

Explanation:

Since, a₂ – a₁ = – 6 – (– 10)
= – 6 + 10 = 4
a₃ – a₂ = – 2 – (– 6)
= – 2 + 6 = 4
a₄ – a₃ = 2 – (– 2)
= 2 + 2 = 4
... ... ... ... ... ... ... ...
... ... ... ... ... ... ... ...
... ... ... ... ... ... ... ...
Each successive terms of given list have same difference i.e., 4.
∴ The given list forms an AP with common difference, d = 4.

4. The 11th term of the AP :
$$-5,\frac{-5}{2},0,\frac{-5}{2},...is$$
(a) – 20
(b) 20
(c) – 30
(d) 30

Sol. (b) 20

Explanation:

$$\text{Given, AP,}-5,\frac{-5}{2},0,\frac{-5}{2},...\\\text{Here,}a=-5,d=\frac{-5}{2}+5=\frac{5}{2}\\∴ a_{11} = a + (11 – 1)d\space[ an = a + (n – 1)d]\\=-5+(10)×\frac{5}{2}\\= – 5 + 25 = 20$$

5. The first four terms of an AP, whose first term is – 2 and the common difference is – 2, are
(a) – 2, 0, 2, 4
(b) – 2, 4, – 8, 16
(c) – 2, – 4, – 6, – 8
(d) – 2, – 4, – 8, – 16

Sol. (a) –2, 0, 2, 4

Explanation:

Let the first four terms of an AP are a, a + d, a + 2d and a + 3d.
Given that first term; a = – 2 and common difference, d = – 2, then
– 2, – 2 + 2, – 2 + 2(2), – 2 + 3(2)
– 2, 0, 2, 4

6. The 21st term of the AP whose first two terms are – 3 and 4 is
(a) 17
(b) 137
(c) 143
(d) – 143

Sol. (b) 137

Explanation:

Given, first two terms of an AP are a = – 3 and a + d = 4.
⇒ – 3 + d = 4
⇒ common difference, d = 7
∴ a21 = a + (21 – 1)d
[.. an = a + (n – 1)d]
= – 3 + (20)7
= – 3 + 140
= 137

7. If the 2nd term of an AP is 13 and the 5th term is 25, what its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38

Sol. (b) 33

Explanation:

Given, a₂ = 13 and a₅ = 25
a + (2 – 1)d = 13 [.. a_n = a + (n – 1)d]
and a + (5 – 1)d = 25
a + d = 13 ...(i)
and a + 4d = 25 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
3d = 25 – 13 = 12
⇒ d = 4
∴ From equation (i),
a = 13 – 4 = 9
∴ a7 = a + (7 – 1)d
= 9 + 6 × 4 = 33

8. Which term of the AP : 21, 42, 63, 84, ... is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th

Sol. (b) 10th

Explanation:

Let nth term of an AP be 210.
Here, first term, a = 21
and common difference, d = 42 – 21 = 21
and an = 210
∴ an = a + (n – 1)d
∴ 210 = 21 + (n – 1) 21
210 = 21 + 21n – 21
210 = 21n ⇒ n = 10
Hence, the 10th term of an AP is 210.

9. If the common difference of an AP is 5, then what is a₁₈ – a₁₃?
(a) 5
(b) 20
(c) 25
(d) 30

Sol. (c) 25

Explanation:

Given, the common difference of AP, i.e., d = 5
Now, we have,
a₁₈ – a₁₃ = a + (18 – 1)d – [a + (13 – 1)d]
[.. a_n = a + (n – 1)d]
= a + 17 × 5 – a – 12 × 5
= 85 – 60 = 25

10. What is the common difference of an AP in which a₁₈ – a₁₄ = 32?
(a) 8
(b) – 8
(c) – 4
(d) 4

Sol. (a) 8

Explanation:

Given, a₁₈ – a₁₄ = 32
[.. a_n = a + (n – 1)d]
⇒ a + (18 – 1) d – [a + (14 – 1)d] = 32
⇒ a + 17d – a – 13d = 32
⇒ 4d = 32
⇒ d = 8
which is the required common difference of an AP.

11. If 7 times the 7th trem of an AP is equal to 11 times its 11th term, then its 18th term will be
(a) 7
(b) 11
(c) 18
(d) 0

Sol. (d) 0

Explanation:

According to question,
7a₇ = 11a₁₁
7[a + (7 – 1)d] = 11[a + (11 – 1)d]
[.. a_n = a + (n – 1)d]
7(a + 6d) = 11(a + 10d)
7a + 42d = 11a + 110d
4a + 68d = 0
2(2a + 34d) = 0
2a + 34d = 0 (.. 2 ≠ 0)
⇒ a + 17d = 0 ...(i)
∴ 18th term of an AP,
a₁₈ = a + (18 – 1)d
= a + 17d = 0 [From Eq. (i)]

12. The 4th term from the end of the AP : – 11, – 8, – 5, ...., 49 is
(a) 37
(b) 40
(c) 43
(d) 58

Sol. (b) 40

Explanation:

We know that, the nth term of an AP from the end is
a_n = l – (n – 1)d
Here, l = last term, given l = 49, common difference,
d = – 8 – (– 11)
= – 8 + 11 = 3
From equation (i), we get
a₄ = 49 – (4 – 1) 3
= 49 – 9 = 40

13. If the first term of an AP is – 5 and the common difference is 2, then the sum of the first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15

Sol. (a) 0

Explanation:

Given, a = – 5, d = 2
$$\text{Given}, a = – 5, d = 2\\S_n=\frac{6}{2}[2a+(6-1)d]\\{[S_n=\frac{n}{2}[2a+(n-1)d]]}\\= 3[2(– 5) + 5(2)]\\= 3(– 10 + 10) = 0$$

14. The sum of first 16 terms of the AP 10, 6, 2, .... is
(a) – 320
(b) 320
(c) – 352
(d) – 400

Sol. (a) – 320

Explanation:

Given, AP is 10, 6, 2, ....
Here, first term, a = 10, common difference,
d = – 4
$$S_n=\frac{16}{2}[2a+(16-1)d]\\{[..S_n=\frac{n}{2}[2a+(n-1)d]]}$$
= 8[2 × 10 + 15(– 14)]
= 8(20 – 60)
= 8(– 40) = – 320

15. In an AP, if a = 1, an = 20 and Sn = 399, then n is
(a) 19
(b) 21
(c) 38
(d) 42

Sol. (c) 38

Explanation:

$${[S_n=\frac{n}{2}[2a+(n-1)d]]}\\399=\frac{n}{2}[2×1+(16-1)d]$$
798 = 2n + n(n – 1)d ...(i)
⇒ a_n = 20
⇒ a + (n – 1)d = 20 [a_n = a + (n – 1)d]
⇒ 1 + (n – 1)d = 20
⇒ (n – 1)d = 19 ...(ii)
Using Eq. (ii) in Eq. (i), we get
798 = 2n + 19n
798 = 21n
n =798/21=38
n = 38

