# NCERT Solutions for Class 11 Maths Chapter 8 - Sequences and Series

Exercise 9.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth term are:

1. an = n(n + 2).

Sol. We have, an = n(n + 2)

Substituting n = 1, 2, 3, 4, 5

At n = 1, a1 = 1(1 + 2) = 3

At n = 2, a2 = 2(2 + 2) = 8

At n = 3, a3 = 3(3 + 2) = 15

At n = 4, a4 = 4(4 + 2) = 24

At n = 5, a5 = 5(5 + 2) = 35

Hence, the first five terms are 3, 8, 15, 24, 35.

$$\textbf{2.}\space \textbf{a}_\textbf{n}=\frac{\textbf{n}}{\textbf{n+1}}\\\textbf{Sol.}\space\text{We have}\space a_n=\frac{1}{n+1}$$

Substituting n = 1, 2, 3, 4, 5

$$\text{At n = 1,}\space a_1=\frac{1}{1+1}=\frac{1}{2}\\\text{At n = 2,}\space a_2=\frac{2}{2+1}=\frac{2}{3}\\\text{At n = 3}\space a_3=\frac{3}{3+1}=\frac{3}{4}\\\text{At n = 4,}\space a_4=\frac{4}{4+1}=\frac{4}{5}\\\text{At n = 5,}\space a_5=\frac{5}{5+1}=\frac{5}{6}\\\text{Hence,thefirstfive terms are}\\\space\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}\text{and}\frac{5}{6}.$$

3. an = 2n.

Sol. We have, an = 2n

Substituting n = 1, 2, 3, 4, 5

At n = 1, a1 = 21 = 2

At n = 2, a2 = 22 = 4

At n = 3, a3 = 23 = 8

At n = 4, a4 = 24 = 16

At n = 5, a5 = 25 = 32

$$\textbf{4.}\space \textbf{a}_\textbf{n}=\frac{\textbf{2n-3}}{\textbf{6}}\textbf{.}\\\textbf{Sol.}\space\text{We have}\space a_n=\frac{2n-3}{6}\\\text{Substituting n = 1, 2, 3, 4, 5}\\\text{At n = 1,}\space a_1=\frac{2×1-3}{6}=\frac{2-3}{6}=-\frac{1}{6}\\\text{At n = 2,}\space a_2=\frac{2×2-3}{6}=\frac{4-3}{6}=\frac{1}{6}\\\text{At n = 3}\space a_3=\frac{2×3-3}{6}=\frac{6-3}{6}\\=\frac{3}{6}=\frac{1}{2}\\\text{At n = 4,}\space a_4=\frac{2×4-3}{6}=\frac{8-3}{6}=\frac{5}{6}$$

$$\text{At n = 5,}\space a_5=\frac{2×5-3}{6}=\frac{10-3}{6}=\frac{7}{6}\\\text{ Hence, the first five terms are}\\\frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$$

5. an = (– 1)n – 1 5n + 1.

Sol. We have, an = (– 1)n – 1 5n + 1

Substituting n = 1, 2, 3, 4, 5

At n = 1, a1 = (– 1)1 – 1 51 + 1 = (– 1)0 52 = 25

At n = 2, a2 = (– 1)2 – 1 + 52 + 1 = (– 1)1 53 = – 125

At n = 3, a3 = (– 1)3 – 1 53 + 1 = (– 1)2 54 = 625

At n = 4, a4 = (– 1)4 – 1 54 + 1 = (– 1)3 55 = – 3125

At n = 3, a5 = (– 1)5 – 1 55 + 1 = (– 1)4 56 = 15625

Hence, the required terms are 25, – 125, 625, – 3125, 15625.

$$\textbf{6.}\space \textbf{a}_\textbf{n}=\frac{\textbf{n}(\textbf{n}^\textbf{2}\textbf{+5})}{\textbf{4}}\textbf{.}\\\textbf{Sol.}\space\text{We have,}\space a_n=\frac{n(n^2+5)}{4}$$

Substituting n = 1, 2, 3, 4, 5

$$\text{At n = 1,}\space a_1=\frac{1(1^2+5)}{4}=\frac{6}{4}=\frac{3}{2}\\\text{At n = 2,}\space a_2=\frac{2(2^2+5)}{4}=\frac{2(4+5)}{4}=\frac{9}{2}\\\text{At n = 3,}\space a_3=\frac{3(3^2+5)}{4}=\frac{3(9+5)}{4}=\frac{21}{2}\\\text{At n = 4,}\space a_4=\frac{4(4^2+5)}{4}=\frac{4(16+5)}{4}=21\\\text{At n = 5,}\space a_5=\frac{5(5^2+5)}{4}=\frac{5(25+5)}{4}=\frac{75}{2}\\\text{Hence, the first five terms are}\\\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}.$$

Find the indicated term in each of the sequence in Exercises 7 to 10 whose nth terms are:

7. an = 4n – 3, a17, a24.

Sol. We have, an = 4n – 3

Substituting n = 17,

a17 = 4 × 17 – 3 = 68 – 3 = 65

Substituting n = 24,

a24 = 4 × 24 –3 = 96 – 3 = 93.

$$\textbf{8.}\space \textbf{a}_\textbf{n}=\frac{\textbf{n}^\textbf{2}}{\textbf{2}^\textbf{n}};\textbf{a}_\textbf{7}\textbf{.}\\\textbf{Sol.}\space\text{We have,}\space a_n=\frac{n^2}{2^n}$$

Substituting n = 7

$$a_7=\frac{7^2}{2^7}=\frac{49}{128}$$

9. an = (– 1)n – 1 n3, a9.

Sol. We have, an = (– 1)n– 1 n3

Substituting n = 9,

a9 = (– 1)9 – 1 93 = (– 1)8 × 9 × 9 × 9 = 729

$$\textbf{10.}\space \textbf{a}_\textbf{n}=\frac{\textbf{n(n-2)}}{\textbf{n+3}};\textbf{a}_{\textbf{20}}\textbf{.}\\\textbf{Sol.}\space\text{We have,}\space a_n=\frac{n(n-2)}{n+3}\\\text{Substituting n = 20,}\\a_{20}=\frac{20(20-2)}{20+3}\\a_{20}=\frac{20×18}{23}=\frac{360}{23}.$$

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series.

11. a1 = 3, an = 3an – 1 + 2 for all n > 1

Sol. We have, a1 = 3, an = 3an – 1 + 2, for all n > 1

Substituting n = 2, a2 = 3a2 – 1 + 2

⇒ a2 = 3a1 + 2 = 3 × (3) + 2 = 9 + 2 = 11

Substituting n = 3, a3 = 3a3 – 1 + 2 = 3a2 + 2 = 3(11) + 2 = 33 + 2 = 35

Substituting n = 4, a4 = 3a4 – 1 + 2 = 3a3 + 2 = 3 × 35 + 2 = 105 + 2 = 107

Substituting n = 5, a5 = 3 × a5 – 1 + 2 = 3a4 + 2 = 3 × 107 + 2 = 321 + 2 = 323

Thus, the required first five term are 3, 11, 35, 107

and 323.

∴ Series is 3 + 11 + 35 + 107 + 323 …

$$\textbf{12.}\space \textbf{a}_\textbf{1}\textbf{=-1},\textbf{a}_\textbf{n}=\frac{\textbf{a}_\textbf{n-1}}{\textbf{n}},\textbf{n}\geq\textbf{2.}\\\textbf{Sol.}\space\text{We have}\space a_1=-1,a_n=\frac{a_n-1}{n}\\\text{Substituting n = 2,}\\a_2=\frac{a_2-1}{2}=\frac{a_1}{2}=\frac{-1}{2}\\\text{Substituting n = 3,}\\a_3=\frac{a_3-1}{3}=\frac{a_2}{3}\\a_3=\bigg(\frac{-1}{2}\bigg).\frac{1}{3}=\frac{-1}{6}\\\text{ Substituting n = 4}\\a_4=\frac{a_4-1}{4}=\frac{a_3}{4}$$

$$a_4=\bigg(\frac{-1}{6}\bigg).\frac{1}{4}=\frac{-1}{24}\\\text{Substituting n = 5,}\\a_5=\frac{a_5-1}{5}=\frac{a_4}{5}\\a_5=\bigg(\frac{-1}{24}\bigg)\frac{1}{5}=\frac{-1}{120}\\\text{Thus first give terms of the sequence are}\\-1,\frac{-1}{2},\frac{-1}{6},\frac{-1}{24},\frac{-1}{120}\\\text{Hence corresponding series is}\\-1+\bigg(\frac{\normalsize-1}{2}\bigg)+\bigg(\frac{-1}{6}\bigg)+\bigg(\frac{-1}{24}\bigg)+\bigg(\frac{-1}{120}\bigg)+...$$

13. a1 = a2 = 2, an = an – 1 – 1, n > 2.

Sol. We have, a1 = a2 = 2

and an = an – 1 – 1, n > 2

Substituting n = 3,

a3 = a3 – 1 – 1 = a2 – 1 = 2 – 1 = 1

Substituting n = 4,

a4 = a4 – 1 – 1

= a3 – 1 = 1 – 1 = 0

Substituting n = 5,

a5 = a5 – 1 – 1 = a4 – 1 = 0 – 1 = – 1

Hence, the first five terms are 2, 2, 1, 0 and – 1

∴ Series is 2 + 2 + 1 + 0 + (– 1) + …

14. The Fibonacci sequence is defined by 1 = a1 = a2 and an = an – 1 + an – 2, n > 2.

$$\textbf{Find}\space\frac{\textbf{a}_{\textbf{n+1}}}{\textbf{a}_\textbf{n}},\textbf{for n= 1, 2, 3, 4, 5}\textbf{.}$$

Sol. Here, 1 = a1 = a2

an = an – 1 + an – 2, n > 2

Substituting n = 3, 4, 5, 6

At n = 3, a3 = a3 – 1 + a3 – 2

= a2 + a1

= 1 + 1 = 2

At n = 4, a4 = a4 – 1 + a4 – 2

= a3 + a2

= 2 + 1 = 3

At n = 5, a5 = a5 – 1 + a5 – 2

= a4 + a3 = 3 + 2 = 5

At n = 6, a6 = a6 – 1 + a6 – 2

= a5 + a4 = 5 + 3 = 8

$$\text{Now,}\frac{a_{n+1}}{a_n},\space\text{for n = 1, 2, 3, 4, 5}\\\text{At n = 1,}\space\frac{a_2}{a_1}=\frac{1}{1}=1\\\text{At n = 2,}\space\frac{a_3}{a_2}=\frac{2}{1}=2\\\text{At n = 3,}\space\frac{a_4}{a_3}=\frac{3}{2}=\frac{3}{2}\\\text{At n = 4,}\space\frac{a_5}{a_4}=\frac{5}{3}\\\text{At n = 5,}\space\frac{a_6}{a_5}=\frac{8}{5}\\\text{Hence, the first five terms are}\\ 1, 2,\frac{3}{2},\frac{5}{3}\text{and}\space\frac{8}{5}.$$

Exercise 9.2

1. Find the sum of odd integers forms 1 to 2001.

Sol. Series of odd integers from 1 to 2001 is 1 + 3 + 5 + 7 + … + 2001

Here, first term a = 1

Common, difference d = 2

We know that nth term of a series,

Tn = a + (n – 1)d

∴ 2001 = 1 + (n – 1) × 2

(∵ a = 1, Tn = 2001 given)

⇒ 2000 = (n – 1) × 2

$$\Rarr\space\text{(n-1)}=\frac{2000}{2}$$

⇒ n – 1 = 1000

⇒ n = 1001

Now, required sum

$$\text{S}_n=\frac{n}{2}[2a+(n-1)d]\\\Rarr\space\text{S}_{1001}=\frac{1001}{2}[2×1+(1001-1)×2]\\\Rarr\space\text{S}_{1001}=\frac{1001}{2}×2[1+1001-1]$$

⇒ S1001 = 1001 × 1001 = 1002001

Hence, the sum of odd integers is 1002001.

2. Find the sum of all natural numbers lying between 100 and 1000, which are multiple of 5.

Sol. The number are 105, 110, 115, …, 995.

Here, first term a = 105

and common difference d = 110 – 105 = 5

Now,nth term,

Tn = a + (n – 1)d

⇒ 995 = 105 + (n – 1)5

(∵ a = 105, Tn = 995)

⇒ 995 – 105 = (n – 1)5

⇒ 890 = (n – 1)5

$$\Rarr\space n-1=\frac{890}{5}$$

⇒ n – 1 = 178

⇒ n = 178 + 1 = 179

Now, required sum

$$\text{S}_n=\frac{n}{2}[2a+(n-1)d]\\\Rarr\space\text{S}_{179}=\frac{179}{2}[2×105+(179-1)5]\\\Rarr\space\text{S}_{179}=\frac{179}{2}[210+178×5]\\\Rarr\space\text{S}_{179}=\frac{179}{2}[210+890]=\frac{179}{2}×1100$$

= 179 × 550 = 98450

Hence, the sum of the numbers lying between 100 and 1000 is 98450.

3. In an A.P. the first term is 2 and the sum of the first five terms is one-fourth of the next five terms, show that 20th term is – 112.

Sol. Let the A.P. is a, a + d, a + 2d, a + 3d, …

Now, given a = 2

According to the given condition,

$$\text{Sum of first five terms}\\=\frac{1}{4}\text{(Sum of next five terms)}$$

a + (a + d) + (a + 2d) + (a + 3d) + a + 4d

$$=\frac{1}{4}\text{[a + 5d + a + 6d + a + 7d}+\\ \text{a + 8d + a + 9d]}\\\Rarr\space\text{5a+10d}=\frac{1}{4}[5a+35d]$$

⇒ 4[5a + 10d] = 5a + 35d

⇒ 20a + 40d = 5a + 35d

⇒ 20a – 5a = 35d – 40d

⇒ 15a = – 5d

⇒ 15 × 2 = – 5d

⇒ 30 = – 5d (∵ a = 2)

$$\Rarr\space d=\frac{-30}{5}=-6$$

Now, Tn = a + (n – 1)d

⇒ T20 = 2 + (20 – 1) (– 6) = 2 + 19(– 6)

= 2 – 19 × 6 = 2 – 114 = – 112

Hence, the 20th term of the A.P. is – 112.

$$\textbf{4. How many terms of the A.P.}\\\space\textbf{-6,}-\frac{\textbf{11}}{\textbf{2}},\textbf{-5,\space...}\\\textbf{are needed to give sum – 25?}$$

$$\textbf{Sol.}\space\text{Given that the sequence}\\-6,-\frac{11}{2},-5,...\text{is in A.P}$$

$$\text{Here, a = – 6,}\\\space d=-\frac{11}{2}-(-6)=-\frac{11}{2}+6\\=-\frac{11}{2}+\frac{6}{1}\\\text{d}=\frac{-11+12}{2}=\frac{1}{2}\\\text{Now,}\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\\\Rarr\space -25=\frac{n}{2}\bigg[2×(-6)+(n-1)\frac{1}{2}\bigg]\\\Rarr\space-25×2=n\bigg[\frac{-12}{1}+\frac{(n-1)}{2}\bigg]\\\Rarr\space -50=n\bigg[\frac{-24+n-1}{2}\bigg]$$

⇒ – 50 × 2 = n(n – 25)

⇒ – 100 = n2 – 25n

⇒ n2 – 25n + 100 = 0

Now, factorizing it by splitting the middle term,

n2 – (20 + 5)n + 100 = 0

⇒ n2 – 20n – 5n + 100 = 0

⇒ n(n – 20) – 5(n – 20) = 0

⇒ n = 5, 20.

