Mathematics 11 (Punjab Board)
Exercise 6.8 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 6: Sequences and Series.
Question 1: The \(n^{th}\) terms of an A.P. \(1, 4, 7, \dots\) and a G.P. \(5, 10, 20, \dots\) are multiplied to form an A.G.P. Find its 8th term.
A.P. ka \(n\)th term \((3n-2)\), G.P. ka \(5 \cdot 2^{n-1}\) — dono multiply karke \(n = 8\) rakho.
\[
T_n = (3n-2) \cdot 5 \cdot 2^{n-1}, \qquad T_8 = 22 \cdot 5 \cdot 128 = 14080
\]
Answer: \(14080\)
Question 2: The \(n^{th}\) terms of an A.P. \(3, 7, 11, \dots\) and a G.P. \(2, 6, 18, \dots\) form an A.G.P. Find its \(n^{th}\) term.
A.P. ka term \((4n-1)\), G.P. ka \(2 \cdot 3^{n-1}\) — multiply karo.
\[
T_n = (4n-1) \cdot 2 \cdot 3^{n-1} = 2(4n-1)3^{n-1}
\]
Answer: \(T_n = 2(4n-1)3^{n-1}\)
Question 3: If \(a_{n+1} = 2n+5\) is the \((n+1)^{th}\) term of an A.P. and \(b_{n+1} = \frac{1}{9}(-3)^n\) is the \((n+1)^{th}\) term of a G.P., find the \(n^{th}\) term of the A.G.P. formed from them, and the sum of its first three terms.
Index shift karo: \(a_n = 2n+3\), \(b_n = \frac{1}{9}(-3)^{n-1}\). Multiply karke \(T_n\), phir \(T_1 + T_2 + T_3\).
\[
T_n = \frac{2n+3}{9}(-3)^{n-1}
\]
\[
T_1 = \frac{5}{9}, \quad T_2 = -\frac{7}{3}, \quad T_3 = 9 \implies \text{Sum} = \frac{5}{9} - \frac{21}{9} + \frac{81}{9} = \frac{65}{9}
\]
Answer: \(T_n = \frac{2n+3}{9}(-3)^{n-1}\); sum of first three terms \(= \frac{65}{9}\)
Question 4: Sum to \(n\) terms:
Yeh A.G.P. hain — \(S_n\) likho, \(rS_n\) se ghatao, bachi G.P. ka sum nikalo.
(i) \(1 \cdot 2 + 3 \cdot 4 + 5 \cdot 8 + 7 \cdot 16 + \dots\)
\[
S_n - 2S_n = 2 + 2(4 + 8 + \cdots + 2^n) - (2n-1)2^{n+1}
\]
\[
S_n = 6 + (2n-3)2^{n+1}
\]
(ii) \(2 \cdot 3 + 4 \cdot 3^2 + 6 \cdot 3^3 + 8 \cdot 3^4 + \dots\)
\[
S_n - 3S_n = 2(3 + 3^2 + \cdots + 3^n) - 2n \cdot 3^{n+1}
\]
\[
S_n = \frac{(2n-1)3^{n+1} + 3}{2}
\]
(iii) \(2 + \frac{5}{4} + \frac{8}{4^2} + \frac{11}{4^3} + \dots\)
\[
\left(1-\frac{1}{4}\right)S_n = 2 + 3\left(\frac{1}{4} + \cdots + \frac{1}{4^{n-1}}\right) - \frac{3n-1}{4^n}
\]
\[
S_n = 4 - \frac{n+1}{4^{n-1}}
\]
(iv) \(1 + \frac{3}{5} + \frac{5}{5^2} + \frac{7}{5^3} + \dots\)
\[
\left(1-\frac{1}{5}\right)S_n = 1 + 2\left(\frac{1}{5} + \cdots + \frac{1}{5^{n-1}}\right) - \frac{2n-1}{5^n}
\]
\[
S_n = \frac{15}{8} - \frac{4n+3}{8 \cdot 5^{n-1}}
\]
Answer: (i) \(6 + (2n-3)2^{n+1}\), (ii) \(\frac{(2n-1)3^{n+1}+3}{2}\), (iii) \(4 - \frac{n+1}{4^{n-1}}\), (iv) \(\frac{15}{8} - \frac{4n+3}{8 \cdot 5^{n-1}}\)
Question 5: Sum to infinity:
\(S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}\) — pehle \(a, d, r\) pehchano.