16. The sum of first five multiples of 3 is
(a) 45
(b) 55
(c) 65
(d) 75

Sol. (a) 45

Explanation:

The first five mumtiples of 3 are 3, 6, 9, 12 and 15.
Here, first term, a = 3, common difference,
d = 6 – 3 = 3 and number of terms, n = 5
$$S_n=\frac{5}{2}[2a+(5-1)d]\\{[..S_n=\frac{n}{2}[2a+(n-1)d]]}\\=\frac{5}{2}[2×3+4×3]\\=\frac{5}{2}(6+12)=5×9=45$$

Exercise 5.2

1. Which of the following form an AP? Justify your answer.
(i) – 1, – 1, – 1, – 1, ...
(ii) 0, 2, 0, 2, ...
(iii) 1, 1, 2, 2, 3, 3, ...
(iv) 11, 22, 33, ...
$$\textbf{(v)}\frac{1}{2},\frac{1}{3},\frac{1}{4},...\\\textbf{(vi)}2, 2^2, 2^3, 2^4\\\textbf{(vii)}\sqrt{3},\sqrt{12},\sqrt{27},\sqrt{48}....$$

Sol. (i) Here, t₁ = – 1, t₂ = – 1, t₃ = – 1, t₄ = – 1
t₂ – t₁ = – 1 + 1 = 0
t₃ – t₂ = 1 + 1 = 0
t₄ – t₃ = – 1 + 1 = 0
Clearly, the difference of successive terms is constant, therefore given list of numbers form an AP.
(ii) Here, t₁ = 0, t₂ = 2, t₃ = 0, t₄ = 2
t₂ – t₁ = 2 – 0 = 2
t₃ – t₂ = 0 – 2 = – 2
t₄ – t₃ = 2 – 0 = 2
Clearly, the difference of successive terms is not constant, therefore given list of numbers does not form an AP.
(iii) Here, t₁ = 1, t₂ = 1, t₃ = 2, t₄ = 2
t₂ – t₁ = 1 – 1 = 0
t₃ – t₂ = 2 – 1 = 1
t₄ – t₃ = 2 –2 = 0
Clearly, the difference of successive terms is not constant, therefore given list of numbers does not form an AP.
(iv) Here, t₁ = 11, t₂ = 22, t₃ = 33
t₂ – t₁ = 22 – 11 = 11
t₃ – t₂ = 33 – 22 = 11
t₄ – t₃ = 33 – 22 = 11
Clearly, the difference of successive terms is constant, therefore given list of numbers form an AP.
$$\textbf{(v)}\frac{1}{2},\frac{1}{3},\frac{1}{4},...\\\text{Here,}\space t_1=\frac{1}{2},t_2=\frac{1}{2},t_3=\frac{1}{2}\\t_2-t_1=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=-\frac{1}{6}\\t_3-t_2=\frac{1}{4}-\frac{1}{3}=\frac{3-4}{12}=-\frac{1}{12}$$
Clearly, the difference of successive terms is not constant, therefore given list of numbers does not form an AP.
(vi) 2, 22, 23, 24, ....
i.e., 2, 4, 8, 16, ...
Here, t₁ = 2, t₂ = 4, t₃ = 8, t₄ = 16
t₂ – t₁ = 4 – 2 = 2
t₃ – t₂ = 8 – 4 = 4
t₄ – t₃ = 16 – 8 = 8
Clearly, the difference of successive terms is not constant, therefore given list of numbers does not form an AP.

$$\textbf{(vii)}\sqrt{3},\sqrt{12},\sqrt{27},\sqrt{48}\\\text{i.e.,}\sqrt{3},\sqrt{12},\sqrt{27},\sqrt{48}$$
$$\text{Here},t_1=\sqrt{3},t_2=2\sqrt{3},t_3=3\sqrt{3},t_4=4\sqrt{3}\\t_2 – t_1 =2\sqrt{3}-\sqrt{3}=\sqrt{3}\\t_3 – t_2 =-3\sqrt{3}-2\sqrt{3}=\sqrt{3}\\t_4 – t_3 =-4\sqrt{3}-3\sqrt{3}=\sqrt{3}$$
Clearly, the difference of successive terms is constant, therefore given list of numbers form an AP.

2. Justify whether it is true to say that
$$-1,\frac{-3}{2},-2,\frac{5}{2},...\text{ forms an AP as}\space a_2 – a_1 = a_3 – a_2.$$

Ans. False. Here,
$$a_1=1,a_2=\frac{-3}{2},a_3,=-2,a_4=\frac{5}{2}\\a_2-a_1=\frac{-3}{2}+1=-\frac{1}{2}\\a_3-a_2=-2+\frac{3}{2}=-\frac{1}{2}\\a_4-a_3=\frac{5}{2}+2=\frac{9}{2}$$
Clearly, the difference of successive terms is not constant, therefore it does not form an AP.

3. For the AP : – 3, – 7, – 11, ...., can we find directly a₃₀ – a₂₀ without actually finding a₃₀ and a₂₀? Give reasons for your answer.

Sol. Yes, .. nth term of a_n AP,
a_n = a + (n – 1)d
∴ a₃₀ = a + (30 – 1)d
= a + 29d
and a₂₀ = a + (20 – 1)d
= a = 19d
Now, a₃₀ – a₂₀ = (a + 29d) – (a + 19d) = 10d
and from given AP common difference,
d = – 7 – (– 3)
= – 7 + 3 = – 4
∴ a₃₀ – a₂₀ = (30 – 20)d = 10d,
10(– 4) = – 40

4. Justify whether it is true to say that the following are the nth terms of an AP.
(i) 2n – 3
(ii) 3n² + 5
(iii) 1 + n + n²

