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

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

Substituting n = 7

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

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

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

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

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

Given, T₄ = (T₂)² (∵ T_n = ar^{n – 1})

⇒ ar³ = (ar)²

⇒ ar³ = a²r²

⇒ r = a

But given, a = – 3

⇒ r = – 3

Therefore T₇ = ar⁶ = (– 3) (– 3)⁶

= (– 3)⁷ = – 2187.

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, …, ar^{2n – 2}, ar^{2n – 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 + … ar^{2n – 1}

= 5 × (a + ar² + ar⁴ + … + ar^{2n – 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 T_n, then sum of last four terms
= 112

T_n + T_{n – 1} + T_{n – 2} + T_{n – 3} = 112

[∵ T_n = 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.