5. In an A.P., if the pth term is

$$\frac{\textbf{1}}{\textbf{q}}\textbf{and q}^{\textbf{th}} \textbf{term is}\space\frac{\textbf{1}}{\textbf{p}}.\\\textbf{Prove that the sum of first pq terms is }\space\\\frac{\textbf{1}}{\textbf{2}}\textbf{(pq+1)},\textbf{where p} ≠ \textbf{q}.$$

Sol. ∵ Tn = a + (n – 1)d

$$\text{Therefore}\space\text{T}_p=a+(p-1)d=\frac{1}{q}\space\text{(given) …(i)}\\\text{and}\space\text{T}_q=a+(q-1)=\frac{1}{p}\space\text{(given)\space...(ii)}$$

Subtracting equation (i) and equation (ii),

$$d(p-1-q+1)=\frac{1}{q}-\frac{1}{p}\\\Rarr\space d(p-q)=\frac{\text{p-q}}{\text{pq}}\\\Rarr\space d=\frac{1}{\text{pq}}$$

Substituting the value of d in equation (i), we get

$$a+\frac{\text{(p-1)}}{\text{pq}}=\frac{1}{q}\\\Rarr\space a=\frac{1}{q}-\frac{\text{p-1}}{\text{pq}}\\\Rarr\space a=\frac{\text{p-p+1}}{\text{pq}}=\frac{1}{\text{pq}}\\\text{Now,}\space\text{S}_{\text{pq}}=\frac{\text{pq}}{2}[2a+(pq-1)d]\\\bigg(\because\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\bigg)\\=\frac{\text{pq}}{2}\bigg[2×\frac{1}{pq}+(pq-1)\frac{1}{pq}\bigg]\\=\frac{\text{pq}}{2}×\frac{\text{1}}{pq}(2+pq-1)$$

$$=\frac{1}{2}(pq+1).\\\textbf{Hence Proved.}$$

6. If the sum of a certain a number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

Sol. Given A.P. is 25, 22, 19, …

Here, a = 25, d = – 3 and Sn = 116

$$\because\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\\\therefore\space 116=\frac{n}{2}[2×25+(n-1)(-3)]$$

⇒ 116 × 2 = n[50 – 3n + 3]

⇒ 232 = n[53 – 3n]

⇒ 232 = 53n – 3n2

⇒ 3n2 – 53n + 232 = 0

Now, factorizing it by splitting the middle term,

⇒ 3n2 – (24 + 29)n + 232 = 0

⇒ 3n2 – 24n – 29n + 232 = 0

⇒ 3n(n –8) – 29(n – 8) = 0

⇒ (3n – 29) (n – 8) = 0

$$\Rarr\space n=\frac{29}{3},\text{n=8}\\n=\frac{29}{3}\text{is useless as n cannot be fraction},\\\text{only n = 8 is valid.}$$

Now    Tn = a + (n – 1)d

T8 = 25 + (8 – 1) (–3)

= 25 + 7 × (– 3) = 25 – 21 = 4

Hence, the last term of A.P. is 4.

7. Find the sum of n terms of the A.P. whose Kth term is 5K + 1.

Sol. Given, Kth term

TK = 5K + 1

Substituting K = 1, 2, 3, 4, …

T1 = 5 × 1 + 1 = 6

T2 = 5 × 2 + 1 = 11

T3 = 5 × 3 + 1 = 16 and so on.

⇒ a = 6, d = 11 – 6 = 5

$$\text{Now}\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\\=\frac{n}{2}[2×6+(n-1)5]\\=\frac{n}{2}[12+5n-5]\\=\frac{n}{2}[5n+7].$$

8. If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.

Sol. Given, Sn = pn + qn2

Now,

Tn = Sn – Sn – 1

⇒ Tn = (pn + qn2) – [p(n – 1) + q(n – 1)2]

= pn + qn2 – [pn – p + q(n2 + 1 – 2n)]

= pn + qn2 – [pn – p + qn2 + q – 2qn]

= pn + qn2 – pn + p – qn2 – q + 2qn

= p – q + 2qn

Now, substituting n = 1, 2, 3, …

⇒ T1 = p – q + 2q × 1 = p – q + 2q = p + q

⇒ T2 = p – q + 2q × 2 = p – q + 4q = p + 3q

⇒ T3 = p – q + 2q × 3 = p – q + 6q = p + 5q
....................
....................
....................
∴ series is p + q, p + 3q, p + 5q ...........

Hence, the common difference of the A.P.

= (p + 3q) – (p + q) = 2q.

9. The sum of n terms of two A.P. are in the ratio of 5n + 4; 9n + 6. Find the ratio of their 18th terms.

Sol. Let the a1, a2 be the first terms of two AP’s and d1, d2 be the common difference of two AP’s respectively. Now, according to the question

$$\frac{\text{S}_{n1}}{\text{S}_{n2}}=\frac{5n+4}{9n+6}\\\Rarr\space\frac{\frac{n}{2}[2a+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}=\frac{5n+4}{9n+4}\\\bigg(\because\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\bigg)\\\Rarr\space\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}$$

(We have to find the ratio of 18th terms, so convert the equation in the form of nth term by taking the 2 as common).

$$\Rarr\space\frac{2[a_1+\frac{n-1}{2}d_1]}{2[a_2+\frac{n-1}{2}d_2]}=\frac{5n+4}{9n+6}\\\Rarr\space\frac{a_1+\bigg(\frac{n-1}{2}\bigg)d_1}{a_2+\bigg(\frac{n-1}{2}\bigg)d_2}=\frac{5n+4}{9n+6}\\\text{...(i)}\\\text{We have to find}\space\frac{a_1+17d_1}{a_2+17d_2}\\\Rarr\space\frac{n-1}{2}=17$$

⇒ n – 1 = 2 × 17

⇒ n = 34 + 1

⇒ n = 35

Hence, from equation (i),

$$\frac{a_1+17d_1}{a_2+17d_2}=\frac{5×35+4}{9×35+6}\\=\frac{175+4}{315+6}=\frac{179}{321}$$

10. If the sum of first p terms of an A.P. is equal to the sum of first q terms, then find the sum of first (p + q) terms.

Sol. Let a and d be the first term and common difference of an A.P.

∵ Sum of the p terms

= Sum of first q terms (given)

∴ Sp = Sq

$$\Rarr\space\frac{p}{2}[2a+(p-1)d]\\=\frac{q}{2}[2a+(q-1)d]\\\bigg(\because\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\bigg)$$

⇒ p[2a + (p – 1)d] = q[2a + (q – 1)d]

⇒ 2ap + p(p – 1)d = 2aq + q(q – 1)d

⇒ 2ap – 2aq = q(q – 1)d – p(p – 1)d

⇒ 2a(p – q) = d(q2 – q – p2 + p)

⇒ 2a(p – q) = d(p – q) + (q2 – p2)]

⇒ 2a(p – q) = d[(p – q) + (q – p) (q + p)]

⇒ 2a(p – q) = d(p – q) [1 – (p + q)]

⇒ 2a = d[1 – (p + q)]

⇒ 2a = – d[p + q) – 1]

⇒ 2a + d(p + q – 1) = 0   …(i)

$$\text{Now,}\space\text{S}_{p+q}=\frac{p+q}{2}[2a+(p+q-1)d]\\=\frac{p+q}{2}×0\space[\text{from equation (i)}]$$

= 0

Hence, the sum of the first (p + q) term of the A.P. is 0.

11. Sum of the first p, q and r terms of an A.P. are a, b and c respectively. Prove that:

$$\frac{\textbf{a}}{\textbf{p}}\textbf{(q-r)}\textbf{+}\frac{\textbf{b}}{\textbf{q}}\textbf{(r-p)}\textbf{+}\frac{\textbf{c}}{\textbf{r}}\textbf{(p-q)}\textbf{=0.}$$

Sol. Let the first term is A and common difference is d.

Given,  Sp = a

$$\Rarr\space\frac{p}{2}[2A+(p-1)d]=a\space\space\text{...(i)}\\\bigg(\because\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\bigg)$$

Sq = b

$$\Rarr\space\frac{q}{2}[2A+(q-1)d]=b\space\text{...(ii)}\\\text{and S}_r = c\\\Rarr\space\frac{r}{2}[2A+(r-1)d]=c\space\text{...(iii)}$$

Now, we are to prove

$$\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0\\\text{L.H.S}=\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)$$

Substituting the values of a, b and c from equation (i), (ii) and (iii) respectively, we get

$$\therefore\space\text{L.H.S}=\frac{1}{p}×\frac{p}{2}[2A+(p-1)d](q-r)\\+\frac{1}{q}×\frac{q}{2}[2A+(q-1)d](r-p)+\\\frac{1}{r}×\frac{r}{2}[2A+(r-1)d](p-q)$$

$$=\frac{1}{2}[\lbrace2A+(p-1)d\rbrace (q-r) + \\\lbrace2A+(q-1)d\rbrace(r-p)+\\\lbrace2A+(r-1)d\rbrace(p-q)]\\=\frac{1}{2}[2A(q-r)+(p-1)d(q-r)+\\2A(r-p)+(q-1)d(r-p)+\\2A(p-q)+(r-1)d(p-q)]$$

$$=\frac{1}{2}2A(q-r+r-p+p-q)+\\d[(p-1)(q-r)+(q-1)(r-p)+\\(r-1)(p-q)]\\=\frac{1}{2}[2A×(0)+d(pq-pr-q+r+qr-\\pq-r+p+rp-rq-p+q)]\\=\frac{1}{2}(0+d×0)=0$$

∴ L.H.S. = R.H.S. Hence Proved.

12. The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1).

Sol. Let the first term is a and common difference is d.

$$\text{Given,\space}\frac{\text{S}_m}{\text{S}_n}=\frac{m^2}{n^2}\\\Rarr\space\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2}\\\bigg(\because\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\bigg)\\\Rarr\space\frac{m[2a+(m-1)d]}{n[2a+(n-1)d]}=\frac{m^2}{n^2}\\\Rarr\space\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$$

⇒ [2a + (m – 1)d]n = [2a + (n – 1)d]m

⇒ 2an + (m – 1)dn = 2am + (n – 1)dm

⇒ 2an – 2am = (n – 1)dm – (m – 1)dn

⇒ 2a(n – m) = d[(n –1)m – (m – 1)n]

⇒ 2a(n – m) = d[mn – m – mn + n]

⇒ 2a(n – m) = d(n – m)

⇒ 2a = d

$$\text{Now,}\space\frac{\text{T}_m}{\text{T}_n}=\frac{a+(m-1)d}{a+(n-1)d}\\=\frac{a+(m-1)2a}{a+(n-1)2a}$$

[∵ Tn = a + (n – 1)d]

Taking a as common in numerator and denominator,

$$\Rarr\space\frac{\text{T}_m}{\text{T}_n}=\frac{1+2(m-1)}{1+2(m-1)}\\=\frac{1+2m-2}{1+2n-2}=\frac{2n-1}{2n-1}$$

13. If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.
Sol. Given, Sn = 3n2 + 5n
∴ Sm = 3m2 + 5m
and Sm – 1 = 3(m – 1)2 + 5(m – 1)

Now, using the formula

Tm = Sm – Sm – 1

⇒ Tm = (3m2 + 5m) – [3(m – 1)2 + 5(m – 1)]

= (3m2 + 5m) – [3(m2 + 1 – 2m) + 5m – 5]

= (3m2 + 5m) – [3m2 + 3 – 6m + 5m – 5]

= 3m2 + 5m – 3m2 – 3 + 6m – 5m + 5

= 6m + 2

But given, Tm = 164

∵ 6m + 2 = 164

⇒ 6m = 164 – 2

⇒ 6m = 162

$$\Rarr\space\text{m}=\frac{162}{6}=27.\\\text{Hence, the value of m is 27.}$$

14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Sol. Let the numbers are A1, A2, A3, A4, A5, then 8, A1, A2, A3, A4, A5, 26 are in A.P.

∵ Tn = a + (n – 1)d

∴ 26 = 8 + (7 –1)d

[∵ n = 7 since, 2 terms and 5 inserted numbers]

⇒ 26 – 8 = 6d

⇒ 18 = 6d

$$\Rarr\space d=\frac{18}{6}$$

⇒ d = 3

Now,  A1 = a + d = 8 + 3 = 11

⇒ A2 = a + 2d = 8 + 2 × 3 = 14

A3 = a + 3d = 8 + 3 × 3 = 17

A4 = a + 4d = 8 + 4 × 3 = 20

and A5 = a + 5d = 8 + 5 × 3 = 23

Hence, the required five numbers between 8 and 26 are 11, 14, 17, 20 and 23.

$$\textbf{15. \space If}\space\frac{\textbf{a}^\textbf{n}\textbf{+}\textbf{a}^\textbf{n}}{\textbf{a}^{\textbf{n-1}}\textbf{+}\textbf{b}^{\textbf{n-1}}}\textbf{is the arithmetic mean}\\\textbf{ between a and b, find the value of n.}$$

Sol. Given, arithmetic mean between a and b is

$$\frac{a^n+b^n}{a^{n-1}+b^{n-1}}\space\text{but we know that the}\\\text{arithmetic mean between a and b is}\\\frac{a+b}{2}.$$

$$\text{Therefore,}\space\frac{a^n+b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}$$

⇒ 2(an + bn) = (an – 1 + bn – 1) (a + b)

⇒ 2an + 2bn = an – 1 a1 + an – 1 b + bn – 1 a1
+ bn – 1 b1

⇒ 2an + 2bn = an + an – 1 b + bn – 1 a + bn

⇒ 2an + 2bn – an – an – 1 b – bn – 1 a – bn = 0

⇒ an + bn – an – 1 b – bn – 1 a = 0

⇒ (an – an – 1 b) + (bn – bn – 1 a) = 0

⇒ an –1 (a – b) – bn – 1 (a – b) = 0

⇒ (a – b) (an – 1 – bn – 1) = 0

⇒ an –1 – bn – 1 = 0 (∵ a – b ≠ 0)

⇒ an – 1 = bn –1

$$\Rarr\space\bigg(\frac{a}{b}\bigg)^{n-1}=1\\\text{On comparing the power of}\space\bigg(\frac{a}{b}\bigg)\\\text{both sides,}\\\Rarr\space\bigg(\frac{a}{b}\bigg)^{n-1}=\bigg(\frac{a}{b}\bigg)^0$$

⇒ n – 1 = 0

⇒ n = 1

Hence, the values of n is 1.

16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m –1)th numbers is 5 : 9. Find the value of m.

Sol. Let A1, A2, A3, …, Am are m numbers inserted between 1 and 31.

i.e., 1, A1, A2, A3, …, Am, 31

Now, Tn = a + (n – 1)d

31 = 1 + (m + 2 – 1)d

[∵ n = m + 2 because 2 terms (1 and 31)
and m numbers]

⇒ 31 – 1 = (m + 1)d

⇒ 30 = (m + 1)d

$$\Rarr\space d=\frac{30}{m+1}\space\text{...(i)}\\\text{Given,\space}\frac{\text{T}_7}{\text{T}_{m-1}}=\frac{5}{9}\\\Rarr\space\frac{a+7d}{a+(m-1)d}=\frac{5}{9}\\\Rarr\space\frac{1+7×\frac{30}{m+1}}{1+(m-1)×\frac{30}{m+1}}=\frac{5}{9}\\\space\text{[from equation (i)]}\\\Rarr\space\frac{\frac{m+1+210}{m+1}}{\frac{(m+1)+30(m-1)}{m+1}}=\frac{5}{9}\\\Rarr\space\frac{m+211}{m+1+30m-30}=\frac{5}{9}$$

⇒ 9(m + 211) = 5(31m – 29)

⇒ 9m + 1899 = 155m – 145

⇒ 1899 + 145 = 155m – 9m

⇒ 146m = 2044

$$\Rarr\space m=\frac{2044}{146}=14$$

Hence, the value of m is 14.