Sol. (i) Yes. Here, a_n = 2n – 3
So, put n = 1, a₁ = 2(1) – 3 = – 1
put n = 2, a₂ = 2(2) – 3 = 1
put n = 3, a₃ = 2(3) – 3 = 3
put n = 4, a₄ = 2(4) – 3 = 5
List of numbers becomes – 1, 1, 3, ....
Here, a₂ – a₁ = 1 – (– 1)
= 1 + 1 = 2
a₃ – a₂ = 3 – 1 = 2
a₄ – a₃ = 5 – 3 = 2
So, it terms an AP,
Hence, 2n – 3 is the nth term of an AP.
(ii) No. Here, a_n = 3n² + 5
put n = 1, a₁ = 3(1)2 + 5 = 8
put n = 2, a₂ = 3(2)2 + 5
= 3(4) + 5 = 17
put n = 3, a₃ = 3(3)2 + 5
= 3(9) + 5
= 27 + 5 = 32
∴ List of number becomes 8, 17, 32, ...
Here,
a₂ – a₁ = 17 – 8 = 9
a₃ – a₂ = 32 – 17 = 15
∴ a₂ – a₁ ≠ a₃ – a₂
Since, the successive difference of the list is not constant. So, it does not form an AP.
(iii) No. Here, a_n = 1 + n + n²
put n = 1, a₁ = 1 + 1 + (1)² = 3
put n = 2, a₂ = 1 + 2 + (2)²
= 1 + 2 + 4 = 7
put n = 3, a₃ = 1 + 3 + (3)²
= 1 + 3 + 9 = 13
List of number becomes 3, 7, 13, ....
Here, a₂ – a₁ = 7 – 3 = 4
a₃ – a₂ = 13 – 7 = 6
∴ a₂ – a₁ ≠ a₃ – a₂
Since, the successive difference of the list is not constant. So, it does not form an AP.

5. Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why ?

Sol. Let the same common difference of two APs is d. Given that, the first term of first AP and second AP are 2 and 7 respectively, then the APs are
2, 2 + d, 2 + 2d, 2 + 3d, ...
and 7, 7 + d, 7 + 2d, 7 + 3d, ....
Here, we observe that the difference between any two corresponding terms of such APs is the same as the difference between their first term i.e., 7 – 2 = 5.

6. Is 0 a term of the AP : 31, 28, 25, .... ? Justify your answer.

Sol. No. Let 0 be the nth term of given AP, i.e., an = 0
Given that, first term, a = 31, common difference, d = 28 – 31 = – 3
The nth term of an AP, is
an = a + (n – 1)d
⇒ 0 = 31 + (n – 1) – 3
⇒ 3(n – 1) = 31
$$⇒n-1=\frac{31}{3}\\⇒n=\frac{31}{3}+1\\⇒n=\frac{34}{3}=11\frac{1}{3}$$
Since, n should be positive integer.
∴ 0 is not a term of the given AP.

7. The taxi fare after each km, when the fare is ₹ 15 for the first km and ₹ 8 for each additional km, does not form an AP as the total fare (in ₹) after each km is 15, 8, 8, 8, .... . Is the statement true? Give reasons.

Sol. No. Because the total fare (in ₹) after each km is
15, (15 + 8), (15 + 2 × 8), (15 + 3 × 8), ... = 15, 23, 31, 39, ....
Let t₁ = 15, t₂ = 23, t₃ = 31, t₄ = 39
Now, t₂ – t₁ = 23 – 15 = 8
t₃ – t₂ = 31 – 23 = 8
t₄ – t₃ = 39 – 31 = 8
Since, all the successive terms of the given list have constant difference i.e., common difference = 8.
Hence, the total fare after each km form an AP.

8. In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.
(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is ₹ 400.
(ii) The fee charged every month by a school from classes I to XII, when the monthly fee for class I is ₹ 250 and it increase by ₹ 50 for the next higher class.
(iii) The amount of money in the account of Varun at the end of every year when ₹ 1000 is deposited at simple interest of 10% per annum.
(iv) The number of bacteria in a certain food item after each second, when they double in every second.

Sol. (i) The fee charged from a student every month by a school for the whole session is
400 × 1, 400 × 2, 400 × 3, 400 × 4
i.e., 400, 800, 1200, 1600, ...
which form an AP, with common difference (d) = 800 – 400 = 400

(ii) The fee charged monthly by a school from I to XII is
250, (250 + 50), (250 + 2 × 50), (250 + 3 × 50),...
i.e., 250, 300, 350, 400,...
which form an AP, with common difference (d) = 300 – 250 = 50
$$\textbf{(iii)}\text{ Simple interest} =\frac{Principal×Rate×Time}{100}\\=\frac{1000×10×1}{100}=100$$
∴ The amount of money in the account of Varun at the end of every year is
1000, (1000 + 100 × 1), (1000 + 100 × 2), (1000 + 100 × 3), ...
i.e., 1000, 1100, 1200, 1300,...
which form an AP, with common difference (d) = 1100 – 1000 = 100

(iv) Let the number of bacteria in a certain food = x
Since, they double in every second.
∴ x, 2x, 2(2x), 2(2.2.x), ...
i.e., x, 2x, 4x, 8x, ...
Now, let t₁ = x, t₂ = 2x, t₃ = 4x, t₄ = 8x
t₂ – t₁ = 2x – x = x
t₃ – t₂ = 4x – 2x = 2x
t₄ – t₃ = 8x – 4x = 4x
Since, the difference between each successive terms is not constant.
So, the list is not form an AP.

Exercise 5.3

1. Match the APs given in column A with suitable common differences given in column B.

	Column A		Column B
(A ₁)	2, – 2, – 6, – 10,...	(B ₁)	2/3
(A ₂)	a = – 18, n = 10, an = 0	(B ₂)	–5
(A ₃)	a = 0, a10 = 6	(B ₃)	4
(A ₄)	a ₂ = 13, a ₄ = 3	(B ₄)	–4
		(B ₅)	2
		(B ₆)	1/2
		(B ₇)	5

Sol. (A₁) 2, – 2, – 6, – 10, ....
Here, Common difference,
d = – 2 – 2 = – 4
(A₂) ..a_n = a + (n – 1)d
0 = – 18 + (10 – 1)d
18 = 9d
⇒ Common difference, d = 2
(A₃) .. a₁₀ = 6 ⇒ a + (10 – 1)d = 6
⇒ 0 + 9d = 6 [.. a = 0 (given)]
⇒ 9d = 6
⇒ d = 23
(A₄) .. a₂ = 13 ⇒ a + (2 – 1)d = 13
[.. a₁₁ = a + (n – 1)d]
⇒ a + d = 13 ...(i)
and a₄ = 3
⇒ a + (4 – 1)d = 3
⇒ a + 3d = 3 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
2d = – 10
⇒ d = – 5
∴ (A₁) → B₄
(A₂) → B₅
(A₃) → B₁
(A₄) → B₂

