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Punjab Board Class 11 Mathematics

Exercise 6.7 Solutions — Sequences and Series

Complete worked solutions for Exercise 6.7 from Chapter 6 (Sequences and Series), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 6.7 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 6: Sequences and Series.

Question 1: Find the sum of first 15 terms of the G.P., \(1, \frac{1}{3}, \frac{1}{9}, \dots\)

\(a_1 = 1, r = \frac{1}{3}\) — formula \(S_n = \frac{a_1(1-r^n)}{1-r}\) lagao.

\[ S_{15} = \frac{1-\left(\frac{1}{3}\right)^{15}}{1-\frac{1}{3}} = \frac{3}{2}\left[1-\left(\frac{1}{3}\right)^{15}\right] \]
Answer: \(\frac{3}{2}\left[1-\left(\frac{1}{3}\right)^{15}\right]\)

Question 2: The \(3^{rd}\) term of a G.P. is 16 and the \(6^{th}\) term is −128. Find the first term and the sum of the first 10 terms.

Dono equations divide karke \(r^3\) nikalo, phir \(a_1\), phir \(S_{10}\).

\[ \frac{a_1 r^5}{a_1 r^2} = \frac{-128}{16} \implies r^3 = -8 \implies r = -2 \]
\[ a_1 = \frac{16}{4} = 4, \qquad S_{10} = \frac{4\big[1-(-2)^{10}\big]}{1-(-2)} = \frac{4(1-1024)}{3} = -1364 \]
Answer: \(a_1 = 4\), \(S_{10} = -1364\)

Question 3: Sum to \(n\) terms the series:

Har term ko 9 se related likh kar geometric series mein badlo.

(i) \(0.2 - 0.22 + 0.222 - \dots\)

\[ S_n = 2\big[0.1 - 0.11 + 0.111 - \cdots\big] = \frac{2}{9}\big[(1-10^{-1}) - (1-10^{-2}) + \cdots\big] \]
\[ S_n = \frac{2}{9}\left[\frac{1-(-1)^n}{2} - \frac{1}{11}\left(1-\left(-\frac{1}{10}\right)^n\right)\right] \]

(ii) \(3 + 33 + 333 + \dots\)

\[ S_n = 3\left[\frac{9}{9} + \frac{99}{9} + \cdots\right] = \frac{1}{3}\big[(10-1) + (10^2-1) + \cdots + (10^n-1)\big] \]
\[ S_n = \frac{1}{3}\left[\frac{10(10^n-1)}{9} - n\right] = \frac{10^{n+1} - 9n - 10}{27} \]
Answer: (i) \(\frac{2}{9}\left[\frac{1-(-1)^n}{2} - \frac{1}{11}\left(1-\left(-\frac{1}{10}\right)^n\right)\right]\), (ii) \(\frac{10^{n+1}-9n-10}{27}\)

Question 4: Sum to \(n\) terms the series:

Terms ko \((a-b)\) se multiply-divide karke do alag G.P.s banao; doosre part mein \(1+k+\cdots+k^{j-1} = \frac{1-k^j}{1-k}\) likh kar do geometric sums mein toro.

(i) \(1 + (a+b) + (a^2+ab+b^2) + \dots\)

\[ S_n = \frac{1}{a-b}\big[(a-b) + (a^2-b^2) + \cdots + (a^n-b^n)\big] \]
\[ S_n = \frac{1}{a-b}\left[\frac{a(1-a^n)}{1-a} - \frac{b(1-b^n)}{1-b}\right] \]

(ii) \(r + (1+k)r^2 + (1+k+k^2)r^3 + \dots\)

\[ T_j = (1 + k + \cdots + k^{j-1})r^j = \frac{1-k^j}{1-k}\,r^j \]
\[ S_n = \frac{1}{1-k}\left[\sum_{j=1}^{n} r^j - \sum_{j=1}^{n}(kr)^j\right] = \frac{1}{1-k}\left[\frac{r(1-r^n)}{1-r} - \frac{kr\big(1-(kr)^n\big)}{1-kr}\right] \]
Answer: (i) \(\frac{1}{a-b}\left[\frac{a(1-a^n)}{1-a} - \frac{b(1-b^n)}{1-b}\right]\), (ii) \(\frac{1}{1-k}\left[\frac{r(1-r^n)}{1-r} - \frac{kr(1-(kr)^n)}{1-kr}\right]\)

Question 5: Sum the series \(2 + (1-i) + \dots\) to 8 terms, where the common ratio is \(r = \frac{1-i}{2}\).

\(S_8 = \frac{2(1-r^8)}{1-r}\) — pehle \(r^8\) nikalo: \(r^2 = -\frac{i}{2}\), \(r^4 = -\frac{1}{4}\), \(r^8 = \frac{1}{16}\).

\[ S_8 = \frac{2\left(1-\frac{1}{16}\right)}{1-\frac{1-i}{2}} = \frac{\frac{15}{8}}{\frac{1+i}{2}} = \frac{15}{4(1+i)} = \frac{15(1-i)}{8} \]
Answer: \(\frac{15(1-i)}{8}\)

Question 6: Show that the ratio of the sum of first \(n\) terms of a G.P. to the sum of terms from \((n+1)^{th}\) to \((2n)^{th}\) term is \(\frac{1}{r^n}\), where \(r\) is the common ratio of the G.P.

Doosra sum \(= S_{2n} - S_n\) — dono ka ratio nikalo, \((1-r^n)\) cancel ho jayega.

\[ \frac{S_n}{S_{2n}-S_n} = \frac{\frac{a(1-r^n)}{1-r}}{\frac{ar^n(1-r^n)}{1-r}} = \frac{1}{r^n} \]
Answer: Proved.