17. A man starts repaying a loan as first instalment of ₹100. If he increases the instalment by ₹5 every month, what amount he will pay in the 30th instalment?

Sol. Given, a = 100, d = 5

∵ Tn = a + (n – 1)d

∴ T30 = 100 + (30 – 1)5 = 100 + 29 × 5

= 100 + 145 = 245

Hence, the amount to be paid in the 30th instalment is ₹245.

18. The difference between any two consecutive interior angles of polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

$$\textbf{Sol.}\space\text{We know that}\space\\\text{S}_n=\frac{n}{2}[2a+(n-1)d]\\\text{But for a polygon}\\\text{S}_n=(n-2)\space180\degree\\\text{Therefore,}\space (n-2)180\degree\\=\frac{n}{2}[2×120\degree+(n-1)(5)]$$

(∵ a = 120°, d = 5)

⇒ (n – 2)180 × 2 = n[240 + 5n – 5]

⇒ (n – 2)360 = n[5n + 235]

⇒ (n – 2)72 = n[n + 47]

⇒ 72n – 144 = n2 + 47n

⇒ n2 + 47n – 72n + 144 = 0

⇒ n2 – 25n + 144 = 0

Now, factorizing it by splitting the middle term,

⇒ n2 – (16 + 9)n + 144 = 0

⇒ n2 – 16n – 9n + 144 = 0

⇒ n(n – 16) – 9(n – 16) = 0

⇒ (n – 16) (n –9) = 0

⇒ n = 9, 16

Hence, the number of sides of the polygon is 9 or 16.

Exercise 9.3

1. Find the 20th and nth terms of the G.P.;

$$\frac{5}{2},\frac{5}{4},\frac{5}{8},...\\\textbf{Sol.}\space\text{The given series is}\space\frac{5}{2},\frac{5}{4},\frac{5}{8},...\\\text{Here, a = first term=}\frac{5}{2}\\\text{and common ratio (r) =}\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{2}{4}=\frac{1}{2}$$

Now,  Tn = arn – 1

$$\Rarr\space\text{T}_{20}=ar^{20-1}=\bigg(\frac{5}{2}\bigg)×\bigg(\frac{1}{2}\bigg)^{20-1}\\=\frac{5}{2}×\bigg(\frac{1}{2}\bigg)^{19}\\=\frac{5}{2}×\frac{1}{2^{19}}=\frac{5}{2^{20}}\\\text{Again,}\text{T}_n = ar^{n – 1} =\bigg(\frac{5}{2}\bigg)×\bigg(\frac{1}{2}\bigg)^{n-1}\\=\frac{5}{2}×\frac{1}{2^{n-1}}\\=\frac{5}{2^{1+n-1}}=\frac{5}{2^n}\\\text{Hence, the 20th and nth term are}\\\frac{5}{2^{20}}\text{and}\frac{5}{2^n}.$$

2. Find the 12th term of a GP whose 8th term is 192 and the common ratio is 2.

Sol. Let a be the first term of G.P.

Given, 8th term

T8 = 192

and common ratio

r = 2

⇒ ar8 – 1 = 192 (∵ Tn = arn – 1)

⇒ a × (2)7 = 192

⇒ a × 128 = 192

$$\Rarr\space a=\frac{192}{128}\\\Rarr\space a=\frac{48}{32}=\frac{3}{2}\\\text{Now,}\space\text{T}_{12}=\text{ar}^{12-1}=\frac{3}{2}×(2)^{11}\\=\frac{3}{2}×2^{11}=3×2^{10}$$

= 3 × 1024 = 3072.

3. The 5th, 8th and 11th terms of a G.P. are p, q and s respectively. Show that q2 = ps.

Sol. Let a and r be the first term and common ratio of the G.P.

Given, T5 = p

⇒ ar4 = p …(i)

T8 = q

⇒ ar7 = q …(ii)

and T11 = s

⇒ ar10 = s …(iii)

Now, multiplying equation (i) and (iii),

ar4 × ar10 = ps

a2r14 = ps

⇒ [(ar)7]2 = ps [from equation (ii)]

⇒ q2 = ps.    Hence Proved.

4. The 4th term of a G.P. is square of its second term and the first term is – 3. Determine its 7th term.

Sol. Let a and r be the first term and common ratio of the G.P.

Given, T4 = (T2)2 (∵ Tn = arn – 1)

⇒ ar3 = (ar)2

⇒ ar3 = a2r2

⇒ r = a

But given, a = – 3

⇒ r = – 3

Therefore T7 = ar6 = (– 3) (– 3)6

= (– 3)7 = – 2187.

5. Which term of the following sequences:

$$\textbf{(a)\space 2, 2}\sqrt{\textbf{2}},\textbf{4}......\textbf{is 128 ?}\\\textbf{(b)\space}\sqrt{\textbf{3}},\textbf{3},\textbf{3}\sqrt{\textbf{3}}\space\textbf{...... is 729 ?}\\\textbf{(c)}\space\frac{\textbf{1}}{\textbf{3}},\frac{\textbf{1}}{\textbf{9}},\frac{\textbf{1}}{\textbf{27}}\space\textbf{...... is}\space\frac{\textbf{1}}{\textbf{19683}}?$$

$$\textbf{Sol.}\space\text{(a) The series is 2,}2\sqrt{2},4,...128\\\text{Here}\space\text{a=2},r=\frac{2\sqrt{2}}{2}=\sqrt{2}$$

and Tn = 128
Now, Tn = arn – 1

$$\Rarr\space 128=2(\sqrt{2})^{\textbf{n-1}}\\\Rarr\space 2^{\frac{n-1}{2}}=\frac{128}{2}\\\Rarr\space 2^{\frac{n-1}{2}}=64\\\Rarr\space 2^{\frac{n-1}{2}}=2^6\\\text{On comparing the power of 2 both sides,}\\\Rarr\space\frac{n-1}{2}=6$$

⇒ n = 12 + 1

n = 13.

$$\text{(b)}\space\sqrt{3},3,3\sqrt{3},...,729\\\text{Here\space}a=\sqrt{3}, r=\frac{3}{\sqrt{3}}=\sqrt{3},$$

Tn = 729

Now, Tn = arn – 1

$$\Rarr\space 729=\sqrt{3}(\sqrt{3})^{n-1}\\\Rarr\space 729=(\sqrt{3})^n\\\Rarr\space 3^{\frac{n}{2}}=3^6$$

On comparing the power of 3 both sides,

$$\Rarr\space\frac{n}{2}=6$$

⇒ n = 12.

$$\text{(c)}\space\frac{1}{3},\frac{1}{9},\frac{1}{27},...,\frac{1}{19683}\\\text{Here,}\space a=\frac{1}{3}, r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{3}{9}=\frac{1}{3}\\\text{T}_n=\frac{1}{19683}$$

Now,  Tn = arn – 1

$$\frac{1}{19683}=\frac{1}{3}×\bigg(\frac{1}{3}\bigg)^{n-1}\\\Rarr\space\frac{1}{6561}=\bigg(\frac{1}{3}\bigg)^{n-1}\\\Rarr\space\bigg(\frac{1}{3}\bigg)^{n-1}=\bigg(\frac{1}{3}\bigg)^8$$

On comparing the power of (1/3) both sides.

⇒ n – 1 = 8

⇒ n = 8 + 1

⇒ n = 9.

6. For what values of x, the numbers

$$-\frac{\textbf{2}}{\textbf{7}},\textbf{x},-\frac{\textbf{7}}{\textbf{2}}\\\textbf{are in GP ?}$$

$$\textbf{Sol.}\space\text{Since}\space-\frac{2}{7},x,-\frac{7}{2}\space\text{are in GP, then}\\\Rarr\space x^2=\bigg(-\frac{2}{7}\bigg)×\bigg(-\frac{7}{2}\bigg)\\\Rarr\space x^2=\frac{2}{7}×\frac{7}{2}$$

⇒ x2 = 1

⇒ x = ± 1

Hence, the numbers x = ± 1 will be in G.P.

Find the sum to indicated number of terms in each of the geometric progression is Exercise 7 to 10:

7. 0.15, 0.015, 0.0015 .......... 20 terms.

Sol. Here, a = 0.15,

$$\text{and}\space r=\frac{0.015}{0.15}\\=\frac{15}{1000}×\frac{1000}{15}\\r=\frac{1}{10}\lt1\space\text{and n=20}\\\text{Now,}\space\text{S}_n=\frac{a(1-r^n)}{1-r}\\\Rarr\space S_{20}=\frac{0.15\bigg[1-\bigg(\frac{1}{10}\bigg)^{20}\bigg]}{1-\frac{1}{10}}\\=\frac{0.15\bigg[1-\frac{1}{10^{20}}\bigg]}{\frac{10-1}{10}}$$

$$=\frac{0.15×10}{9}\bigg[1-\frac{1}{10^{20}}\bigg]\\=\frac{15×10}{900}\bigg[1-\bigg(\frac{1}{10}\bigg)^{20}\bigg]\\=\frac{1}{6}[1-(0.1)^{20}]$$

$$\textbf{8.}\space\sqrt{\textbf{7}},\sqrt{\textbf{21}},\textbf{3}\sqrt{\textbf{7}}\space\textbf{........}\textbf{n terms.}$$

$$\textbf{Sol.}\space\text{Here}\space a=\sqrt{7},\text{and}\\r=\frac{\sqrt{21}}{\sqrt{7}}=\frac{\sqrt{7×3}}{\sqrt{7}}\\=\sqrt{3}\gt1\\\because\space\text{S}_a=\frac{a(r^n-1)}{r-1}\\\Rarr\space\text{S}_n=\frac{\sqrt{7}[(\sqrt{3})^n-1]}{\sqrt{3}-1}\\=\frac{\sqrt{7}[(3^{1/2})^n-1]}{\sqrt{3}-1}×\frac{\sqrt{3}+1}{\sqrt{3}+1}\\=\frac{\sqrt{7}(\sqrt{3}+1)(3^{n/2}-1)}{3-1}$$

[∵ a2 – b2 = (a + b) (a – b)]

$$=\frac{\sqrt{7}}{2}(\sqrt{3}+1)(3^{n/2}-1)$$

9. 1, – a, a2, – a3, …, n terms (if a ≠ – 1).

$$\textbf{Sol.}\space\text{Here, a = 1 and r =}-\frac{a}{1}=-a\lt1\\\therefore\space\text{S}_n=\frac{a(1-r^n)}{1-r}\space\space (\because r\lt 1)\\\text{S}_n=\frac{1[1-(-a)^n]}{1-(-a)}\\=\frac{1-(-a)^n}{1+a}$$

10. x3, x5, x7, …, n terms (if x ≠ ± 1).

$$\textbf{Sol.}\space\text{Here,}\space a=x^2,r=\frac{x^5}{x^3}=x^2\\\because\space\text{S}_n=\frac{a(1-r^n)}{1-r}\space(\because r\lt1)\\\therefore\space\text{S}_n=\frac{x^3[1-(x^2)^n]}{1-x^2}\\=\frac{x^3[1-x^{2n}]}{1-x^2}$$

11. Evaluate:

$$\displaystyle\sum^{\textbf{11}}_{\textbf{k=1}}\textbf{(2+3}^\textbf{k}\textbf{).}$$

$$\textbf{Sol.}\space\displaystyle\sum^{11}_{k=1}(2+3^k)=\\(2+3^1)+(2+3^2)+(2+3^3)+...+(2+3^11)\\\text{2×11+}(3^1+3^2+3^3+...+3^{11})\\\bigg[\because\space\text{S}_n=\frac{a(r^n-1)}{r-1},r\gt1\bigg]\\=22+\frac{3(3^{11}-1)}{3-1}\\=22+\frac{3(3^{11}-1)}{2}$$

12. The sum of first three terms of a GP is

$$\frac{\textbf{39}}{\textbf{10}}\space\textbf{and their product is 1.}\\\textbf{Find the common ratio and the terms.}$$

Sol. Let the first three terms of GP be

$$\frac{a}{r},\text{a and ar.}$$

∴ According to the question, the sum of first three terms are

$$\frac{a}{r}+a+ar=\frac{39}{10}\space\text{...(i)}\\\text{and}\space\bigg(\frac{a}{r}\bigg)×(a)×(ar)=1$$

⇒ a3 = 1

⇒ a = 1

Putting the value of a = 1 in equation (i), we get

$$\frac{1}{r}+1+r=\frac{39}{10}\\\Rarr\space \frac{1}{r}+\frac{1}{1}+\frac{r}{1}=\frac{39}{10}\\\Rarr\space \frac{1+r+r^{2}}{r}=\frac{39}{10}$$

⇒ 10 + 10r + 10r2 = 39r

⇒ 10r2 + 10r – 39r + 10 = 0

⇒ 10r2 – 29r + 10 = 0

Now, factorizing it by splitting the middle term,

⇒ 10r2 – 25r – 4r + 10 = 0

⇒ 5r(2r – 5) – 2(2r – 5) = 0

⇒ (5r – 2) (2r – 5) = 0

⇒ 5r – 2 = 0 and 2r – 5 = 0

$$\Rarr\space r=\frac{2}{5}\space\text{and} r=\frac{5}{2}\\\text{When}\space a=1\space\text{and}\space r=\frac{2}{5},\text{then number are}\\\frac{a}{r}=\frac{1}{\frac{2}{5}}=\frac{5}{2}; a=1\\\text{and}\space ar=1×\frac{2}{5}=\frac{2}{5}\\\frac{5}{2},1,\frac{2}{5}\\\text{When a = 1 and r}=\frac{5}{2},\text{then numbers are}\\\frac{a}{r}=\frac{1}{\frac{5}{2}}=\frac{2}{5}; a=1$$

$$\text{ar = 1}×\frac{5}{2}=\frac{5}{2}\\\therefore\space\frac{2}{5},1,\frac{5}{2}$$

13. How many terms of a GP 3, 32, 33, … are needed to give the sum 120?

Sol. Here, a = 3, r = 3, Sn = 120. The given G.P. is 3, 32, 33, …

$$\text{Now,}\space\text{S}_n=\frac{a(a^r-1)}{r-1},r\gt1\\\Rarr\space 120=\frac{3(3^n-1)}{3-1}\\\Rarr 120=\frac{3(3^n-1)}{2}\\\Rarr\space 120×2=3(3^n-1)\\\Rarr\space\frac{240}{3}=3^n-1$$

⇒ 3n – 1 = 80

⇒ 3n = 80 + 1

⇒ 3n = 81

⇒ 3n = 34

On comparing the power of 3 both sides,

⇒ n = 4

Hence, the number of terms is 4.

14. The sum of the first three terms of a GP is 16 and the sum of next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the GP.