2. Verify that each of the following is an AP and then write its next three terms.
$$\textbf{(i) }0,\frac{1}{4},\frac{1}{2},\frac{3}{4}....\\\textbf{(ii)}5,\frac{14}{3},\frac{13}{3},4....\\\textbf{(iii)}\sqrt{3},2\sqrt{3},\sqrt{3},....\\\textbf{(iv)}a + b, (a + 1) + b, (a + 1) + (b + 1),...\\\textbf{(v)} a, 2a + 1, 3a + 2, 4a + 3, ...$$

$$\textbf{Sol. (i)} \text{Here,}a_1=0,a_2=\frac{1}{4},a_3=\frac{1}{2},a_4=\frac{3}{4}\\a_2 – a_1=\frac{1}{4}\\a_3 – a_2=\frac{1}{2}-\frac{1}{4}=\frac{1}{4},\\a_4 – a_3 =\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\\a_2 – a_1 = a_3 – a_2 = a_4 – a_3$$

Since, the each successive terms of the given list have the same difference. So, it form an AP.
The next three terms are, a₅ = a₁ + 4d
$$=a+4(\frac{1}{4})=1\\a_6 = a_1 + 5d\\= a+5(\frac{1}{4})=\frac{5}{4}\\a_7 = a_1 + 6d\\=0+\frac{6}{4}=\frac{3}{2}$$

$$\textbf{(ii)}\text{Here,}a_1=5,a_2=\frac{14}{3},a_3=\frac{13}{3},a_4=4\\a_2 – a_1 = \frac{14-15}{3}=\frac{-1}{3}\\a_3 – a_2 =\frac{13}{4}-\frac{14}{3}\\=-\frac{1}{3} \\a_4 – a_3 = 4- \frac{13}{3}\\=\frac{12-13}{3}=\frac{-1}{3}\\ a_2 – a_1 = a_3 – a_2 = a_4 – a_3$$
Since, the each successive terms of the given list have same difference, it form an AP.
The next three terms are
$$a_5 = a_1 + 4d\\= 5+4(-\frac{1}{3})\\5-\frac{4}{3}=\frac{11}{3}\\a_6 = a_1 + 5d\\= 5+5(-\frac{1}{3})\\=5-\frac{5}{3}=\frac{10}{3}\\a_7 = a_1 + 6d\\= 5+6(-\frac{1}{3})\\= 5 – 2 = 3$$

$$\textbf{(iii)}\text{Here},a_1=\sqrt{3},a_2=2\sqrt{3},a_3=3\sqrt{3}\\a_2-a_1=2\sqrt{3}-\sqrt{3}=\sqrt{3},a_3-a_2=3\sqrt{3}-2\sqrt{3},a_3=\sqrt{3}$$
.. a₂ – a₁ = a₃ – a₂ = √3= common difference
Since, the each successive terms of the given list have same difference.
∴ It form an AP.
The next three terms are
a₄ = a₁ + 3d
$$\sqrt{3}+3(\sqrt{3})=4\sqrt{3}\\a_5 = a_1 + 4d\\\sqrt{3}+4\sqrt{3}=5\sqrt{3}\\a_6 = a_1 + 5d\\=\sqrt{3}+5\sqrt{3}=6\sqrt{3}$$

(iv) Here, a₁ = a + b, a₂ = (a + 1) + b, a₃ = (a + 1) + (b + 1)
a₂ – a₁ = (a + 1) + b – (a + b)
= a + 1 + b – a – b = 1
a₃ – a₂ = (a + 1) + (b + 1) – [(a + 1) + b]
= a + 1 + b + 1 – a – 1 – b
= 1
... a₂ – a₁ = a₃ – a₂
= 1 = common difference
Since, the each successive terms of the given list have same difference.
∴ It form an AP.
The next three terms are
a₄ = a₁ + 3d
= a + b + 3(1)
= (a + 2) + (b + 1)
a₅ = a₁ + 4d
= a + b + 4(1)
= (a + 2) + (b + 2)
So, a₆ = a₁ + 5d
= a + b + 5(1)
= (a + 3) + (b + 2)

(v) Here, a₁ = a, a₂ = 2a + 1, a₃ = 3a + 2, a₄ =
4a + 3
a₂ – a₁ = 2a + 1 – a = a + 1
a₃ – a₂ = 3a + 2 – 2a – 1 = a + 1
a₄ – a₃ = 4a + 3 – 3a – 2 = a + 1
... a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = a + 1 = common difference
Since, the each successive terms of the given list have same difference.
∴ It form an AP.
The next three terms are
a₅ = a + 4d
= a + 4(a + 1)
= 5a + 4
a₆ = a + 5d
= a + 5(a + 1)
= 6a + 5
a₇ = a + 6d
= a + 6(a + 1)
= 7a + 6

3. Find a, b and c such that the following numbers are in AP : a, 7, b, 23 and c.

Sol. .. a, 7, b, 23 and c are in AP.
7 – a = b – 7 = 23 – b = c – 23 = common difference
I II III IV
Taking II and III terms, we get
b – 7 = 23 – b
⇒ 2b = 30
⇒ b = 15
Taking I and II terms, we get
7 – a = b – 7
7 – a = 15 – 7 (.. b = 15)
7 – a = 8
⇒ a = – 1
Taking III and IV terms, we get
23 – b = c – 23
⇒ 23 – 15 = c – 23 (.. b = 15)
⇒ 8 = c – 23
⇒ 8 + 23 = c
⇒ c = 31
Hence, a = – 1
b = 15
c = 31

4. Determine the AP whose fifth term is 19 and the difference of the eight term from the thirteenth term is 20.

Sol. Let the first term of an AP be a and common difference d.
Given, a₅ = 19 and a₁₃ – a₆ = 20 (Given)
⇒ a + (5 – 1)d = 19
and a + (13 – 1)d – [a + (8 – 1)d] = 20
[ a_n = a + (n – 1)d]
⇒ a + 4d = 19 ...(i)
and a + 12d – a – 7d = 20
⇒ 5d = 20
⇒ d = 4
Putting d = 4 in equation (i), we get
a + 4(4) = 19
a + 16 = 19
a = 19 – 16 = 3
∴ The required AP is a, a + d, a + 2d, a + 3d, ....
i.e., 3, 3 + 4, 3 + 2(4), 3 + 3(4), ....
i.e., 3, 7, 11, 15, ....

5. The sum of the 5th and the 7th terms of an AP is 52 and the 10th term is 46. Find the AP.

Sol. Let the first term and common difference of AP are a and d, respectively.
According to the question,
a₅ + a₇ = 52 and a₁₀ = 46
a + (5 – 1)d + a + (7 – 1)d = 52
[.. a_n = a + (n – 1)d]
and a + (10 – 1)d = 46
a + 4d + a + 6d = 52
and a + 9d = 46
2a + 10d = 52
and a + 9d = 46
a + 5d = 26 ...(i)
a + 9d = 46 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
4d = 20 ⇒ d = 5
∴ From Eq. (i), we get a = 26 – 5(5) = 1
∴ The required AP is a, a + d, a + 2d, a + 3d, ...
i.e., 1, 1 + 5, 1+ 2(5), 1 + 3(5), ....
i.e., 1, 6, 11, 16, ...

