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Punjab Board Class 11 Mathematics

Exercise 6.6 Solutions — Sequences and Series

Complete worked solutions for Exercise 6.6 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 6.6 Solutions

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

Question 1: Find G.M. between:

Do numbers ka geometric mean \(\pm\sqrt{ab}\) hota hai — sign ka khayal rakho.

(i) \(-2\) and \(8\)

\[ G = \pm\sqrt{(-2)(8)} = \pm\sqrt{-16} = \pm 4i \]

(ii) \(-2i\) and \(8i\)

\[ G = \pm\sqrt{(-2i)(8i)} = \pm\sqrt{-16i^2} = \pm\sqrt{16} = \pm 4 \]

(iii) \(6\) and \(9\)

\[ G = \pm\sqrt{54} = \pm 3\sqrt{6} \]
Answer: (i) \(\pm 4i\), (ii) \(\pm 4\), (iii) \(\pm 3\sqrt{6}\)

Question 2: Insert four real geometric means between 3 and 96.

4 means hon to \(r = \left(\frac{96}{3}\right)^{1/5} = 2\) — phir terms nikalo.

\[ r = (32)^{1/5} = 2 \implies 6, \; 12, \; 24, \; 48 \]
Answer: \(6, 12, 24, 48\)

Question 3: If both \(x\) and \(y\) are positive distinct real numbers, show that the geometric mean between \(x\) and \(y\) is less than their arithmetic mean.

\((\sqrt{x} - \sqrt{y})^2 > 0\) expand karo — yahi A.M. \(\ge\) G.M. ki proof hai.

\[ (\sqrt{x}-\sqrt{y})^2 > 0 \implies x + y - 2\sqrt{xy} > 0 \implies \frac{x+y}{2} > \sqrt{xy} \]
Answer: Proved.

Question 4: For what value of \(n\), is \(\frac{a^n + b^n}{a^{n-1} + b^{n-1}}\) the positive geometric mean between \(a\) and \(b\)?

Expression ko \(\sqrt{ab}\) ke barabar rakho, cross-multiply karke \((a-b)\) factor nikalo.

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

Question 5: The A.M. of two positive integral numbers exceeds their (positive) G.M. by 2 and their sum is 20, find the numbers.

Sum \(= 20\) se A.M. \(= 10\), to G.M. \(= 8\) — phir product \(= 64\) aur quadratic banao.

\[ A = 10, \; G = 8 \implies ab = 64, \; a + b = 20 \]
\[ t^2 - 20t + 64 = 0 \implies (t-4)(t-16) = 0 \]
Answer: \(4\) and \(16\)

Question 6: The A.M. between two numbers is 5 and their (positive) G.M. is 4. Find the numbers.

Sum \(= 10\), product \(= 16\) — quadratic ke roots hi numbers honge.

\[ t^2 - 10t + 16 = 0 \implies (t-2)(t-8) = 0 \]
Answer: \(2\) and \(8\)

Question 7: The arithmetic mean between two positive numbers \(a\) and \(b\) is double their geometric mean. Prove that \(a:b = 2+\sqrt{3} : 2-\sqrt{3}\).

\(\frac{a+b}{2} = 2\sqrt{ab}\) — dono taraf square karke \((a-b)^2\) nikalo, phir ratio.

\[ a + b = 4\sqrt{ab} \implies (a+b)^2 = 16ab \implies (a-b)^2 = 12ab \]
\[ \frac{a-b}{2\sqrt{ab}} = \sqrt{3} \implies \frac{a+b+ (a-b)}{a+b-(a-b)} = \frac{4\sqrt{ab}+2\sqrt{3ab}}{4\sqrt{ab}-2\sqrt{3ab}} \]
\[ \frac{a}{b} = \frac{2+\sqrt{3}}{2-\sqrt{3}} \implies a:b = 2+\sqrt{3} : 2-\sqrt{3} \]
Answer: Proved.

Question 8: If one geometric mean \(G\) and two arithmetic means \(p\) and \(q\) are inserted between two positive numbers, show that \(G^2 = (2p-q)(2q-p)\).

Numbers \(a, b\) lo: \(G^2 = ab\); \(p, q\) A.P. mein hain to \(p+q = a+b\) aur \(q - p = \frac{b-a}{3}\).

\[ p = \frac{2a+b}{3}, \qquad q = \frac{a+2b}{3} \]
\[ 2p - q = a, \qquad 2q - p = b \implies (2p-q)(2q-p) = ab = G^2 \]
Answer: Proved.