Sol. Let the GP is a, ar, ar2, ar3, …

Given,

a + ar + ar2 = 16 …(i)

ar3 + ar4 + ar5 = 128 …(ii)

Dividing equation (i) by equation (ii),

$$\frac{a+ar+ar^2}{ar^3+ar^4+ar^5}=\frac{16}{128}\\\Rarr\space\frac{a(1+r+r^2)}{ar^3(1+r+r^2)}=\frac{1}{8}\\\Rarr\space\frac{1}{r^3}=\bigg(\frac{1}{2}\bigg)^3$$

On comparing the base of the power 3 on both sides,

⇒ r = 2

Putting r = 2 in equation (i),

⇒ a + 2a + 4a = 16

⇒ 7a = 16

$$\Rarr\space a=\frac{16}{7}\\\text{Now,}\space\text{S}_n=\frac{a(a^r-1)}{r-1}\space\\(\because r=2\gt1)\\=\frac{\frac{16}{7}(2^n-1)}{2-1}\\=\frac{16}{7}(2^n-1)\\(\because r\gt1)\\\text{Hence,}\space a=\frac{16}{7},r=2\\\text{and}\space \text{S}_n=\frac{16}{7}(2^n-1).$$

15. Given a GP with a = 729 and 7th term 64, determine S7.

Sol. Given, a = 729, T7 = 64

⇒ T7 = ar7 – 1 = 64

⇒ 729r6 = 64

$$\Rarr\space r^6=\frac{64}{729}=\bigg(\frac{2}{3}\bigg)^6$$

On comparing base of the power 6 on the both sides,

$$\Rarr\space r=\frac{2}{3}\lt1\\\text{Now,}\space\text{S}_n=\frac{a(1-r^n)}{1-r}\\\therefore\space \text{S}_7=\frac{729\bigg[1-\bigg(\frac{2}{3}\bigg)^7\bigg]}{1-\frac{2}{3}}\\=\frac{729\bigg[1-\frac{2^7}{3^7}\bigg]}{\frac{1}{1}-\frac{2}{3}}\\=\frac{729\bigg[\frac{1}{1}-\frac{128}{2187}\bigg]}{\frac{3-2}{3}}$$

$$=\frac{729×3}{1}×\frac{2187-128}{2187}\\\frac{1}{1}×2059=2059.$$

16. Find a GP for which sum of the first two terms is – 4 and fifth term is 4 times the third term.

Sol. Let the GP is a, ar, ar2, ar3, …

Given, a + ar = – 4 …(i)

and T5 = 4T3

⇒ ar5 – 1 = 4ar3 – 1

⇒ r4 = 4r2

⇒ r2 = 4

⇒ r = ± 2

If r = 2, then from equation (i),

a + a(2) = – 4

3a = – 4

$$\Rarr \space a=-\frac{4}{3}$$

Hence, GP is

$$-\frac{4}{3},\bigg(-\frac{4}{3}\bigg)(2),\bigg(-\frac{4}{3}\bigg)(2)^2\space...\\\text{or}\space-\frac{4}{3},-\frac{8}{3},-\frac{16}{3},...$$

If r = – 2, then from equation (i),

a + a(– 2) = – 4

⇒ – a = – 4

⇒ a = 4

Hence, GP is 4, 4(– 2), 4(– 2)2 or 4, – 8, 16, …

17. If the 4th, 10th and 16th terms of a GP are x, y and x, respectively, prove that x, y, z are in GP.

Sol. According to given question,

T4 = x ⇒ ar4 – 1 = x ⇒ ar3 = x …(i)

T10 = y ⇒ ar10 – 1 = y ⇒ ar9 = y …(ii)

T16 = z ⇒ ar16 – 1 = z ⇒ ar15 = z …(iii)

Now, multiplying equation (i) by equation (iii),

⇒ ar3 × ar15 = x × z

⇒ a2r3 + 15 = xz

⇒ a2r18 = xz

⇒ (ar9)2 = xz

⇒ y2 = xz [from equation (ii)]

Therefore, x, y, z are in G.P.

18. Find the sum of n terms of the sequence 8, 88, 888, 8888, …

Sol. Let S = 8 + 88 + 888 + 8888 + … + n terms

⇒ S = 8[1 + 11 + 111 + 1111 + … + n terms)

$$=\frac{8}{9}[9+99+999+9999+...+\text{n terms}]\\\frac{8}{9}[10+100+1000+...+ \text{n terms}]-\\(1+1+1+...+\text{n terms})]\\=\frac{8}{9}\bigg[10\frac{(10^n-1)}{10-1}-n\bigg]\\\bigg[\because\text{S}_n=\frac{a(r^n-1)}{r-1}\text{as}\space r\gt1\bigg]\\=\frac{8}{9}\bigg[\frac{10(10^n-1)}{9}-n\bigg]\\=\frac{8}{9}×\frac{10}{9}×(10^n-1)-\frac{8}{9}×n\\=\frac{80}{81}(10^n-1)-\frac{8n}{9}$$

19. Find the sum of the products of the corresponding terms of the sequence 2, 4, 8, 16, 32 and 128, 32, 8, 2, $$\frac{\textbf{1}}{\textbf{2}}\textbf{.}$$

Sol. Given, sequence are 2, 4, 8, 16, 32.    …(i)

and 128, 32, 8, 2,$$\frac{\text{1}}{\text{2}}.$$ ...(ii)

Multiplying the corresponding terms of equations (i) and (ii) to obtain a new sequence,

256, 128, 64, 32, 16

Let S = 256 + 128 + 64 + 32 + 16

$$\text{Here,}\space a=256, r=\frac{1}{2}\\\therefore\space\text{Required sum S}_5\\=\frac{256\bigg[1-\bigg(\frac{1}{2}\bigg)^5\bigg]}{1-\frac{1}{2}}\\=256×2\bigg(1-\frac{1}{2^5}\bigg)\\\bigg[\because\space \text{S}_n=\frac{a(1-r^n)}{1-r}, r\lt 1\bigg]\\=512×\bigg(1-\frac{1}{32}\bigg)$$

$$=512\bigg(\frac{32-1}{32}\bigg)$$

= 16 × 31 = 496.

20. Show that the products of the corresponding terms of the sequences a, ar, ar2, … arn – 1 and A, AR, AR2, …, ARn – 1 form a GP and find the common ratio.

Sol. Given, sequence are a, ar, ar2, …, arn – 1 …(i)

and A, AR, AR2, …, ARn – 1 …(ii)

Now, multiplying the corresponding terms of equations (i) and (ii),

aA + arAR + ar2 AR2 + … + arn – 1 ARn – 1

$$\therefore\space\text{Common ratio =}\frac{2^{\text{nd}}\space\text{term}}{1^{\text{st}\space}\text{term}}\\=\frac{\text{arAR}}{\text{aA}}=\text{aR}.$$

21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9 and the second term is greater than 4th by 18.

Sol. Let the first term is a and common ratio is r.

and the GP is a, ar, ar2, ar3

Given, third term = first term + 9

T3 = a + 9

⇒ ar2 = a + 9

ar2 – a = 9 …(i)

Again, second term = fourth term + 18

T2 = T4 + 18

⇒ ar = ar3 + 18

⇒ ar – ar3 = 18 …(ii)

Dividing equation (i) by equation (ii), we get

$$\frac{ar^2-a}{ar-ar^3}=\frac{9}{18}\\\Rarr\space\frac{a(r^2-1)}{ar(1-r^2)}=\frac{1}{2}\\\Rarr\space\frac{-1(1-r^2)}{r(1-r^2)}=\frac{1}{2}\\\Rarr\space-\frac{1}{r}=\frac{1}{2}$$

⇒ r = – 2

Putting r = – 2 in equation (ii),

a(– 2) – a(– 2)3 = 18

⇒ – 2a + 8a = 18

⇒ 6a = 18

⇒ a = 3

GP is 3, 3(– 2), 3(– 2)2, 3(– 2)3

⇒ 3, – 6, 12, – 24, …

22. If the pth, qth, rth terms of a GP are a, b and c, respectively prove that aq – r br – p cp – q = 1.

Sol. Let the first term of GP is A and common ratio is R.

Given, pth term = Tp = a

⇒ ARp – 1 = a …(i)

qth term = Tq = b

⇒ ARq – 1 = b …(ii)

rth term = Tr = c

⇒ ARr – 1 = c …(iii)

Now, we are to prove

aq – r br – p cp – q = 1

L.H.S. = aq – r br – p cp – q

Putting the values of a, b and c from equation (i), (ii) and (iii), we get

L.H.S. = (ARp – 1)q – r (ARq – 1)r – p (ARr – 1)p – q

= Aq – r R(p – 1)(q – r) Ar – p R(q – 1) (r – p) Ap – q R(r – 1) (p – q)

= Aq – r + r – p + p – q R(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)

= A0 Rpq – pr – q + r – qr – pq + r + p + rp – rq + p + q

= A0 R0 = 1 × 1 = R.H.S.   Hence Proved.

23. If the first and nth term of a GP are a and b, respectively and if P is the product of n terms, prove that P2 = (ab)n.

Sol. Let the GP is A, AR, AR2, AR3

Here, first term A = a …(i)

nth term ARn – 1 = b …(ii)

Now,

p = product of n terms

P = A × AR1 × AR2 × AR3 × … × n terms

p = A1 + 1 + 1 + … + n terms  R1 + 2 + 3 + … + (n – 1)

$$\text{p}=\text{A}^n\text{R}^\frac{n(n-1)}{2}\\\bigg[\because\space\text{Sum of first n natural numbers}=\frac{n(n+1)}{2}\bigg]$$

Squaring on both sides,

p2 = A2n Rn(n – 1)

⇒ p2 = An An Rn(n – 1)

= An (ARn – 1)n

⇒ p2 = anbn

[using equations (i) and (ii)]

⇒ p2 = (ab)n.   Hence Proved.

24. Show that the ratio of the sum of the first n terms of a GP to the sum of terms from (n + 1)th  to (2n)th terms is
$$\frac{\textbf{1}}{\textbf{r}^\textbf{n}}\textbf{.}$$

Sol. Let Q be the first term and r be the common ratio, the GP is

$$\underbrace{ar,ar^2,ar^3,ar^4,ar^5,...ar^n-1}_{n\text{ terms}}\\\space\underbrace{ar^n,ar^{n+1},...ar^{2n-1}}_{n\text{ terms}}$$

Now, required ratio

$$=\frac{\text{sum of first n terms}}{\text{sum from (n+1)th terms to (2n)th terms}}\\=\frac{\frac{a(r^n-1)}{r-1}}{\frac{ar^n(r^n-1)}{r-1}}=\frac{1}{r^n}\\\space\textbf{Hence Proved}$$

25. If a, b, c and d are in GP show that

(a2 + b2 + c2) (b2 + c2 + a2) = (ab + bc + cd)2.

Sol. ∵ a, b, c, d are in G.P.

$$\therefore\space \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=r\space\text{(say)}$$

⇒ b = ar, c = br, d = cr

⇒ b = ar, c = (ar)r, d = (br)r

⇒ b = ar, c = ar2, d = br2

⇒ b = ar, c = ar2, d = (ar)r2 = ar3 …(i)

Now, we have to prove that

(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2

Taking L.H.S. = (a2 + b2 + c2) (b2 + c2 + d2)

= (a2 + a2r2 + a2r4) (a2r2 + a2r4 + a2r6)

= a2 (1 + r2 + r4)a2 r2 (1 + r2 + r4)

= a4r2 (1 + r2 + r4)2

= [a2r (1 + r2 + r4)]2

= [a2r + a2r3 + a2r5]2

= [a·ar + ar·ar2 + ar2·ar3]2

= [ab + bc + cd]2     [from (i)]

= R.H.S.      Hence Proved.

26. Insert two numbers between 3 and 81 so that resulting sequence is GP.

Sol. Let the two numbers are a and b, then 3, a, b, 81 are in GP.

∵ nth term Tn = ARn – 1

∴ 81 = 3R4 – 1

$$\Rarr\space \text{R}^3=\frac{81}{3}$$

⇒ R3 = 27

⇒ R3 = 33

⇒ R = 3

On comparing the base of power 3 on both sides,

⇒ a = AR = 3 × 3 = 9, b = AR2 = 3 × 32 = 27

Hence, the required two numbers are 9 and 27.

27. Find the value of n so that

$$\frac{\textbf{a}^{\textbf{n+1}}\textbf{+}\textbf{b}^{\textbf{n+1}}}{\textbf{a}^\textbf{n}\textbf{+}\textbf{b}^\textbf{n}}$$

may be the geometric mean between a and b.

Sol. According to question,

$$\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{ab}\\\Rarr\space\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\frac{a^{\frac{1}{2}}b^{\frac{1}{2}}}{1}\\\Rarr\space a^{n+1}+b^{n+1}=(a^n+b^n)(a^{\frac{1}{2}}b^{\frac{1}{2}})\\\Rarr\space a^{n+1}+b^{n+1}=a^{n+\frac{1}{2}}b^{\frac{1}{2}}+a^{\frac{1}{2}}b^{n+\frac{1}{2}}\\\Rarr\space a^{n+1}+b^{n+1}-a^{n+\frac{1}{2}}b^{\frac{1}{2}}-a^{\frac{1}{2}}b^{n+\frac{1}{2}}=0\\\Rarr\space(a^{n+1}-a^{n+\frac{1}{2}}b^{\frac{1}{2}})+(b^{n+1}-a^{\frac{1}{2}}b^{n+\frac{1}{2}})=0\\\Rarr\space a^{n+\frac{1}{2}}[a^{\frac{1}{2}}-b^{\frac{1}{2}}]-b^{n+\frac{1}{2}}[a^{\frac{1}{2}}-b^{\frac{1}{2}}]=0\\\Rarr\space(a^{n+\frac{1}{2}}-b^{n+\frac{1}{2}})(a^{\frac{1}{2}}-b^\frac{1}{2})=0\\\Rarr\space a^{n+\frac{1}{2}}-b^{n+\frac{1}{2}}=0\\\space(\text{as}\space a^{\frac{1}{2}}-b^{\frac{1}{2}}\neq 0)$$

$$\Rarr\space a^{n+\frac{1}{2}}=b^{n+\frac{1}{2}}\\\Rarr\space\bigg(\frac{a}{b}\bigg)^{n+\frac{1}{2}}=1\\\Rarr\space \bigg(\frac{a}{b}\bigg)^{n+\frac{1}{2}}=\bigg(\frac{a}{b}\bigg)^0$$

On comparing the power of base (a/b) on both sides,

$$\Rarr\space n+\frac{1}{2}=0\\\Rarr\space n=-\frac{1}{2}$$

28. The sum of two numbers is 6 times their geometirc mean, show that the numbers in the ratio

$$\textbf{(3+2}\sqrt{\textbf{2}}\textbf{):}\textbf{(3-2}\sqrt{\textbf{2}}\textbf{).}$$

Sol. Let the numbers are a and b, whose geometric mean is $$\sqrt{ab}.$$

$$\text{Given,}\space a+b=6\sqrt{ab}\\\Rarr\space\frac{a+b}{2\sqrt{ab}}=\frac{3}{1}$$

Now, applying compendo and dividendo,

$$\Rarr\space\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{3+1}{3-1}\\\Rarr\space \frac{(\sqrt{a})^2+(\sqrt{b})^2+2\sqrt{ab}}{(\sqrt{a})^2+(\sqrt{b})^2-2\sqrt{ab}}=\frac{4}{2}\\\Rarr\space\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^{2}}=\frac{2}{1}\\\Rarr\space \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{2}}{1}$$

Again applying componendo and dividendo,

$$\Rarr\space\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}\space- (\sqrt{a}-\sqrt{b})}=\frac{\sqrt{2}+1}{\sqrt{2}-1}\\\Rarr\space\frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}\\\Rarr\space \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}\\\text{Now squaring both sides,}\\\Rarr\space \frac{a}{b}=\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2}\\\Rarr\space\frac{a}{b}=\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}$$

$$\begin{bmatrix}\because\space(a+b)^2=a^2+b^2+2ab\space \text{and}\\(a-b)^2=a^2+b^2-2ab\end{bmatrix}\\\Rarr\space\frac{a}{b}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\\\Rarr\space a:b=(3+2\sqrt{2}):(3-2\sqrt{2})\\\textbf{Hence Proved.}$$

29. If A and G be AM and GM respectively between two positive numbers prove that the number are

$$\textbf{A}\pm\sqrt{\textbf{(A+G)(A-G)}}.$$

Sol. Let the numbers are a and b.

$$\text{Given, sum of roots,}\space\frac{a+b}{2}=\text{A}\\(\text{Arithmetic mean})$$

⇒ a + b = 2A

and product of the roots.

$$\sqrt{ab}=\text{(Geometric mean)}$$

⇒ ab = G2

Now, quadratic equation having roots a and b is

x2 – (a + b)x + ab = 0

⇒ x2 – 2Ax + G2 = 0

$$\Rarr\space x=\frac{2A\pm\sqrt{4A^{2}-4×1×G^{2}}}{2×1}\\\bigg[\because\space x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\bigg]\\=\frac{2A\pm2\sqrt{\text{A}^{2}-\text{G}^{2}}}{2}\\=\text{A}\pm\sqrt{(A+G)(A-G)}$$

[∵ a2 – b2 = (a + b) (a – b)]

Hence Proved.