6. Determine k so that k² + 4k + 8, 2k² + 3k + 6, 3k² + 4k + 4 are three consecutive terms of an AP.

Sol. Since, k² + 4k + 8, 2k² + 3k + 6 and 3k² + 4k + 4 are consecutive terms of an AP.
∴ 2k² + 3k + 6 – (k² + 4k + 8) = (3k² + 4k + 4 – (2k² + 3k + 6) = common difference
⇒ 2k² + 3k + 6 – k² – 4k – 8 = 3k² + 4k + 4 – 2k²– 3k – 6
⇒ k² – k – 2 = k² + k – 2
⇒ – k = k ⇒ 2k = 0 ⇒ k = 0

7. If sum of the 3rd and the 8th terms of an AP is 7 and the sum of the 7th and the 14th terms is –3, find the 10th term.

Sol. Let the first term and common difference of an AP are a and d, respectively.
According to question,
a₃ + a₈ = 7 and a₇ + a₁₄ = – 3
a + (3 – 1)d + a + (8 – 1)d = 7
[.. a_n = a + (n – 1)d]
and a + (7 – 1)d+ a + (14 – 1)d = – 3
a + 2d + a + 7d = 7
and a + 6d + a + 13d = – 3
2a + 9d = 7 ...(i)
and 2a + 19d = – 3 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
10d = – 10 ⇒ d = –1
∴ From equation (i), we get
2a + 9(– 1) = 7
⇒ 2a – 9 = 7
⇒ 2a = 16
⇒ a = 8
∴ a₁₀ = a + (10 – 1)d
= 8 + 9 (–1)
= 8 – 9 = – 1

8. The 26th, 11th and the last terms of an AP are 0, 3 and (-1/5) respectively. Find the common difference and the number of terms.

Sol. Let a and d be the first and common difference of an AP. Since, the nth term of an AP is
an = a + (n – 1)d ...(i)
Given that, a20 = 0 = a + (26 – 1)d
⇒ a + 25d = 0 ...(ii)
and a11 = a + (11 – 1)d = 3
⇒ a + 10d = 3 ...(iii)
$$\text{Last term (l)}=-\frac{1}{5}\\\text{On subtracting Eq. (iii) from Eq. (ii), we get}\\15d = – 3\\⇒d=-\frac{1}{5}\\\text{From Eq. (ii), we get}\\a + 25(-\frac{1}{5})= 0 ⇒ a = 5\\ a_n = l = a + (n – 1)d\\⇒-\frac{1}{5}= 5 +(n-1)(-\frac{1}{5})\\⇒ –1 = 25 – (n – 1)\\⇒ (n – 1) = 26 ⇒ n = 27$$

9. If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term.

Sol. Let a and d be the first term and common difference of an AP, respectively.
Given that, a₉ = 0
⇒ a + (9 – 1)d = 0 [.. a_n = a + (n – 1)d]
⇒ a+ 8d = 0 ...(i)
Now, a₂₉ = a + (29 – 1)d = a + 28d
= – 8d + 28d [From Eq. (i)]
= 20d ...(ii)
and a₁₉ = a + (19 – 1)d = a + 18d
= – 8d + 18d = 10d ...(iii)
From Eqs. (ii) and (iii), we get
a₂₉ = 2 × a₁₉

10. Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.

Sol. Let the three parts of 204 are a – d, a, a + d, also which lies in AP with common difference d.
Now, according to the question,
(a – d) + a + (a + d) = 207
⇒ 3a = 207
⇒ a = 69
Also, product of two smaller parts = 4623
⇒ a (a – d) = 4623
⇒ 69(69 – d) = 4623
⇒ 69 – d = 67
⇒ d = 69 – 67 = 2
∴ The required three parts are (69 – 2), 69,
(69 + 2) or 67, 69, 71.

11. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

Sol. Let the angles of a triangle are a – d, a and a + d lie in AP.
Since, sum of all interior angles of a triangle
= 180°
⇒ (a – d) + a + (a + d) = 180°
⇒ 3a = 180°
⇒ a = 60°
Also, by condition,
Greatest angle = 2 × Least angle
⇒ (a + d) = 2(a – d)
⇒ 60° + d = 2(60° – d)
⇒ 60° + d = 120° – 2d
⇒ 3d = 60°
⇒ d = 20°
∴ The required angles of the triangle are 60°
– 20°, 60°, 60° + 20°.
i.e., 40°, 60°, 80°

12. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?

Sol. Here, the first number is 11, which divided by 4 leave remainder 3 between 10 and 300. Last term before 300 is 299, which divided by 4 leave remainder 3.
∴ 11, 15, 19, 23, ...., 299
Here, first term (a) = 11, common difference
d = 15 – 11 = 4
... n^th term, an = a + (n – 1)d = l (last term)
⇒ 299 = 11 + (n – 1)4
⇒ 299 – 11 = (n – 1) 4
⇒ 4(n – 1) = 299
⇒ (n – 1) = 72
⇒ n = 73

13. Find the sum of the two middle most terms of the AP :$$-\frac{4}{3},-1,-\frac{2}{3},....4\frac{1}{3}.$$

Sol. Here, first term
$$(a)=-\frac{4}{3},\text{common difference}\\(d) =-1+\frac{4}{3}=\frac{1}{3},\text{and the last term (l)}=4\frac{1}{3}=\frac{1}{3}\\\text{ The nth term of an AP is}\\l = a_n = a + (n – 1)d\\⇒\frac{13}{3}=-\frac{4}{3}+(n-1)\frac{1}{3}\\⇒ 13 = – 4 + (n – 1)\\⇒ n – 1 = 17\\⇒ n = 18 \text{(Even)}\\\text{∴ The two middle most term is}(\frac{n}{2})th\space\text{and}(\frac{n}{2}+1)th\space\text{terms.}\\i.e.,(\frac{18}{2})th \space and(\frac{n}{2}+1)th\space term\\\text{i.e., 9th and 10th terms}\\a_9 = a + 8d\\=-\frac{4}{3}+8(\frac{1}{3})\\=\frac{8-4}{3}=\frac{4}{3}\\\text{and}\space a_{10} = a + 9d\\=\frac{-4}{3}+9(\frac{1}{3})\\=\frac{9-4}{3}=\frac{5}{3}\\\text{∴ Sum of the two middle most terms} = a_9 + a_{10}\\=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3$$

