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Punjab Board Class 11 Mathematics

Exercise 6.3 Solutions — Sequences and Series

Complete worked solutions for Exercise 6.3 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 6.3 Solutions

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

Question 1: Find A.M. between the given numbers:

Do numbers ka A.M. unka average hai: \(\frac{\text{sum}}{2}\).

(i) \(2 + \sqrt{3}\,i\), \(2 - \sqrt{3}\,i\)

\[ \text{A.M.} = \frac{(2+\sqrt{3}\,i) + (2-\sqrt{3}\,i)}{2} = \frac{4}{2} = 2 \]

(ii) \((a+b)^2\), \((a-b)^2\)

\[ \text{A.M.} = \frac{(a+b)^2 + (a-b)^2}{2} = \frac{2a^2 + 2b^2}{2} = a^2 + b^2 \]
Answer: (i) \(2\), (ii) \(a^2 + b^2\)

Question 2: If 6, 11, 16 are three A.Ms. between \(a\) and \(b\), find \(a\) and \(b\).

Yeh A.P. hai: \(a, 6, 11, 16, b\) — \(d = 5\), isliye \(a = 6 - 5\) aur \(b = 16 + 5\).

\[ d = 11 - 6 = 5 \implies a = 6 - 5 = 1, \quad b = 16 + 5 = 21 \]
Answer: \(a = 1\), \(b = 21\)

Question 3: Insert five A.Ms. between \(\sqrt{2}\) and \(\frac{15}{\sqrt{2}}\).

5 means hon to \(d = \frac{b-a}{6}\) — phir \(a+d, a+2d, \dots, a+5d\) nikalo.

\[ d = \frac{\frac{15}{\sqrt{2}} - \sqrt{2}}{6} = \frac{\frac{13}{\sqrt{2}}}{6} = \frac{13}{6\sqrt{2}} = \frac{13\sqrt{2}}{12} \]
\[ \sqrt{2} + \frac{13\sqrt{2}}{12} = \frac{25\sqrt{2}}{12}, \quad \frac{19\sqrt{2}}{6}, \quad \frac{17\sqrt{2}}{4}, \quad \frac{16\sqrt{2}}{3}, \quad \frac{77\sqrt{2}}{12} \]
Answer: \(\frac{25\sqrt{2}}{12}, \frac{19\sqrt{2}}{6}, \frac{17\sqrt{2}}{4}, \frac{16\sqrt{2}}{3}, \frac{77\sqrt{2}}{12}\)

Question 4: The A.M. of two numbers is 7 and their product is 45. Find the numbers.

Sum \(= 14\), product \(= 45\) — yeh quadratic \(t^2 - 14t + 45 = 0\) ke roots honge.

\[ t^2 - 14t + 45 = 0 \implies (t-5)(t-9) = 0 \implies t = 5, 9 \]
Answer: \(5\) and \(9\)

Question 5: If \(n\) arithmetic means are inserted between \(a\) and \(b\), prove that \(d = \frac{b-a}{n+1}\), where \(d\) is the common difference.

Total terms \(n+2\) honge: \(a, a+d, \dots, a+(n+1)d = b\) — aakhri equation se \(d\) nikalo.

\[ b = a + (n+1)d \implies d = \frac{b-a}{n+1} \]
Answer: Proved.

Question 6: If \(A\) is the A.M. between \(a\) and \(b\), prove that \((a-A)^2 + (A-b)^2 = \frac{1}{2}(a-b)^2\).

\(A = \frac{a+b}{2}\) rakho aur dono taraf simplify karo.

\[ \text{LHS} = \left(a - \frac{a+b}{2}\right)^2 + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a-b}{2}\right)^2 + \left(\frac{a-b}{2}\right)^2 \]
\[ = 2 \cdot \frac{(a-b)^2}{4} = \frac{1}{2}(a-b)^2 = \text{RHS} \]
Answer: Proved.

Question 7: For what value of \(n\), \(\frac{a^{n+1} + b^{n+1}}{a^n + b^n}\) is the A.M. between \(a\) and \(b\), where \(a \ne b\)?

Expression ko \(\frac{a+b}{2}\) ke barabar rakho, cross-multiply karke factorize karo.

\[ \frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \frac{a+b}{2} \implies 2(a^{n+1}+b^{n+1}) = (a+b)(a^n+b^n) \]
\[ a^{n+1} + b^{n+1} - a^nb - ab^n = 0 \implies (a-b)(a^n - b^n) = 0 \]
\[ a \ne b \implies a^n = b^n \implies n = 0 \]
Answer: \(n = 0\)