30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?

Sol. Given, the Number of bacteria doubles every hour in the culture, form a GP whose first term a = 30 and common ratio r = 2.

Bacteria present after 2nd hour = ar2 = 30 × (2)2 = 120 (∵ Tn = arn – 1)

Bacteria present after 4th hour = ar4 = 30 × (2)4 = 30 × 16 = 480

Bacteria present after nth hour = arn = 30 × 2n = (30)·(2n).

31. What will ₹500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Sol. Let A be the amount, P be principal, r be the rate of interest and t be the time period in years. Then, amount A is given by

$$\text{A}=\text{P}\bigg(1+\frac{r}{100}\bigg)^r\\=500\bigg[1+\frac{10}{100}\bigg]^{10}\\=500\bigg[1+\frac{1}{10}\bigg]^{10}\\=500[1+0.1]^{10}$$

= 500 × (1·1)10 = ₹500(1·1)10

∴ The amount at the end of 10 years is ₹500(1·1)10.

32. If AM and GM of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Sol. Let the roots of the quadratic equation are α and β, then (Arithmetic mean)

$$\frac{a+b}{2}=8\space\text{and}\\\text{(Geomgeric mean)}\sqrt{ab}=5$$

⇒ a + b = 16 and ab = 25

Now, if roots are a and b, then quadratic equation is

x2 – (Sum of roots)x + Product of roots = 0

⇒ x2 – (a + b)x + ab = 0

⇒ x2 – 16x + 25 = 0

Hence, the quadratic equation is x2 – 16x + 25 = 0.

Exercise 9.4

Find the sum to n terms of each series in Exercises 1 to 7:

1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Sol. Hence, the given series is

S = 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

First, we will split the series into two parts which are 1, 2, 3, 4, … and 2, 3, 4, 5, … and find the nth term of the given series.

Tn = (nth term of 1, 2, 3, …) × [nth term of 2, 3, 4, …]

= [1 + (n – 1)1] × [2 + (n – 1)1]

[∵ Tn = a + (n – 1)d]

= [1 + n – 1] [2 + n – 1]

⇒ Tn = n(n + 1)

Now  S = ΣTn = Σ(n2 + n) = Σn2 + Σn

$$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$$

$$\begin{bmatrix}\because\space\Sigma n^2=\frac{n(n+1)(2n+1)}{6}\\\Sigma n=\frac{n(n+1)}{2}\end{bmatrix}\\=\frac{n(n+1)}{2}\bigg[\frac{2n+1}{3}-\frac{1}{1}\bigg]\\=\frac{n(n+1)}{2}\bigg(\frac{(2n+1+3)}{3}\bigg)\\=\frac{n(n+1)}{2}\bigg(\frac{2n+4}{3}\bigg)\\=\frac{n(n+1)(n+2)}{3}$$

2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Sol. Here, the given series is S = 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

First, we will split the given series into three parts which are 1, 2, 3, … separately to find the nth term of the given series.

i.e., Tn = (nth term of 1, 2, 3, …) × (nth term of 2, 3, 4, …) × (nth term of 3, 4, 5, …)

= [1 + (n – 1)1] [2 + (n – 1)1] [3 + (n – 1)1]

[∵ Tn = a + (n – 1)d]

= (1 + n – 1) (2 + n – 1) (3 + n – 1)

⇒ Tn = n(n + 1) (n + 2)

= n(n2 + 2n + n + 2) = n(n2 + 3n + 2)

⇒ Tn = n3 + 3n2 + 2n

Now,

S = ΣTn = S(n3 + 3n2 + 2n)

= Σn3 + 3Σn2 + 2Σn

$$=\bigg[\frac{n(n+1)}{2}\bigg]^{2}+\frac{3n(n+1)(2n+1)}{6}+\\\frac{2n(n+1)}{2}\\=\frac{n(n+1)}{2}\bigg[\frac{n(n+1)}{2}+(2n+1)+2\bigg]\\\bigg[\because\space\Sigma n=\frac{n(n+1)}{2}\bigg]\\=\frac{n(n+1)}{2}\bigg[\frac{n(n+1)}{2}+\frac{2n+1}{1}+\frac{2}{1}\bigg]\\\bigg[\Sigma n^2=\frac{n(n+1)(2n+1)}{6}\bigg]\\$$

$$=\frac{n(n+1)}{2}\bigg[\frac{n^2+n+4n+2+4}{2}\bigg]\\=\bigg[\Sigma n^3=\bigg[\frac{n(n+1)}{2}\bigg]^2\bigg]\\=\frac{n(n+1)(n^2+5n+6)}{4}\\=\frac{n(n+1)(n^2+2n+3n+6)}{4}\\=\frac{n(n+1)[n(n+2)+3(n+2)]}{4}\\=\frac{n(n+1)(n+2)(n+3)}{4}$$

3. 3 × 12 + 5 × 22 + 7 × 32 + …

Sol. Here, the given series is

S = 3 × 12 + 5 × 22 + 7 × 32 + …

First, we will split the given series into two parts which are 3, 5, 7, … and 12, 22, 32, … and find the nth term of the given series.

Tn = (nth term of 3, 5, 7, …) × (nth term of 1, 2, 3, …)2

= [3 + (n – 1)2] [1 + (n – 1)1]2

= (3 + 2n – 2) (n)2

= (2n + 1)n2 = 2n3 + n2

Now,

S = ΣTn = 2Σn3 + Σn2

$$=2\bigg[\frac{n(n+1)}{2}\bigg]^2+\frac{n(n+1)(2n+1)}{6}\\\begin{bmatrix}\because\space \Sigma n^3=\bigg[\frac{n(n+1)}{2}\bigg]^2\\\Sigma n^2=\frac{n(n+1)(2n+1)}{6}\end{bmatrix}\\=\frac{n(n+1)}{2}\bigg[2×\frac{n(n+1)}{2}+\frac{2n+1}{3}\bigg]\\=\frac{n(n+1)}{2}\bigg[\frac{3n(n+1)+2n+1}{3}\bigg]\\=\frac{n(n+1)}{6}×(3n^2+3n+2n+1)\\=\frac{n(n+1)(3n^2+5n+1)}{6}$$

$$\textbf{4.}\space\frac{\textbf{1}}{\textbf{1×2}}+\frac{\textbf{1}}{\textbf{2×3}}+\frac{\textbf{1}}{\textbf{3×4}}\textbf{+...}\\\textbf{Sol.}\space\text{Let the given series is}\\\text{S}=\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+...$$

Here, we will split the denominator of the series into two parts i.e.,

$$\text{T}_n=\\\frac{1}{\text{(nth term of 1,2,3 .....)×(nth term of 2,3,4, .....)}}\\=\frac{1}{[1+(n-1)1]+[2+(n-1)1]}\\\lbrack\because\space\text{T}_m=a+(m-1)d\rbrack\\=\frac{1}{n(n+1)}\\=\frac{(n+1)-n}{n(n+1)}\\=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}\\=\frac{1}{n}-\frac{1}{n+1}$$

Now, putting n = 1, 2, 3, 4 …

$$\text{T}_1=\frac{1}{1}-\frac{1}{2}\\\text{T}_2=\frac{1}{2}-\frac{1}{3}\\\text{T}_3=\frac{1}{3}-\frac{1}{4}\\.................\\.................\\\text{T}_n=\frac{1}{n}-\frac{1}{n+1}$$

Now, adding these terms, we get

S = T1 + T2 + T3 + … + Tn

$$=\bigg(\frac{1}{1}-\frac{1}{2}\bigg)+\bigg(\frac{1}{2}-\frac{1}{3}\bigg)+...+\\\bigg(\frac{1}{n}+\frac{1}{n+1}\bigg)\\=1-\frac{1}{n+1}\\=\frac{1}{1}-\frac{1}{n+1}\\=\frac{n+1-1}{n+1}\\\Rarr\space\text{S}=\frac{n}{n+1}$$

5. 52 + 62 + 72 + … + 202

Sol. Here, the given series is

S = 52 + 62 + 72 + … + 202

= (12 + 22 + 32 + 42 + 52 + 62 + … + 202) – (12 + 22 + 32 + 42)

$$=\displaystyle\sum^{20}_{n=1}n^2-\displaystyle\sum^{4}_{n=1}n^2\\=\frac{20×(20+1)×(2×20+1)}{6}-\\\frac{4(4+1)(2×4+1)}{6}\\\bigg(\because\Sigma n^2=\frac{n(n+1)(2n+1)}{6}\bigg)\\=\frac{20×21×41}{6}-\frac{4×5×9}{6}\\=\frac{17220}{6}-\frac{180}{6}\\=\frac{17220-180}{6}=\frac{17040}{6}$$

= 2840.

6. 3 × 8 + 6 × 11 + 9 × 14 + …

Sol. Here, the given series is

S = 3 × 8 + 6 × 11 + 9 × 1 4 + …

Tn = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14 …)

Tn = [3 + (n – 1)3] [8 + (n – 1)3]

[∵ Tm = a + (m – 1)d]

= (3 + 3n – 3) (8 + 3n – 3)

= 3n(3n + 5) = 9n2 + 15n

Now,

S = ΣTn

= Σ(9n2 + 15n) = 9Σn2 + 15Σn

$$=\frac{9n(n+1)(2n+1)}{6}+\frac{15n(n+1)}{2}\\\begin{bmatrix}\because\space\Sigma n^2=\frac{n(n+1)(2n+1)}{6}\\\Sigma n=\frac{n(n+1)}{2}\end{bmatrix}\\=\frac{n(n+1)}{2}\bigg[\frac{9(2n+1)}{3}+15\bigg]\\=\frac{n(n+1)}{2}[3(2n+1)+15]\\=\frac{n(n+1)3[2n+1+5]}{2}\\=\frac{3n(n+1)(2n+6)}{2}\\=3n(n+1)(n+3).$$

7. 12 + (12 + 22) + (12 + 22 + 32) + …

Sol. Here, the given series is

S = 12 + (12 + 22) + (12 + 22 + 32) + …

Here, first term has 1 term, second term has 2 terms, third terms has 3 and so on. Therefore, nth term will have n terms.

i.e., Tn = 12 + 22 + … + n2

Tn = Σn2

$$=\frac{n(n+1)(2n+1)}{6}\\\Rarr\space\text{T}_n=\frac{n(2n^2+n+2n+1)}{6}\\\Rarr\space\text{T}_n=\frac{n(2n^2+3n+1)}{6}\\\Rarr\space\text{T}_n=\frac{2n^3+3n^2+n}{6}\\\text{Now,}\space\text{S}=\Sigma \text{T}_n=\frac{1}{6}(2n^3+3n^2+n)\\=\frac{1}{6}[2\Sigma n^3+3\Sigma n^2+\Sigma n]$$

$$=\frac{1}{6}\begin{bmatrix}2\begin{Bmatrix}\frac{n(n+1)}{2}\end{Bmatrix}^2+\\3\frac{n(n+1)(2n+1)}{6}+\\\frac{n(n+1)}{2}\end{bmatrix}$$

$$\begin{Bmatrix}\because\Sigma n=\frac{n(n+1)}{2},\Sigma n^2=\frac{n(n+1)(2n+1)}{6},\\\Sigma n^2=\bigg[\frac{n(n+1)}{2}\bigg]^2\end{Bmatrix}\\=\frac{1}{6}×\frac{n(n+1)}{2}\bigg[\frac{2n(n+1)}{2}+\frac{2n+1}{1}+1\bigg]$$

$$=\frac{n(n+1)}{12}×[n^2+n+2n+1+1]\\=\frac{n(n+1)}{12}[n^2+3n+2]\\=\frac{n(n+1)(n^2+2n+n+2)}{12}\\=\frac{n(n+1)(n+2)(n+1)}{12}\\=\frac{n(n+1)^2(n+2)}{12}$$

Find the sum to n terms of the series is Exercises 8 to 10 whose nth terms is given by:

8. n(n + 1) (n + 4).

Sol. The nth term

Tn = n(n + 1) (n + 4)

= n(n2 + 4n + n + 4)

= n(n2 + 5n + 4)

= n3 + 5n2 + 4n

Now,

S = ΣTn = Σ(n3 + 5n2 + 4n)

= Σn3 + 5Σn2 + 4Σn

$$=\bigg[\frac{n(n+1)}{2}\bigg]^2+\frac{5n(n+1)(2n+1)}{6}+\\\frac{4n(n+1)}{2}\\=\frac{n(n+1)}{2}\bigg[\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+\\\frac{4}{1}\bigg]\\=\frac{n(n+1)}{2}\bigg[\frac{3n(n+1)+10(2n+1)+24}{6}\bigg]\\=\frac{n(n+1)[(3n^2+3n+20n+1)+24]}{12}\\=\frac{n(n+1)(3n^2+23n+34)}{12}$$

9. n2 + 2n.

Sol. The nth term

Tn = n2 + 2n

⇒ S = ΣTn = Σ(n2 + 2n) = Σn2 + Σ2n

$$=\frac{n(n+1)(2n+1)}{6}+\\(2^1+2^2+2^3+2^4+....+2^n)\\=\frac{n(n+1)(2n+1)}{6}+\frac{2(2^n-1)}{2-1}\\\bigg[\because\space S_n=\frac{a(r^n-1)}{r-1},r\gt1\bigg]\\\text{S}=\frac{n(n+1)(2n+1)}{6}+\\2(2^n-1)$$

10. (2n – 1)2

The nth term

Tn = (2n – 1)2

⇒ Tn = 4n2 + 1 – 4n

Now,

S = ΣTn = Σ(4n2 + 1 – 4n)

= 4Σn2 + Σ1 – 4Σn

$$=\frac{4n(n+1)(2n+1)}{6}+n-\frac{4n(n+1)}{2}\\(\because \Sigma1=n)\\=n\bigg[\frac{2(n+1)(2n+1)}{3}+\frac{1}{1}-\frac{2(n+1)}{1}\bigg]\\=n\bigg[\frac{2(2n^2+n+2n+1)+3-6(n+1)}{3}\bigg]\\=\frac{n[4n^2+6n+2+3-6n-6]}{3}\\=\frac{n(4n^2-1)}{3}\\=\frac{n}{3}(2n+1)(2n-1)$$

[∵ (a2 – b2) = (a – b) (a + b)}

Miscellaneous Exercise

1. Show that the sum of (m + n)th and (m – n)th terms of an AP is equal to twice the mth term.

Sol. Let a be the first term and d be the common difference of the A.P.

Now, LHS = Tm + n + Tm – n

= a + (m + n – 1)d + a + (m – n – 1)d

[∵ Tn = a + (n – 1)d]

= 2a + d(m + n – 1 + m – n – 1)

= 2a + d(2m – 2)

= 2[a + d(m – 1)] = 2Tm = R.H.S.

Hence Proved.

2. If the sum of three numbers in AP is 24 and their product is 440, find the numbers.

Sol. Let the three numbers are a – d, a, a + d.