14. Find the sum
$$(i)(4-\frac{1}{n})+(4-\frac{2}{n})+(4-\frac{3}{n})+..\text{upto terms.}\\(ii)\frac{a-b}{a+b}+\frac{3a-2b}{a+b}+\frac{5a-3b}{a+b}+..\text{to 11 times.}$$

Sol. (i) Here, first term,
$$4-\frac{1}{n},\\\text{Common difference, d}=(4-\frac{1}{n})-(4-\frac{1}{n})\\=-\frac{-2}{n}+\frac{1}{n}=\frac{-1}{n}\\\text{ Sum of n terms of an AP is}\\S_n=\frac{n}{2}[2a+(n-1)d]\\⇒S_n=\frac{n}{2}[2(4-\frac{1}{n})+(n-1)(-\frac{1}{n})]\\=\frac{n}{2}[8-\frac{2}{n}-1+\frac{1}{n}]\\=\frac{n}{2}(7-\frac{1}{n})\\=\frac{n}{2}×(\frac{7n-1}{n})\\=\frac{7n-1}{2}$$

(ii) Here, first term (A)
$$=\frac{a-b}{a+b}\text{and the common difference.}\\D=\frac{3a-2b}{a+b}-\frac{a-b}{a+b}=\frac{2a-b}{a+b}\\\text{ Sum of n terms of an AP is}\\S_n=\frac{n}{2}[2A+(n-1)D]\\\frac{n}{2}[2\frac{(a-b)}{(a+b)}+(n-1)\frac{(a-b)}{(a+b)}]\\=\frac{n}{2}[\frac{2a-2b+2an-2a-bn+b}{a+b}]\\=\frac{n}{2}[\frac{2an+bn-b}{a+b}]\\S_{11}=\frac{11}{2}\frac{[2a(11)-b(11)-b}{a+b}]\\=\frac{11}{2}\frac{[(22a-12b)}{(a+b)}]\\=\frac{11(11a-6b)}{a+b}$$

15. In an AP, if S_n = n(4n + 1), find the AP.

Sol. We know that, the nth term of an AP is
a_n = S_n – S_{n – 1}
a_n = n(4n + 1) – (n – 1) [4(n – 1) + 1]
[Given, S_n = n(4n + 1)]
⇒ a_n = 4n² + n – (n – 1)(4n – 3)
= 4n² + n – 4n² + 3n + 4n – 3
= 8n – 3
∴ Put n = 1, a₁ = 8(1) – 3 = 5
n = 2, a₂ = 8(2) – 3 = 16 – 3 = 13
n = 3, a₃ = 8(3) – 3 = 24 – 3 = 21
Hence, required AP is 5, 13, 21, ....

16. In an AP, if S_n = 3n² + 5n and a_k = 164, find the value of k.

Sol. The nth term of an AP is
a_n = S_n – S_{n – 1}
= 3n² + 5n –3(n – 1)2 – 5(n – 1)
[ S_n = 3n² + 5n (Given)]
= 3n² + 5n – 3n² – 3 + 6n – 5n + 5
a_n = 6n + 2 ...(i)
or a_k = 6k + 2 = 164 [(Given) a_k = 164]
⇒ 6k = 164 – 2 = 162
⇒ k = 27

17. If S_n denotes the sum of first n terms of an AP, prove that S₁₂ = 3(S₈ – S4).

Sol. .. Sum of n terms of an AP,

$$S_n=\frac{n}{2}[2a+(n-1)d]...(i)\\S_n=\frac{8}{2}[2a+(8-1)d]\\= 4(2a + 7d)\\= 8a + 28d\\\text{and}\space S_4=\frac{4}{2}[2a+(4-1)d]\\= 2(2a + 3d)\\= 4a + 6d\\\text{Now,}\\S_8 – S_4 = 8a + 28d – 4a – 6d\\= 4a + 22d ...(ii)\\\text{and}\space S_{12}=\frac{12}{2}[2a+(12-1)d]\\= 6(2a + 11d)\\= 3(4a + 22d)\\= 3(S_8 – S_4) \text{[From Eq. (ii)]}$$

18. If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Sol. Let a and d be the first term and common difference respectively of an AP.
.. Sum of n terms of an AP,
$$S_n=\frac{n}{2}[2a+(n-1)d]...(i)\\\text{Now, given},\space S_8 = 36\\⇒\frac{6}{2}[2a+(6-1)d]=36\\⇒ 2a + 5d = 12 ...(ii)\\\text{and}\space S_{16} = 256\\⇒\frac{16}{2}[2a+(16-1)d]=256\\⇒ 2a + 15d = 32 ...(iii)\\\text{On subtracting Eq. (ii) from Eq. (iii), we get}\\10d = 20\\d = 2\\\text{From equation (ii), we get}\\2a + 5(2) = 12\\⇒ 2a = 12 – 10 = 2\\⇒ a = 1\\S_{10}=\frac{10}{2}[2a+(10-1)d]\\= 5[2(1) + 9(2)]\\= 5(2 + 18)\\= 5 × 20\\= 100$$

19. Find the sum of all the 11 terms of an AP whose middle most term is 30.

Sol. Since, the total number of terms (n) = 11 (Odd)
$$\text{∴ The middle most term}=\frac{(n+1)}{2}th\\=\frac{(11+1)}{2}\text{th\space term}\\\text{Given that}, \space a_6 = 30\\⇒ a + (6 – 1)d = 30\\⇒ a + 5d = 30 ...(i)\\\text{ The sum of n terms of an AP,}\\S_n=\frac{n}{2}[2a+(n-1)d]\\\frac{11}{2}[2a+10d]\\= 11(a + 5d) [From Eq. (i)]\\= 11 × 30 = 330$$

20. Find the sum of last ten terms of the AP 8, 10, 12, ...., 126.

Sol. For finding, the sum of last ten terms, we write the given AP in reverse order.
i.e., 126, 124, 122, ....., 12, 10, 8
Here, first term (a) = 126, common difference,
d = 124 – 126 = – 2
$$S_{10}=\frac{10}{2}[2a+(10-1)d]\\{[..S_n=\frac{n}{2}[2a+(n-1)d]]}\\= 5[2(126) + 9(– 2)]\\= 5(252 – 18)\\= 5 × 234 = 1170$$

21. Find the sum of first seven numbers which are multiples of 2 as well as of 9.

Sol. For finding, the sum of first seven numbers which are multiples of 2 as well as of 9. Take LCM of 2 and 9 which is 18.
So, the series becomes 18, 36, 54, ....
Here, first term (a) = 18, common difference (d) = 36 – 18 = 18
$$S_n=\frac{n}{2}[2a+(n-1)d]\\=\frac{7}{2}[2(18)+(7-1)18]\\\frac{7}{2}[36+6(18)]\\= 7(18 + 3 × 18)\\= 7(18 + 54)\\= 7 × 72 = 504$$