(i) \(1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \dots\)
\[
S_\infty = \frac{1}{1-\frac{1}{2}} + \frac{2 \cdot \frac{1}{2}}{(1-\frac{1}{2})^2} = 2 + 4 = 6
\]
(ii) \(2 + \frac{5}{3} + \frac{8}{9} + \frac{11}{27} + \dots\)
\[
S_\infty = \frac{2}{1-\frac{1}{3}} + \frac{3 \cdot \frac{1}{3}}{(1-\frac{1}{3})^2} = 3 + \frac{9}{4} = \frac{21}{4}
\]
Answer: (i) \(6\), (ii) \(\frac{21}{4}\)
Question 6: Show that \(2^{1/2} \cdot 4^{1/4} \cdot 8^{1/8} \cdot 16^{1/16} \cdots \infty = 4\).
Sab ko base 2 mein likho — powers ka sum A.G.P. banega. (Note: some printed keys list 16 here, but the verified sum of the exponents is 2, so the product is \(2^2 = 4\).)
\[
= 2^{\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \cdots}, \qquad E = \sum_{k=1}^{\infty}\frac{k}{2^k}
\]
\[
E - \frac{E}{2} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1 \implies E = 2
\]
\[
\text{Product} = 2^2 = 4
\]
Answer: \(4\)
Question 7: Show that \(\sqrt{4} \cdot \sqrt[4]{16} \cdot \sqrt[8]{64} \cdot \sqrt[16]{256} \cdots \infty = 16\).
Har term ko \(2\) ki power mein likho — exponent \(\frac{2n}{2^n}\) ka sum nikalo.
\[
= 2^{1 + 1 + \frac{3}{4} + \frac{1}{2} + \cdots}, \qquad \sum_{n=1}^{\infty}\frac{2n}{2^n} = 2\sum_{n=1}^{\infty}\frac{n}{2^n} = 4
\]
\[
\text{Product} = 2^4 = 16
\]
Answer: Proved.
Question 8: Sum to \(n\) terms the series \(2 + 4x + 6x^2 + 8x^3 + \dots\)
A.G.P. hai (\(a = 2, d = 2, r = x\)). Standard formula lagao aur simplify karo.
\[
S_n = \frac{2}{1-x} + \frac{2x(1-x^{n-1})}{(1-x)^2} - \frac{2nx^n}{1-x} = \frac{2 - 2(n+1)x^n + 2nx^{n+1}}{(1-x)^2}
\]
Answer: \(S_n = \frac{2 - 2(n+1)x^n + 2nx^{n+1}}{(1-x)^2}\)
Question 9: Sum to \(n\) terms the series \(1 \cdot r + 3r^2 + 5r^3 + \dots + (2n-1)r^n\), where \(r = \frac{2n+1}{2n-1}\). Show that the sum is \(n(2n+1)\).
\((1-r)S_n\) nikalo — \(r^{n+1}\) wale terms \((2n-1)r = 2n+1\) ki wajah se cancel ho jayenge.
\[
(1-r)S_n = r + 2(r^2 + \cdots + r^n) - (2n-1)r^{n+1}
\]
\[
(1-r)^2 S_n = r + r^2 - (2n+1)r^{n+1} + (2n-1)r^{n+2} = r(1+r)
\]
\[
1 - r = -\frac{2}{2n-1} \implies S_n = \frac{r(1+r)}{(1-r)^2} = n(2n+1)
\]
Answer: Proved — the sum is \(n(2n+1)\).
Question 10: Sum to \(n\) terms: \(1 + 2\left(1+\frac{1}{n}\right) + 3\left(1+\frac{1}{n}\right)^2 + \dots\) Show that the sum is \(n^2\).
\(r = 1 + \frac{1}{n}\) rakho. \((1-r)^2 S_n\) nikalne par \(r^n\) wale terms cancel — sirf 1 bachta hai.
\[
(1-r)^2 S_n = 1 - r^n\big[1 + n(1-r)\big] = 1 - r^n(1 - 1) = 1
\]
\[
S_n = \frac{1}{(1-r)^2} = \frac{1}{(-1/n)^2} = n^2
\]
Answer: Proved — the sum is \(n^2\).
Question 11: Sum to \(n\) terms the series \(2 + 5x + 8x^2 + 11x^3 + \dots\). Also find the sum to infinity if \(|x| < 1\).
A.G.P. hai (\(a = 2, d = 3, r = x\)). Finite sum formula, phir \(n \to \infty\).
\[
S_n = \frac{2}{1-x} + \frac{3x(1-x^{n-1})}{(1-x)^2} - \frac{(3n-1)x^n}{1-x}
\]
\[
S_\infty = \frac{2}{1-x} + \frac{3x}{(1-x)^2} = \frac{2+x}{(1-x)^2} \quad (|x| < 1)
\]
Answer: \(S_n\) as above; \(S_\infty = \frac{2+x}{(1-x)^2}\)