Given, sum of numbers = 24

∴ a – d + a + a + d = 24

⇒ 3a = 24

⇒ a = 8

And product of the numbers = 440

∴ (a – d) a(a – d) = 440

⇒ a(a2 – d2) = 440

⇒ 64 – d2 = 55

⇒ d2 = 64 – 55

⇒ d2 = 9

⇒ d = ± 3

When, a = 8 and d = 3, then numbers are

a – d = 8 – 3 = 5

and a = 8

and a + d = 8 + 3 = 11

⇒ Numbers are 5, 8, 11

When a = 8 and d = – 3, then numbers are

a – d = 8 + 3 = 11

a = 8

a + d = 8 – 3 = 5

⇒ Numbers aer 11, 8, 5

Hence, the nunbers are 5, 8, 11 or 11, 8, 5.

3. Let the sum of n, 2n, 3n terms of an AP be S1, S2 and S3, respectively. Show that S3 = 3(S2 – S1).

Sol. Let a and d be the first term and common difference respectively of an AP.

Given, S1 = Sum of n terms

$$=\frac{n}{2}[2a+(n-1)d]\space\text{…(i)}\\\text{S}_2=\text{Sum of 2n terms}\\=\frac{2n}{2}[2a+(2n-1)d]\space\text{...(ii)}$$

and S3 = Sum of 3n terms

$$=\frac{3n}{2}[2a+(3n-1)d]\space\text{...(iii)}$$

Now, we have to prove

S3 = 3(S2 – S1)

Now, R.H.S. = 3(S2 – S1)

$$=3\bigg[\frac{2n}{2}[2a+(2n-1)d]-\frac{n}{2}[2a+(n-1)d]\bigg]\\\text{[from Equations (i) and (ii)]}\\=\frac{3n}{2}[2\lbrace2a+(2n-1)d\rbrace-\lbrace2a+(n-1)d\rbrace]\\=\frac{3n}{2}[4a+2(2n-1)d-2a-(n-1)d]\\=\frac{3n}{2}[(4a-2a)+d(4n-2-n+1)]\\=\frac{3n}{2}[2a+(3n-1)d]$$

= S3 = L.H.S.    [from equation (iii)]

∴ L.H.S. = R.H.S. Hence Proved.

4. Find the sum of all numbers between 200 and 400 which are divisible by 7.

Sol. The number which are divisible by 7 between 200 and 400 is 203, 210, 217, …, 399.

Clearly, they form an AP.

where, a = 203, d = 7 and Tn = 399

Now, Tn = a + (n – 1)d

⇒ 399 = 203 + (n – 1)7

⇒ (n – 1)7 = 399 – 203

⇒ (n – 1)7 = 196
$$\Rarr\space\text{n-1}=\frac{196}{7}$$

⇒ n = 28 + 1

n = 29

$$\text{Now,}\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\\\therefore\space\text{S}_n=\frac{29}{2}[2×203+(29-1)7]\\\Rarr\space\text{S}_{29}=\frac{29}{2}[406+28×7]\\=\frac{29}{2}[406+196]\\=\frac{29}{2}×602\\\text{29 × 301 = 8729}$$

Hence, the required sum is 8729.

5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Sol. The integers from 1 to 100 which are divisible by 2 are 2, 4, 6, 8, …, 100.

Clearly they are in AP, where a = 2, d = 4 – 2 = 2

∴ Tn = a + (n – 1)d ⇒ 100 = 2 + (n – 2)2

⇒ 100 – 2 = (n – 1)2 ⇒ 98 = (n – 1)2

⇒ 49 = n – 1 ⇒ n = 50

Therefore, sum of 50 numbers,

$$\text{S}_{50}=\frac{50}{2}[2×2+(50-1)2]\\\bigg[\because\text{S}_n=\frac{n}{2}[2a+(n-1)d]\bigg]$$

= 25[4 + 49 × 2]

= 25[4 + 98] = 25 × 102

⇒ S50 = 2550 …(i)

Now, the integers from 1 to 100 which are divisible by 5 are

5, 10, 15, 20, … 100

Clearly, they are in AP where a = 5, d = 10 – 5 = 5

∴ Tn = a + (n – 1)d

100 = 5 + (n – 1)5

⇒ 100 –5 = (n – 1)5

$$\Rarr\space(n-1)=\frac{95}{5}$$

⇒ n – 1 = 19

⇒ n = 19 + 1 = 20

$$\text{Now,\space}\text{S}_n=\frac{n}{2}[2a+(n-1)d]\\\Rarr\space\text{S}_{20}=\frac{20}{2}[2×5+(20-1)5]$$

= 10[10 + 19 × 5]

= 10[10 + 95] = 10(105)

⇒ S20 = 1050 …(ii)

Now, the integers from 1 to 100 which are divisible by 10 are 10, 20, 30, …, 100.

Clearly Clearly, they can AP where a = 10, d =20 – 10 = 10 and n = 10

$$\because\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\\\therefore\space\text{S}_{10}=\frac{10}{2}[2×10+(10-1)10]$$

= 5[20 + 9 × 10]

= 5[20 + 90]

= 5 × 110 = 550 …(iii)

Hence, the required sum of integers from 1 to 100

which are divisible by 2 or 5 = 2550 + 1050 – 550 (using Equations (i), (ii) and (iii))) = 3600 – 550 = 3050.

6. Find the sum of all two digit numbers which when divided by 4, yield 1 as remainder.

Sol. The two digit numbers which when divided by 4, leave 1 as remainder, the sequences forms an A.P. are

13, 17, 21, …, 97

where, a = 13, d = 4, Tn = 97

∵ Tn = a + (n – 1)d

∴ 97 = 13 + (n – 1)4

⇒ 97 – 13 = (n – 1)4

⇒ 84 = (n – 1)4

⇒ 21 = n – 1

⇒ n = 21 + 1 = 22

$$\text{Now,}\space\text{S}_{22}=\frac{22}{2}[2×13+(22-1)14]$$

= 11(26 + 21 × 4)

= 11(26 + 84)

= 11 × 110 = 1210

Hence, the required sum is 1210.

7. If f is a function satisfying f(x + y) = f(x) (y) for all x, y ∈ N such that f(1) = 3 and

$$\displaystyle\sum^{\textbf{n}}_{\textbf{x=1}}\textbf{f(x)=120, \space find the value of n.}$$

Sol. Given, f(x + y) = f(x) f(y) for all x, y ∈ N …(i)

Putting x = y = 1 in equation (i),

f(1 + 1) = f(1) f(1)

⇒ f(2) = 3 × 3 [∵ f(1) = 3]

⇒ f(2) = 9

Putting x = 2, y = 1 in equation (i)

f(2 + 1) = f(2) f(1)

⇒ f(3) = 9 × 3 [∵ f(2) = 9, f(1) = 3]

⇒ f(3) = 27

Putting x = 3, y = 1 in equation (i),

f(3 + 1) = f(3) f(1)

⇒ f(4) = 27 × 3 [∵ f(3) = 27]

⇒ f(4) = 81

$$\text{Now,\space}\displaystyle\sum^{n}_{x=1}\space\text{f(x)}=120$$

⇒ f(1) + f(2) + f(3) + … + f(n) = 120

⇒ 3 + 32 + 33 + … + 3n = 120

⇒ 3 + 9 + 27 + … + n terms = 120

$$\text{Here,\space}a=3,r=\frac{9}{3}=3\\\because\space\frac{a(r^n-1)}{r-1}=\frac{3(3^n-1)}{3-1}\\=120\space(\therefore\space r\gt1)$$

⇒ 3(3n – 1) = 120 × 2

⇒ 3(3n – 1) = 240

$$\Rarr\space 3^n-1=\frac{240}{3}$$

⇒ 3n – 1 = 80

⇒ 3n = 80 + 1

On comparing the power of 3 on both sides,

⇒ 3n = 81

⇒ 3n = 34

⇒ n = 4

Hence, the value of n is 4.

8. The sum of the some terms of a GP is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.

Sol. Let GP of n terms is a, ar2, ar3, …, arn – 1

Given, a = 5, r = 2 and Sn = 315

$$\text{Therefore,\space} 315=\frac{5(2^n-1)}{2-1}\\\Rarr\space\frac{315}{5}=2^{n}-1\\\bigg[\because\space\text{S}_n=\frac{a(r^{n}-1)}{r-1}\bigg]$$

⇒ 2n – 1 = 63

⇒ 2n = 63 + 1

⇒ 2n = 64

On comparing the power of 2 both sides,

⇒ 2n = 26

⇒ n = 6

Again, for last term use

Tn = arn – 1 = 5 × (2)6 – 1

= 5 × 25 = 5 × 32 = 160

Hence, the last term of the G.P. is 160 and the number of terms is 6.

9. The first term of a GP is 1. The sum of the third term and fifth term is 90. Find the common ratio of the GP.

Sol. Let the GP is a, ar, ar2, ar3

Given, a = 1 and T3 + T5 = 90

∴ ar2 + ar4 = 90 (·.· Tn = arn – 1)

⇒ r2 + r4 = 90 (as a = 1)

⇒ r4 + r2 – 90 = 0

Now, factorizing it by splitting the middle term,

⇒ r4 + 10r2 – 9r2 – 90 = 0

⇒ r2(r2 + 10) – 9(r2 + 10) = 0

⇒ (r2 + 10) (r2 – 9) = 0

∵ r2 + 10 ≠ 0 ⇒ r2 – 9 = 0

⇒ r2 = 9 ⇒ r = ± 3

Hence, the common ratio of the G.P. is ± 3.

10. The sum of three numbers in GP is 56. If we subtract the 1, 7, 21 from these numbers in that order, we obtain an Arithmetic Progression (AP). Find the numbers.

Sol. Let three numbers in GP are a, ar, ar2.

Given, the sum of three numbers in G.P. is 56.

a + ar + ar2 = 56 …(i)

Again, a – 1, ar – 7, ar2 – 21 are in A.P.

⇒ 2(ar – 7) = (a – 1) + (ar2 – 21)

(∵ If a, b, c are in AP, then 2b = a + c)

⇒ 2ar – 14 = a + ar2 – 22

⇒ a + ar2 – 2ar = – 14 + 22

⇒ a + ar2 – 2ar = 8 …(ii)

Dividing equation (i) by equation (ii), we get

$$\frac{a+ar+ar^2}{a+ar^2-2ar}=\frac{56}{8}\\\Rarr\space\frac{1+r+r^2}{1+r^2-2r}=\frac{7}{1}$$

⇒ 1 + r + r2 = 7 + 7r2 – 14r

⇒ 6r2 – 15r + 6 = 0

Dividing by 3

⇒ 2r2 – 5r + 2 = 0

Factorizing it by splitting the middle term,

⇒ 2r2 – (4 + 1)r + 2 = 0

⇒ 2r2 – 4r – r + 2 = 0

⇒ 2r(r – 2) – (r – 2) = 0

⇒ (r – 2) (2r – 1) = 0

$$\Rarr\space r=2,\frac{1}{2}$$

If r = 2, then from equation (i)

a + 2a + 4a = 56

⇒ 7a = 56

a = 8

Then, numbers are

a = 8

ar = 8 × 2 = 16

⇒ ar2 = 8 × 4 = 32

8, 16, 32

$$\text{If}\space r=\frac{1}{2},\text{then from equation (i),}\\\frac{a}{1}+\frac{a}{2}+\frac{a}{4}=56\\\frac{4a+2a+1}{4}=56\\\Rarr\space\frac{7a}{4}=56$$

⇒ a = 32

Then, numbers are

a = 32

$$\text{ar}=32×\frac{1}{2}=16\\\text{or\space} ar^2=32×\frac{1}{4}=8$$

⇒ 32, 16, 8

Hence, required numbers are 8, 16, 32 or 32, 16, 8.

11. A GP consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Sol. Let the GP is a, ar, ar2, ar3, ar4, …, ar2n – 2, ar2n – 1

where a, ar2, ar6 … occupy odd places and ar, ar3, ar5, ar7 … occupy even places.

Given, sum of all terms = 5 × Sum of terms occupying odd places

i.e., a + ar + ar2 + … ar2n – 1

= 5 × (a + ar2 + ar4 + … + ar2n – 2)

$$\Rarr\space\frac{a(r^{2n}-1)}{r-1}=\frac{5a[(r^2)^n-1]}{r^2-1}\\\bigg(\because\space\text{S}_n=\frac{a(r^n-1)}{r-1},r\gt1\bigg)\\\Rarr\space\frac{r^{2n}-1}{r-1}=\frac{5(r^{2n}-1)}{(r-1)(r+1)}\\\Rarr\space 1=\frac{5}{5+1}$$

⇒ r + 1 = 5

⇒ r = 4

Hence, the common ratio of G.P. is 4.

12. The sum of the first four terms of an AP is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Sol. Let the AP is a, a + d, a + 2d, …

Given, sum of first four terms = 56

i.e., T1 + T2 + T3 + T4 = 56

⇒ a + a + d + a + 2d + a + 3d = 56

⇒ 4a + 6d = 56

⇒ 4 × 11 + 6d = 56  (∵ a = 11)

⇒ 6d = 56 – 44 = 12

$$\Rarr\space d=\frac{12}{6}$$

⇒ d = 2

If last term is Tn, then sum of last four terms
= 112

Tn + Tn – 1 + Tn – 2 + Tn – 3 = 112

[∵ Tn = a + (n – 1)d]

∴ a + (n – 1)d + a + (n – 1 – 1)d + a + (n – 2 – 1)d + a + (n – 3 – 1)d = 112

⇒ 4a + d(n – 1 + n – 2 + n – 3 + n – 4) = 112

⇒ 4 × 11 + 2[4n – 10] = 112 (∵ a = 11, d = 2)

⇒ 44 + 8n – 20 = 112

⇒ 8n = 112 – 44 + 20

⇒ 8n = 132 – 44

⇒ 8n = 88

$$\Rarr\space n=\frac{88}{8}=11$$

Hence, total numbers of terms in the series = 11.

13. If $$\frac{\textbf{a+bx}}{\textbf{a-bx}}=\frac{\textbf{b+cx}}{\textbf{b-cx}}\\=\frac{\textbf{c+dx}}{\textbf{c-dx}}\space\textbf{(x}\neq \textbf{0)},\\\textbf{then show that a, b, c, d are in GP.}$$

$$\textbf{Sol.}\space\text{Given that,}\space \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}\\=\frac{c+dx}{c-dx}$$

Applying componendo and dividendo,

$$\Rarr\space\frac{a+bx+a-bx}{a+bx-(a-bx)}=\frac{b+cx+b-cx}{b+cx-(b-cx)}\\=\frac{c+dx+c-dx}{c+dx-(c-dx)}$$

$$\Rarr\space\frac{2a}{2bx}=\frac{2b}{2cx}=\frac{2c}{2dx}\\\Rarr\space\frac{a}{bx}=\frac{b}{cx}=\frac{c}{dx}\\\text{Multiplying each term by x,}\\\Rarr\space\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$$

a, b, c and d are in G.P.

14. Let S be the sum, P be the product and R be the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn.

Sol. Let the GP is a, ar, ar2, ar3 … arn – 1

Given that,

S = Sum of n terms

= a + ar + ar2 + ar3 + … + arn – 1

$$=\frac{a(r^n-1)}{r-1}(\text{let}\space r\gt1)\space\text{...(i)}$$

and R = Sum of the reciprocals of n terms

$$=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}+...+\\\frac{1}{ar^{n-1}}\bigg(\frac{1}{r}\lt1\bigg)\\=\frac{\frac{1}{a}\bigg[1-\bigg(\frac{1}{r}\bigg)^n\bigg]}{1-\frac{1}{r}}\\=\frac{1}{a}\bigg[\frac{1}{1}-\frac{1}{r^n}\bigg]×\frac{1}{\frac{r-1}{r}}\\=\frac{1}{a}\bigg[\frac{r^n-1}{r^n}\bigg]×\frac{r}{r-1}\\\Rarr\space\text{R}=\frac{(r^n-1)r}{ar^n(r-1)}\space\text{...(ii)}$$

and P = Product of n terms

= a × ar × ar2 × ar3 × … × arn – 1

= a1 + 1 + 1 + … + n terms

r1 + 2 + 3 + … + (n – 1) terms

$$= a^nr^{\frac{n(n--1)}{2}}\\\bigg[\because\space \text{Sn}=\frac{n(n+1)}{2}\bigg]$$

⇒ P2 = a2n rn(n – 1) …(iii)

Now, we have to prove

P2Rn = Sn

$$\text{or}\space\text{P}^2=\frac{\text{S}^n}{\text{R}^n}\text{or P}^2=\bigg(\frac{\text{S}}{\text{R}}\bigg)^n\\\text{Taking RHS}=\bigg(\frac{\text{S}}{\text{R}}\bigg)^{n}=\\$$

$$\bigg[\frac{a(r^n-1)}{r-1}×\frac{ar^n(r-1)}{(r^n-1)r}\bigg]^n$$

[using Equations (i) and (ii)]

= (a2rn – 1)n

= [a2n rn(n – 1)]

= P2 = LHS    [from equation (iii)]

Hence Proved.