22. How many terms of the AP : – 15, – 13, – 11, .... are needed to make the sum – 55?

Sol. Let n number of terms are needed to make the sum – 55.
Here, first term a = – 15, common difference
d = – 13 + 15 = 2
.. Sum of n terms of an AP,
$$S_n=\frac{n}{2}[2a+(n-1)d]\\-56=\frac{n}{2}[2(-15)+(n-1)2]\\⇒ – 55 = – 15n + n(n – 1)\\⇒ n^2 – 16n + 55 = 0\\⇒ n^2 – 11n – 5n + 55 = 0\\\text{(By factorisation method)}\\⇒ n(n – 11) – 5(n – 11) = 0\\⇒ (n – 11)(n – 5) = 0\\⇒ n = 5, 11$$
Here, if we take n = 5, i.e., – 55 is the sum of only negative terms. Whereas at n = 11, – 55 is the sum of some negative and some positive terms, that is why two answer is possible for the sum – 55.

23. The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is – 30 and the common difference is 8. Find n.

Sol. Given that, first term of first AP (a₁) = 8
and common difference of first AP (d₁) = 20
First term of second AP(a₂) = 8
and common difference of second AP (d₂) = 8
Now, sum of first n terms of first AP is
$$S_n=\frac{n}{2}[2(8)+(n-1)20]\\S_n=\frac{n}{2}[2(8)+(n-1)20]\\S_n = n(8 + 10n – 10)\\S_n = 10n^2 – 2n ...(i)\\\text{and sum of first 2n terms of second AP is}\\S_{2n}=\frac{n}{2}[2a_2+(2n-1)d_2]\\⇒ S_{2n} = n[2(– 30) + (2n – 1)(8)]\\⇒ S_{2n} = n[– 60 + 16n – 8]\\⇒ S_{2n} = 16n^2 – 68n ...(ii)$$
By condition,
Sum of first n terms of first AP (S_n) = Sum of first 2n terms of second AP (S_2n)
$$⇒ 10n^2 – 2n = 16n^2 – 68n\\⇒ 6n^2 – 66n = 0\\⇒ 6n(n – 11) = 0\\⇒ n = 11 ( n ≠ 0)$$

Exercise 5.4

1. Find the
(i) sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
(ii) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
(iii) sum of those integers from 1 to 500 which are multiples of 2 or 5.

Sol. (i) Since, multiples of 2 as well as of 5 = LCM of (2, 5) = 10
∴ Multiples of 2 as well as of 5 between 1 and 500 is 10, 20, 30, ...., 490 which form an AP with first term (a) = 10
and common difference (d) = 20 – 10 = 10
nth term an = last term (l) = 490
∴ Sum of n terms between 1 and 500
$$S_n=\frac{n}{2}[a+l]...(i)\\ 10 + (n – 1)10 = 490\\⇒ 10 + (n – 1)10 = 480\\⇒ (n – 1)10 = 480\\⇒ n – 1 = 48\\⇒ n = 49\\\text{Now, from equation (i), we get}\\S_{49}=\frac{49}{2}(10+490])\\=\frac{49}{2}×500\\= 49 × 250 = 12250$$

(ii) Same as part (i);
but multiples of 2 as well as of 5 from 1 to 500 is 10, 20, 30, ...., 500.
∴ a = 10, d = 10, a₁₁ = l = 500
∴ a_n = a + (n – 1)d = l
⇒ 500 = 10 + (n – 1)10
⇒ 490 = (n – 1)10
⇒ n – 1 = 49
⇒ n = 50
$$S_n=\frac{n}{2}(a+l)\\⇒S_{50}=\frac{50}{2}(10+500)\\\frac{50}{2}×510\\= 50 × 255\\= 12750$$

(iii) Since, multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (2, 5) i.e., 10.
∴ Multiples of 2 or 5 from 1 to 500
= List of multiples of 2 from 1 to 500 + List of multiples of 5 from 1 to 500 – List of multiples of 10 from 1 to 500
= (2, 4, 6, ....., 500) + (5, 10, 15, ....., 500)
– (10, 20, ....., 500)
All of these list form an AP.
Now, number of terms in first list
500 = 2 + (n₁ – 1) 2
⇒ 498 = (n₁ – 1)²
⇒ n₁ – 1 = 249
⇒ n₁ = 250
Number of terms in second list
500 = 5 + (n₂ – 1) 5
⇒ 495 = (n₂ – 1)5
⇒ 99 = (n₂ – 1) ⇒ n₂ = 100
and number of terms in third list
500 = 10 + (n₃ – 1)10
⇒ 490 = (n3 – 1)10
⇒ n3 = 50
Now, from equation (i), we get
∴ Sum of multiples of 2 or 5 from 1 to 500
= Sum of (2, 4, 6, ....., 500) + Sum of (5, 10, ......., 500) – Sum of (10, 20, ....., 500)
$$=\frac{n}{2}[2+500+\frac{n_2}{2}[5+500]-\frac{n_3}{2}[10+500]]\\{[..S_n=\frac{n}{2}(a+l)]}\\=\frac{250}{2}×502+\frac{100}{2}×505-\frac{50}{2}×510\\= 250 × 251 + 505 × 50 – 25 × 510\\= 62750 + 26250 – 12750\\= 88000 – 12750\\= 75250$$

2. The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term.

Sol. Let a and d be the first term and common difference of an AP, respectively.
Now, by condition,
$$a_8=\frac{1}{2}a_2\\⇒ a + 7d =\frac{1}{2}(a+d)\\⇒ 2a + 14d = a + d ...(i)\\⇒ a + 13d = 0\\\text{and}\space a_{11}=\frac{1}{3}a_4+1\\⇒ a + 10d =\frac{1}{3}[a+3d]+1\\⇒ 3a + 30d = a + 3d + 3\\⇒ 2a + 27d = 3 ...(ii)\\\text{From Eqs. (i) and (ii), we get}\\2(– 13d) + 27d = 3\\⇒ – 26d + 27d = 3\\⇒ d = 3\\\text{From equation (i), we get}\\a + 13 (3) = 0\\⇒ a = – 39\\∴ a15 = a + 14d = – 39 + 14(3)\\= – 39 + 42 = 3$$

3. An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.

Sol. Since, total number of terms (n) = 37 (odd)
∴ Middle term =((37+1)/(2))th term = 19th term
So, the three middle most terms 18th, 19th and 20th terms.
By condition,
Sum of the three middle most terms = 225
⇒ a₁₈ + a₁₉ + a₂₀ = 225
⇒ (a + 17d) + (a + 18d) + (a + 19d) = 225
⇒ 3a + 54d = 225
⇒ a + 18d = 75 ...(i)
and the sum of the last three terms = 429
⇒ a₃₅ + a₃₆ + a₃₇ = 429
⇒ (a + 34d) + (a + 35d) + (a + 36d) = 429
⇒ 3a + 105d = 429
⇒ a + 35d = 143 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
17d = 68 ⇒ d = 4
From equation (i), we get
a + 18(4) = 75
⇒ a = 75 – 72 ⇒ a = 3
∴ Required AP is : a, a + d, a + 2a, a + 3d,...
i.e., 3, 3 + 4, 3 + 2(4), 3 + 3(4), ....
i.e., 3, 7, 3 + 8, 3 + 12,....
i.e., 3, 7, 11, 15, ...