15. The pth, qth and rth terms of an AP are a, b, c respectively. Show that:
(q – r)a + (r – p)b + (p – q)c = 0.

Sol. Let A be the first term and D is the common difference then the AP is A, A + D, A + 2D, A + 3D, …

Given, pth term = A + (p – 1)D = a …(i)

qth term = A (q – 1)D + b …(ii)

rth term = A + (r – 1)D = c …(iii)

Now, we have to prove

(q – r)a + (r – b)p + (p – q)c = 0

Taking L.H.S. = (q – r)a + (r – p)b + (p – q)c

Putting the values of a, b and c from equations (i), (ii) and (iii) in equation (iv),

L.H.S. = (q – r) [A + (p – 1)D] + (r – p)

[A + (q – 1)D] + (p – q) [A + (r – 1)D]

= (q – r)A + (q – r) (p – 1)D + (r – p)A

+ (r – p) (q – 1)D + (p – q)A + (p – q) (r – 1)D

= A(q – r + r – p + p – q) + D(q – r) (p – 1)

+ (r – p) (p – 1) + (p – q) (r – 1)

= A(0) + D[qp – q – rp + r + rq – r – pq

+ p + pr – p – qr + q]

= 0 + 0 = 0 = R.H.S.   Hence Proved.

$$\textbf{16. If a}\bigg(\frac{\textbf{1}}{\textbf{b}}+\frac{\textbf{1}}{\textbf{c}}\bigg),\textbf{b}\bigg(\frac{\textbf{1}}{\textbf{c}}+\frac{\textbf{1}}{\textbf{a}}\bigg),\\\textbf{c}\bigg(\frac{\textbf{1}}{\textbf{a}}+\frac{\textbf{1}}{\textbf{b}}\bigg)\\\textbf{are in AP, prove that a, b, c are in AP.}$$

$$\textbf{Sol.}\space\text{Given that},\\\space a\bigg(\frac{1}{b}+\frac{1}{c}\bigg),b\bigg(\frac{1}{c}+\frac{1}{a}\bigg),\\\text{c}\bigg(\frac{1}{a}+\frac{1}{b}\bigg)\space\text{are in AP.}\\\Rarr\space a\bigg(\frac{b+c}{bc}\bigg),b\bigg(\frac{a+c}{ac}\bigg),\\\text{c}\bigg(\frac{b+c}{ab}\bigg)\space\text{are in AP.}\\\Rarr\space\frac{ab+ac}{bc},\frac{ba+bc}{ac},\\\frac{cb+ca}{ab}\space\text{are in AP.}\\\text{Adding 1 to each term,}\\$$

$$\Rarr\space\frac{ab+ac}{bc}+1,\frac{ba+bc}{ac}+1,\\\frac{cb+ca}{ab}\space\text{are in A.P.}\\\Rarr\space\frac{ab+ac+bc}{bc},\frac{ba+bc+ac}{ac},\\\frac{bc+ac+ab}{ab}\space\text{are in AP.}$$

Dividing each term by ab + bc + ac (property…)

$$\Rarr\space\frac{1}{bc},\frac{1}{ac},\frac{1}{ab}\space\text{are in AP.}\\\text{Multiplying each term by abc,}\\\Rarr\frac{abc}{bc},\frac{abc}{ac},\frac{abc}{ab}\space\text{are in AP.}$$

⇒ a, b, c in AP.    Hence Proved.

17. If a, b, c, d are in GP, prove that

an + bn, bn + cn, cn + dn are in G.P.

Sol. Let r be the common ratio,

∵ a, b, c, d are in GP.

⇒ b = ar, c = ar2, d = ar3 …(i)

Now, we are to prove tthat, an + bn, bn + cn, cn + dn are in GP.

⇒ (bn + cn)2 = (an + bn) (cn + dn)

Now, taking,

R.H.S. = (an + bn) (cn + dn)

= [an + anrn] × [anr2n + anr3n]

[from equation (i)]

= an(1 + rn)an r2n (1 + rn)

= a2nr2n (1 + rn)2 = (anrn(1 + rn)]2

= [anrn + anr2n]2

= [(ar)n + (ar2)n]2 = [bn + cn]2

[from equation (i)]

= L.H.S.

∴ L.H.S. = R.H.S.   Hence Proved.

18. If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d form a GP prove that (q + p) : (q – p) = 17 : 15.

Sol. Given, a and b are roots of x2 – 3x + p = 0.

∴ Sum of roots,

$$a+b=-\frac{(-3)}{1}=3\\\Rarr\space\text{a + b = 3 …(i)}$$

and product of roots,

ab = p …(ii)

Again, given that c and d are roots of x2 – 12x + q = 0.

⇒ Sum of roots,

$$\text{c+d}=\frac{(-12)}{1}=12$$

⇒ c + d = 12 …(iii)

and product of roots,

cd = q …(iv)

Again, if is given that a, b, c, d are in GP.

⇒ b = ar, c = ar2, d = ar3

Substituting these values in equations (i) and (iii), then dividing equation (i) by equation (iii), we get

$$\frac{a+ar}{ar^2+ar^3}=\frac{3}{12}\\\Rarr\space\frac{a(1+r)}{ar^2(1+r)}=\frac{1}{4}\\\Rarr\space\frac{1}{r^2}=\frac{1}{4}$$

⇒ r2 = 4

⇒ r = ± 2

Again from equation (i),

a + ar = 3

⇒ a + 2a = 3 (·.· r = 2)

⇒ 3a = 3

⇒ a = 1

So, the GP is

a = 1, b = ar = 1 × 2 = 2, c = ar2 = 1 × 22 = 4, d = ar3

= 1 × 23 = 8

From equation (ii),

p = ab = 1 × 2 = 2

From equation (iv),

q = cd = 4 × 8 = 32

$$\therefore\space\frac{\text{q+p}}{\text{q-p}}=\frac{32+2}{32-2}\\=\frac{34}{30}=\frac{17}{15}$$

Hence, (q + p) : (q – p) = 17 : 15

Again, from equation (i)

a + ar = 3 (∵ r = – 2)

⇒ a – 2a = 3

⇒ a = – 3

So, the GP is

a = – 3, b = ar = (– 3) (– 2) = 6

c = ar2 = (– 3) (– 2)2 = – 12

d = ar3 = (– 3) (– 2)3 = 24

From equation (ii),

p = ab = (– 3) · (+ 6) = – 18

From equation (iv),

q = cd = (– 12) (24) = – 288

$$\therefore\space\frac{q+p}{q-r}=\frac{-288-18}{-288+18}\\=\frac{-306}{-270}=\frac{17}{15}$$

Hence, (q + p) : (q – p) = 17 : 15.

19. The ratio of the AM and GM of two positive numbers a and b is m : n. Show that

$$\textbf{a:b\space =}\space(\textbf{m}\textbf{+}\sqrt{\textbf{m}^\textbf{2}\textbf{-n}^\textbf{2}})\textbf{:}\textbf{(m-}\sqrt{\textbf{m}^\textbf{2}\textbf{-n}^\textbf{2}}\textbf{).}$$

Sol. Let the AM of the number a and b is A and GM of a and b is G then,

$$\text A=\frac{a+b}{2}\space\text{and}\space \text{G}=\sqrt{ab}\\\text{Given, A : G = m : n}\\\text{i.e,}\space\frac{\text{A}}{\text{G}}=\frac{m}{n}\\\Rarr\space\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}\\\text{Applying componendo and dividendo rule,}\\\Rarr\space \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{m+n}{m-n}\\\Rarr\space\frac{(\sqrt{a})^2+(\sqrt{b})^2+2\sqrt{ab}}{(\sqrt{a})^2+(\sqrt{b})^2-2\sqrt{ab}}=\frac{m+n}{m-n}$$

$$\Rarr\space\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{m+n}{m-n}\\\Rarr\space\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{m+n}}{\sqrt{m-n}}$$

Again, applying componendo and dividendo rule

$$\Rarr\space\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}\\=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\\\Rarr\space\frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\\\Rarr\space\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{m+m}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\\\text{Now, squaring both sides,}\\\Rarr\space\frac{a}{b}=\frac{(\sqrt{m+n}+\sqrt{m-n})^2}{(\sqrt{m+n}-\sqrt{m-n})^2}\\\begin{bmatrix}\because\space (a+b)^2=a^2+b^2+2ab\\(a-b)^2=a^2+b^2-2ab\end{bmatrix}$$

$$\Rarr\\\space\frac{a}{b}=\frac{m+n+m-n+2\sqrt{\text{m+n}}\sqrt{\text{m-n}}}{m+n+m-n-2\sqrt{\text{m+n}}\sqrt{\text{m-n}}}\\\Rarr\space\frac{a}{b}=\frac{\text{2m+2}\sqrt{\text{m}^2-\text{n}^2}}{\text{2m-2}\sqrt{\text{m}^2-\text{n}^2}}$$

[∵ (a + b) (a – b) = a2 – b2]

$$\Rarr\space\frac{a}{b}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$$

$$\Rarr\space a:b\\=(m+\sqrt{m^2-n^2}):(m-\sqrt{m^2-n^2})$$

20. If a, b, c are in AP and b, c, d are in GP and $$\frac{1}{c},\frac{1}{d},\frac{1}{e}\space\text{are in AP}$$

prove that a, c, e are in GP.

Sol. Given that, a, b, c are in AP.

⇒ 2b = a + c     …(i)

Again, b, c, d are in GP.

⇒  c2 = bd      …(ii)

$$\text{Similarly,}\space\frac{1}{c},\frac{1}{d},\frac{1}{e}\space\text{are in AP.}\\\Rarr\space \frac{2}{d}=\frac{1}{c}+\frac{1}{e}\\\Rarr\space\frac{2}{d}=\frac{e+c}{ce}\\\Rarr\space d=\frac{2ce}{c+e}\space\text{...(iii)}$$

Substituting the values of b and d from equation (i) and (iii), in equation (ii), we get

$$\text{c}^2=\bigg(\frac{a+c}{2}\bigg)×\bigg(\frac{2ce}{c+e}\bigg)\\\Rarr\space c^2=\frac{a+c}{2}×\frac{2ce}{c+e}$$

⇒ c2(c + e) = (a + c)ce

⇒ c(c + e) = (a + c)e

⇒ c2 + ce = ae + ce

⇒ c2 = ae

Therefore, a, c, e are in GP.

21. find the sum of the following series upto n terms:

(i) 5 + 5 + 555 + …

(ii) 0.6 + 0.66 + 0.666 + …

Sol. (i) Let S = 5 + 55 + 555 + … + n terms

= 5(1 + 11 + 111 + … + n terms)

$$=\frac{5}{9}(9+99+999+...+n terms)\\=\frac{5}{9}[(10-1)+(100-1)+(1000-1)+...\\n \text{terms}]\\=\frac{5}{9}[(10+100+1000+...+ n\space \text{terms})]\\=\frac{5}{9}[ (10+100+1000+...+\text{n terms})\\-(1+1+1+...+\text{n terms})]\\=\frac{5}{9}[(10^1+10^2+10^3+...+10^n)-n]\\=\frac{5}{9}\bigg[\frac{10(10^n-1)}{10-1}-n\bigg]\\\bigg(\because\space\text{Sum of GP}=\frac{a(r^n-1)}{r-1}, r\gt1\bigg)$$

$$=\frac{5}{9}\bigg[\frac{10(10^n-1)}{9}-n\bigg]\\=\frac{50}{81}(10^n-1)-\frac{5}{9}n\\(\because\space\Sigma1=n)$$

(ii) Let S = 0.6 + 0.66 + 0.666 + … + n terms

S = 6(0.1 + 0.11 + 0.111 + … + n terms)

$$=\frac{6}{9}(0.9+0.99+0.999+...+\text{n terms})\\=\frac{2}{3}\bigg[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+...+\text{n}\space \text{terms}\bigg]\\=\frac{2}{3}\begin{bmatrix}\bigg(1-\frac{1}{10}\bigg)+\bigg(1-\frac{1}{100}\bigg)+\bigg(1-\frac{1}{1000}\bigg)\\+...+\text{n terms}\end{bmatrix}\\=\frac{2}{3}\bigg[(1+1+1+...+n\space\text{terms})-\\\bigg(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...+ \text{n terms}\bigg)\bigg]\\=\frac{2}{3}\bigg[(n)-\bigg(\frac{1}{10^1}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^n}\bigg)\bigg]$$

$$=\frac{2}{3}\bigg[n-\frac{\frac{1}{10}\lbrace 1-\frac{1}{10}\rbrace^{n}}{1-\frac{1}{10}}\bigg]\\\bigg(\because\text{Sum of GP}=\frac{a(1-r^n)}{1-r}, r\lt1\bigg)\\=\frac{2}{3}\bigg[n-\frac{\frac{1}{10}\lbrace1-\bigg(\frac{1}{10}\bigg)^n\rbrace}{\frac{9}{10}}\bigg]\\=\frac{2}{3}\bigg[n-\frac{1}{9}\begin{Bmatrix}1-\bigg(\frac{1}{10}\bigg)^n\end{Bmatrix}\bigg]\\=\frac{2}{3}n-\frac{2}{27}[1-(0.1)^n].$$

22. Find the 20th term of the series

2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

Sol. Given series is 2 × 4 + 4 × 6 + 6 × 8 + … + n terms

Here, we will split the series into two parts which are 2, 4, 6, … and 4, 6, 8, … and then find nth term of each part separately to find the nth term of the given series i.e.,

Tn = (nth term of 2, 4, 6, …)
× (nth term of 4, 6, 8, …)

= [2 + (n – 1)2] [4 + (n – 1)2]

[∵ Tn = a + (n – 1)d]

= (2 + 2n – 2) (4 + 2n – 2) = 2n(2n + 2)

Substituting n = 20,

T20 = 2 × 20(2 × 20 + 2) = 40(40 + 2)

= 40 × 42 = 1680

Hence, the 20th term of the series is 1680.