4. Find the sum of the integers between 100 and 200 that are not divisible by 9.

Sol. Since, the sum of the integers 100 and 200 which is not divisible by 9 = (sum of total numbers between 100 and 200) – (sum of total numbers between 100 and 200 which is divisible by 9).
...(i)
Now, the number of terms between 100 and 200.
Here, a = 101, d = 102 – 101 = 1 and a_n = l = 199
.. a_n = l = a + (n – 1)d
⇒ 199 = 101 + (n – 1)1
⇒ (n – 1) = 98
⇒ n = 99
∴ Sum of terms between 100 and 200,
$$S_n=\frac{n}{2}[2a+(n-1)d]\\⇒S_{99}=\frac{99}{2}[2(101)+(99-1)1]\\=\frac{99}{2}[202+96]\\=\frac{99}{2}×300\\= 99 × 150 = 14850\\\text{Now, the number of terms between} \\\text{100 and 200 which is divisible by 9.}\\Here, a = 108, d = 117 – 108 = 9 \space and\space a_n = l = 198\\ a_n = l = a + (n – 1)d\\⇒ 198 = 108 + (n – 1)9\\⇒ 90 = (n – 1)9\\⇒ n – 1 = 10\\⇒ n = 11\\∴ \text{Sum of terms between 100 and 200 which is divisible by 9.}\\S_n=\frac{n}{2}[2a+(n-1)d]\\S_{11}=\frac{11}{2}[2(108)+(11-1)9]\\=\frac{11}{2}[216+90]\\=\frac{11}{2}×306=11×153=1683$$
Now, from equation (i), we get
the sum of the integers between 100 and 200 which is not divisible by 9 = 14850 – 1683
= 13167.

5. The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

Sol. Let a and d be the first term and common difference of an AP.
Given that, a11 : a₁₈ = 2 : 3
$$⇒\frac{a+10d}{a+17d}=\frac{2}{3}\\⇒ 3a + 30d = 2a + 34d\\⇒ a = 4d ...(i)\\Now, a_5 = a + 4d\\= 4d + 4d\\= 8d [From, Eq. (i)]\\and\space a_{21} = a + 20d\\= 4d + 20d\\= 24d [From, Eq. (i)]\\∴ a_5 : a_{21} = 8d : 24d = 1 : 3\\\text{Now, sum of the first five terms,}\\S_n=\frac{5}{2}[2a+(5-1)d]\\S_n=\frac{5}{2}[2(4d)+4d]\space [From Eq. (i)]\\=\frac{5}{2}[2(4d)+4d\\=\frac{5}{2}(8d+4d)\\=\frac{5}{2}×12d=30d\\\text{and sum of the first 21 terms,}\\S_{21}=\frac{21}{2}[2a+(21-1)d]\\S_{21}=\frac{n}{2}[2(4d)+20d]\space [From Eq. (i)]\\=\frac{21}{2}(28d)=294d\\\text{∴ Ratio of the sum of the first five terms to the sum of the first 21 terms}\\S_5 : S_{21} = 30d : 294d = 5 : 49$$

6. Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to
$$=\frac{(a+c)(b+c-2a)}{2(b-a)}$$

Sol. Given that, the AP is a, b, ......, c.
Here, first term = a, common difference = b – a
and last term l = a_n = c
.. a_n = l = a + (n – 1)d
⇒ c = a + (n – 1)(b – a)
$$⇒ (n – 1) =\frac{(c-a)}{(b-a)}\\⇒n =\frac{c-a}{b-a}+1\\n=\frac{c-a+b-a}{b-a}\\=\frac{c+b-2a}{b-a}...(i)\\\text{∴ Sum of an AP,}\\S_n=\frac{n}{2}[2a+(n-1)d]\\=\frac{b+c-2a}{2(b-a)}[2a+\frac{b+c-2a}{b-a}-1](b-a)\\=\frac{b+c-2a}{2(b-a)}[2a+\frac{c-a}{b-a}.(b-a)]\\=\frac{b+c-2a}{2(b-a)}.(a+c)\space\textbf{Hence proved.}$$

7. Solve the equation – 4 + (–1) + 2 + ... + x = 437.

Sol. Given that, 4 – 1 + 2 + ... + x = 437 ...(i)
Here, – 4 – 1 + 2 .... + x forms an with first term = – 4, common difference = 3, a_n = l = x
.. nth term of an AP,
a_n = l = a + (n – 1)d
⇒ x = – 4 + (n – 1) 3
$$⇒ \frac{x+4}{3}=n-1\\⇒ n=\frac{x+7}{3}...(ii)\\\text{∴ Sum of an AP,}\\S_n=\frac{n}{2}[2a+(n-1)d]\\S_n=\frac{x+7}{2×3}[2(-4)+\frac{(x+4)}{3}3]\\=\frac{x+7}{2×3}(-8+x+4)\\=\frac{(x+7)(x-4)}{2×3}\\\text{Now, from equation (i), we get}\\S_n = 437\\=\frac{(x+7)(x-4)}{2×3}=437\\⇒ x^2 + 7x – 4x – 28 = 874\\⇒ x^2 + 3x – 2650 = 0\\x=\frac{-3±\sqrt{(3)^2-4(-2650)}}{2}\\\text{(By quadratic formula)}\\=\frac{-3±\sqrt{9+10600}}{2}\\=\frac{-3±103}{2}=\frac{100}{2},\frac{-106}{2}\\= 50, – 53\\\text{Here}, x = 50\\\text{( x cannot be negative} i.e., x ≠ 53)$$

NCERT Solutions for Class 10 Mathematics Chapter 5 Free PDF Download

Please Click on Free PDF Download link to Download the NCERT Solutions for Class 10 Mathematics Chapter 5 Arithmetic Progressions

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NCERT Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions

NCERT Solutions for Class 10 Mathematics Chapter 5 Free PDF Download

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