23. Find the sum upto n terms of the series

3 + 7 + 13 + 21 + 31 + …

Sol. Given series is

S = 3 + 7 + 13 + 21 + … + Tn

S = 3 + 7 + 13 + … + Tn

On subtracting both sides we get

0 = (3 + 4 + 6 + 8 + … + n terms) – Tn

⇒ Tn = 3 + [4 + 6 + 8 + … (n – 1) terms]

$$=3+\frac{\text{n-1}}{2}[2×4+(n-1-1)2]\\\bigg(\because\space\text{S}_n=\frac{n}{2}[2a+(n-1)d]\bigg)\\\text{[Here, number of terms = n – 1]}\\=3+\frac{n-1}{2}[8+(n-2)2]\\=3+\frac{n-1}{2}[2n+4]\\=3+\frac{(n-1)}{2}×2(n+2)$$

= 3 + (n – 1) (n + 2)

= 3 + n2 + 2n – n – 2

⇒ Tn = n2 + n + 1

Now,

S = ΣTn = Σ(n2 + n + 1) = Σn2 + Σn + Σ1

$$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n\\=n\bigg[\frac{2n^2+n+2n+1}{6}+\frac{n+1}{2}+\frac{1}{1}\bigg]\\=n\bigg[\frac{2n^2+3n+1+3n+3+6}{6}\bigg]\\\begin{Bmatrix}\because\space\Sigma n^2=\frac{n(n+1)(2n+1)}{6}\\\Sigma n=\frac{n(n+1)}{2},\Sigma1=n\end{Bmatrix}\\=n\bigg[\frac{2n^2+6n+10}{6}\bigg]\\=\frac{2n(n^2+3n+5)}{6}\\=\frac{n(n^2+3n+5)}{3}$$

24. If S1, S2, S3 are the sum of first n natural numbers, their square and their cubes, respectively, show that 9S22 = S3(1 + 8S1).

Sol. Given, S1 = Sum of first n natural numbers = Σn

$$\Rarr\space\text{S}_1=\frac{n(n+1)}{2}\space\text{...(i)}$$

and S2 = Sum of square of first n natural numbers = Σn2

$$\Rarr\space\text{S}_2=\frac{n(n+1)(2n+1)}{6}\space\text{...(ii)}$$

and S3 = Sum of cubes of first n natural numbers = Σn3

$$\Rarr\space\text{S}_3=\bigg[\frac{n(n+1)}{2}\bigg]^2\space\text{...(iii)}\\\text{Taking R.H.S. = S}_3 (\text{1 + 8S}_1)\\=\bigg[\frac{n(n+1)}{2}\bigg]^{2}\bigg[1+8×\frac{n(n+1)}{2}\bigg]\\\text{[Using equation (i) and (iii)]}\\=\bigg[\frac{n(n+1)}{2}\bigg]^2\space[1+4n(n+1)]\\=\bigg[\frac{n(n+1)}{2}\bigg]^2[4n^2+4n+1]\\=\bigg[\frac{n(n+1)}{2}\bigg]^2(2n+1)^2$$

Multiplying and dividing by 9,

$$= 9×\frac{n^2(n+1)^2}{4}×\frac{(2n+1)^2}{9}\\=9\bigg[\frac{n(n+1)(2n+1)}{6}\bigg]$$

= 9×S22=9S22

[Using equation (ii)]

= L.H.S.

∴ L.H.S. = R.H.S. Hence Proved.

25. Find the sum of the following series upto n terms

$$\frac{\textbf{1}^\textbf{3}}{\textbf{1}}\textbf{+}\frac{\textbf{1}^\textbf{3}\textbf{+}\textbf{2}^\textbf{3}}{\textbf{1+3}}\textbf{+}\frac{\textbf{1}^\textbf{3}\textbf{+}\textbf{2}^\textbf{3}\textbf{+}\textbf{3}^\textbf{3}}{\textbf{1+3+5}}\textbf{+...}$$

$$\textbf{Sol.}\space\text{T}_n=\frac{\text{nth term of numerator}}{\text{nth term of denominator}}\\=\frac{1^3+2^3+3^3+...+n^3}{1+3+5+...+ n\space \text{terms}}\\=\frac{\Sigma n^3}{\frac{n}{2}[2×1+(n-1)2]}\\\begin{bmatrix}\because\space\Sigma_n=\frac{n}{2}[2a+(n-1)d]\\\Sigma n^3=\bigg[\frac{n(n+1)}{2}\bigg]^2\end{bmatrix}\\=\frac{\Sigma n^3}{\frac{n}{2}[2+2n-2]}=\frac{\Sigma n^3}{\frac{n}{2}×2n}$$

$$\Rarr\space\text{T}_n=\frac{\Sigma n^3}{n^2}\\=\frac{\bigg[\frac{n(n+1)}{2}\bigg]^2}{n^2}=\frac{n^2(n+1)^2}{4n^2}\\\Rarr\space\text{T}_n=\frac{1}{4}(n+1)^2\\\Rarr\space\text{T}_n=\frac{1}{4}(n^2+1+2n)\\\Rarr\space S=\Sigma T_n=\frac{1}{4}\Sigma(n^2+1+2n)\\=\frac{1}{4}(\Sigma n^2+\Sigma1+2\Sigma n)\\=\frac{1}{4}\bigg[\frac{n(n+1)(2n+1)}{6}+n+\frac{2×n(n+1)}{2}\bigg]$$

$$\begin{Bmatrix}\because\space\Sigma n^2=\frac{n(n+1)(2n+1)}{6}\\\Sigma n=\frac{n(n+1)}{2},\Sigma1=n\end{Bmatrix}\\=\frac{n}{4}\bigg[\frac{2n^2+n+2n+1}{6}+1+n+1\bigg]\\=\frac{n}{4}\bigg[\frac{2n^2+3n+1}{6}+\frac{n+2}{1}\bigg]\\=\frac{n}{4}\bigg[\frac{2n^2+3n+1+6n+12}{6}\bigg]\\=\frac{n}{24}[2n^2+9n+13]$$

26. Show that:

$$\frac{\textbf{1×2}^\textbf{2}\textbf{+2×3}^\textbf{2}\textbf{+...+n×}\textbf{(n+1)}^\textbf{2}}{\textbf{1}^\textbf{2}\textbf{×2+2}^\textbf{2}\textbf{×3+...+n}^\textbf{2}×\textbf{(n+1)}}\\=\frac{\textbf{3n+5}}{\textbf{3n+1}}$$

Sol. For numerator, 1 × 22 + 2 × 32 + … + n × (n + 1)2

Tn = n(n + 1)2

= n(n2 + 1 + 2n)

= n3 + 2n2 + n

⇒ Sn = ΣTn = Σ(n3 + 2n2 + n)

= Σn3 + 2Σn2 + Σn

$$\begin{bmatrix}\because\space\Sigma n^2=\bigg[\frac{n(n+1)}{2}\bigg]^2,\\\Sigma n^2=\frac{n(n+1)(2n+1)}{6},\\\Sigma n=\frac{n(n+1)}{2}\end{bmatrix}\\=\bigg[\frac{n(n+1)}{2}\bigg]^2+\frac{2n(n+1)(2n+1)}{6}\\+\frac{n(n+1)}{2}$$

$$=\frac{n(n+1)}{2}\bigg[\frac{n(n+1)}{2}+\frac{2(2n+1)}{3}+\frac{1}{1}\bigg]\\=\frac{n(n+1)}{2}\bigg[\frac{3n(n+1)+4(2n+1)+6}{6}\bigg]\\=\frac{n(n+1)}{2}\bigg[\frac{3n^2+3n+8n+4+6}{6}\bigg]\\=\frac{n(n+1)(3n^2+11n+10)}{12}\\=\frac{n(n+1)(3n^2+6n+5n+10)}{12}\\=\frac{n(n+1)[3n(n+2)+5(n+2)]}{12}\\=\frac{n(n+1)(n+2)(3n+5)}{12}\space\text{...(i)}$$

Again, for denominator,

12 × 2 + 22 × 3 + … + n2 × (n + 1)

Let

Tn′ = n2(n + 1)

= n3 + n2

Now,

Sn′ = ΣTn′ = Σ(n3 + n2) = Σn3 + Σn2

$$\begin{Bmatrix}\because\space \Sigma n^3+\bigg[\frac{n(n+1)}{2}\bigg]^2,\\\Sigma n^2=\frac{n(n+1)(2n+1)}{6}\end{Bmatrix}\\=\bigg[\frac{n(n+1)}{2}\bigg]^2+\frac{n(n+1)(2n+1)}{6}\\=\frac{n(n+1)}{2}\bigg[\frac{n(n+1)}{2}+\frac{2n+1}{3}\bigg]\\=\frac{n(n+1)(3n^2+3n+4n+2)}{12}\\=\frac{n(n+1)(3n^2+7n+2)}{12}\\=\frac{n(n+1)(3n^2+6n+n+2)}{12}$$

$$=\frac{n(n+1)[3n(n+2)+1(n+2)]}{12}\\\text{T}_n'=\frac{n(n+1)(3n+1)(n+2)}{12}\space\text{...(ii)}$$

Hence, required sum of the series become

$$\frac{\text{T}_n}{\text{T}_n'}=\frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(3n+1)(n+2)}{12}}\\\text{[from equations (i) and (ii)]}\\=\frac{3n+5}{3n+1}\space\textbf{Hence Proved.}$$

27. A farmer buys a used tractor of ₹12000. He pays ₹6000 cash and agrees to pay the balance in annual instalment of ₹500 plus 12% interest on the upaid amount. How much will the tractor cost him?

Sol. Tractor cost = ₹12000

Down payment = ₹6000

Balance amount = ₹6000

$$\text{Interest on Ist instalment }\\=\frac{6000×12×1}{100}\\\bigg(\because\space\text{I}=\frac{\text{P×R×T}}{100}\bigg)$$

= ₹720

Now, unpaid amount = 6000 – 500 = ₹5500

$$\text{Interest in IInd instalment}\\=\frac{5500×12×1}{100}=₹660$$

Again, unpaid amount = 5500 – 500

= ₹5000

$$\text{Interest on IIIrd instalment} \\=\frac{5000×12×1}{100}=₹600$$

Total interest paid by him = 720 + 660 + 600 + … + 12 terms

which is an AP with a = 720, d = 660 – 720 = – 60

Therefore, total interest

$$=\frac{12}{2}[2×720+(12-1)(-60)]$$

= 6[1440 – 11 × 60] = 6[1440 – 660]

= 6 × 780 = ₹4680

Hence, total amount or actual cost = 12000 + 4680 = ₹16680.

28. Shamshad Ali buys a scooter for ₹22000. He pays ₹4000 cash and agrees to pay the balance in annual instalments of ₹1000 plus 10% interest on the unpaid amount. How much will scooter cost him?

Sol. Scooter cost = ₹22000

Down payment = ₹4000

Balance payment = ₹18000

Now, interest on Ist instalment

$$=\frac{18000×10×1}{100}=₹1800\\\bigg(\because \text{I}=\frac{\text{P×R×T}}{100}\bigg)$$

Unpaid amount = 18000 – 1000 = 17000

Interest on IInd instalment

$$=\frac{17000×10×1}{100}=₹1700$$

Unpaid amount = 17000 – 1000 = 16000

Interest on IInd instalment

$$=\frac{16000×10×1}{100}=₹1600$$

..................................................................

....................................................................

∴ Total interest paid by him = 1800 + 1700 + 1600 + … + 18 terms

which is an AP with a = 1800, d = 1700 –1800 =
– 100

Therefore, total interest = [2 × 1800 + (18 –1)
(– 100)]

(∵  Sum of AP = Sn = [2a + (n – 1)d])

= 9(3600 – 1700) = 9 × 1900 = 17100

Hence, total amount or actual cost = 22000 + 17100 = ₹39100.

29. A person writes a letter to four of his friends. He asks each one of them, to copy the letter and mail to four different person with instructions that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on postage when 8th set of letter is mailed.

Sol. Firstly, one person post four letters to four of his friends. Then those four persons post 4-4 letters i.e., a total 4 × 4 = 16 letters. Similarly, in next step 4 × 4 × 4 = 64 letters are posted.

Thus, there will be a GP of series

i.e., 4, 16, 64, 256, …

where, a = 4 and r = 4

Now, total number of letters till 8th set

$$\text{S}_n=\frac{a(r^n-1)}{r-1}\\\Rarr\space\text{S}_8=\frac{4(4^8-1)}{4-1}\\\Rarr\space\text{S}_8=\frac{4}{3}(4^8-1)\\\bigg[\because \text{S}_n=\frac{a(r^n-1)}{(r-1)},r\gt1\bigg]$$

Cost for one letter = ₹0.50

$$\text{Hence, total cost} =\frac{4}{3}(4^8-1)×0.50\\=\frac{4}{3}×(65536-1)×0.50\\=\frac{4}{3}×65535×0.50=₹43690.$$

Hence, the amount spent on postage when 8th set of letter is mailed is ₹43690.

30. A man deposited ₹10000 in a bank at the rate of 5% simple interest to annually. Find the amount is 15th years, since he deposited the amount and also calculate the amount after 20 years.

Sol. Interst on ₹10000 after 1 year

$$=\frac{10000×5×1}{100}\space\\\bigg(\because\space \text{I}=\frac{\text{P×R×T}}{100}\bigg)$$

= ₹500

Therefore, amount after 1 years = 10000 + 500 = 10500

Now, interest on ₹10000 after 2 years

$$=\frac{10000×5×2}{100}=₹1000$$

∴ Amount after 2 years = 10000 + 1000 = ₹11000

Similarly, interst on ₹10000 after 3 years

$$=\frac{10000×5×3}{100}=1500$$

∴ Amount after 3 years = 10000 + 1500 = ₹11500

Hence, the amount in the amount of man in first, second and third year are ₹10000, 10500, 11000, …

It is an AP, where a = 10000 and d = 500
Amount in 15th years

= T15

= a + 14d

= 10000 + 14 × 500

= 10000 + 7000 = ₹17000

Amount after 20 years

= 10000 + 20 × 500

= 10000 + 10000 = ₹20000.

31. A manufacturer reckons that the value of machine, which costs him ₹15625 will depreciate each year by 20%. Find the estimated value at the end of 5 years.

$$\bigg[\because\space\textbf{A}=\textbf{P}\bigg(\textbf{1}-\frac{\textbf{r}}{\textbf{100}}\bigg)^\textbf{n}\bigg]$$

Sol. Here, series is

$$15625\bigg(1-\frac{20}{100}\bigg),\\15625\bigg(1-\frac{20}{100}\bigg)^2,\\15625\bigg(1-\frac{20}{100}\bigg)^3,...$$

Depreciated values at the end of 5 years = T5

$$=15625\bigg(1-\frac{20}{100}\bigg)^5\\=15625×\bigg(\frac{4}{5}\bigg)^5\\=15625×\bigg(\frac{5-1}{5}\bigg)^5\\=15625×\bigg(\frac{4}{5}\bigg)^5\\=\frac{15625×1024}{625×5}\\=\frac{25×1024}{5}$$

= 5 × 1024 = 5120.

32. 150 workers were engaged to finish a job in a certain number of days, 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Sol. Let the number of days in which the work is completed, is n.

Now, according to question, 4 workers dropped on everyday i.e., the devils of number of workers is 150, 146, 142, 138…

Clearly, work done in both conditions is same.

[Had the workers not dropped then the work would have finished in (n – 8) days with 150 workers working on each day. Hence, the total number of workers who would hae worked for all the n days is 150 (n – 8).

Putting a = 150, d = – 4, Sn = 150(n – 8)

$$\Rarr\space 150(n-8)\\=\frac{n}{2}[2×150+(n-1)(-4)]\\\bigg(\text{Using S}_n=\frac{n}{2}[2a+(n-1)d]\bigg)\\\Rarr\space 150n-1200=\frac{n}{2}×2(150-2n+2)\\\Rarr\space 150n-1200=\frac{n}{2}×2(152-2n)$$

Dividing each term by 2,

75n – 600 = n(76 – n)

⇒ 75n – 600 = 76n – n2

n2 – 76n + 75n – 600 = 0

⇒ n2 – n – 600 = 0

Now factorizing it by splitting the middle term,

⇒ n2 – (25n – 24n) – 600 = 0

⇒ n2 – 26n + 24n – 600 = 0

⇒ n(n – 25) + 24(n – 25) = 0

⇒ (n – 25) (n + 24) = 0

⇒ n = 25

and n = – 24 because it is not possible.

Hence, work will be completed in 25 days.