Mathematics 11 (Punjab Board)
Exercise 6.9 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 6: Sequences and Series.
Question 1: Find the 9th term of each of the following H.P.s: (i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \dots\) (ii) \(-\frac{1}{5}, -\frac{1}{3}, -1, \dots\)
Reciprocals lo — woh A.P. banenge. A.P. ka 9th term nikaal kar reciprocal lo.
(i) Reciprocals: \(3, 5, 7, \dots\)
\[
a_9 = 3 + 8(2) = 19 \implies \text{H.P. 9th term} = \frac{1}{19}
\]
(ii) Reciprocals: \(-5, -3, -1, \dots\)
\[
a_9 = -5 + 8(2) = 11 \implies \text{H.P. 9th term} = \frac{1}{11}
\]
Answer: (i) \(\frac{1}{19}\), (ii) \(\frac{1}{11}\)
Question 2: Insert five harmonic means between \(-\frac{2}{5}\) and \(\frac{13}{2}\).
Reciprocals \(-\frac{5}{2}\) aur \(\frac{13}{2}\) ke darmiyan 5 A.M.s (\(d = \frac{3}{2}\)), phir unke reciprocals lo.
\[
\text{A.M.s: } -1, \; \frac{1}{2}, \; 2, \; \frac{7}{2}, \; 5 \implies \text{H.M.s: } -1, \; 2, \; \frac{1}{2}, \; \frac{2}{7}, \; \frac{1}{5}
\]
Answer: \(-1, 2, \frac{1}{2}, \frac{2}{7}, \frac{1}{5}\)
Question 3: The first and fifth terms of an H.P. are \(-\frac{1}{3}\) and \(\frac{1}{5}\). Find the 9th term.
Reciprocals ki A.P. mein \(b_1 = -3, b_5 = 5\) — \(d = 2\), phir \(b_9\) ka reciprocal.
\[
d = \frac{5-(-3)}{4} = 2, \quad b_9 = -3 + 8(2) = 13 \implies \frac{1}{13}
\]
Answer: \(\frac{1}{13}\)
Question 4: The harmonic mean between 2 and \(b\) is 5. Find \(b\).
Formula \(H = \frac{2ab}{a+b}\) mein values rakho aur solve karo.
\[
\frac{2(2)(b)}{2+b} = 5 \implies 4b = 10 + 5b \implies b = -10
\]
Answer: \(b = -10\)
Question 5: If \(\frac{1}{k}, \frac{1}{2k+1}, \frac{1}{4k-1}\) are in H.P., find \(k\).
Reciprocals \(k, 2k+1, 4k-1\) A.P. mein honge — darmiyani ka double sum ke barabar.
\[
2(2k+1) = k + (4k-1) \implies 4k + 2 = 5k - 1 \implies k = 3
\]
Answer: \(k = 3\)
Question 6: For what value of \(n\) is \(\frac{a^{n+1} + b^{n+1}}{a^n + b^n}\) the harmonic mean between \(a\) and \(b\)?
Expression ko \(\frac{2ab}{a+b}\) ke barabar rakho, simplify karke factorize karo.
\[
(a^{n+1}+b^{n+1})(a+b) = 2ab(a^n+b^n) \implies (a-b)(a^{n+1}-b^{n+1}) = 0
\]
\[
a \ne b \implies n + 1 = 0 \implies n = -1
\]
Answer: \(n = -1\)
Question 7: If \(a^2, b^2, c^2\) are in A.P., prove that \(b+c, c+a, a+b\) are in H.P.
Reciprocals ka A.P. hona prove karna hai: \(\frac{2}{c+a} = \frac{1}{b+c} + \frac{1}{a+b}\).
\[
\frac{1}{b+c} + \frac{1}{a+b} - \frac{2}{c+a} = \frac{(a+2b+c)(c+a) - 2(b+c)(a+b)}{(b+c)(a+b)(c+a)}
\]
\[
\text{Numerator} = (a+c)^2 + 2b(a+c) - 2\big[b^2 + b(a+c) + ac\big] = a^2 + c^2 - 2b^2 = 0
\]
Answer: Proved.
Question 8: If the A.M. between two numbers is \(\frac{9}{2}\) and the H.M. is 4, find the numbers.
Sum \(= 9\), aur \(H = \frac{2ab}{a+b}\) se product \(= 18\) — quadratic banao.
\[
a + b = 9, \quad ab = 18 \implies t^2 - 9t + 18 = 0 \implies t = 3, 6
\]
Answer: \(3\) and \(6\)
Question 9: If the G.M. between two numbers is 4 and the H.M. is \(\frac{16}{5}\), find the numbers.
\(ab = 16\), aur H.M. se \(a + b = 10\) — quadratic banao.
\[
\frac{2(16)}{a+b} = \frac{16}{5} \implies a + b = 10 \implies t^2 - 10t + 16 = 0 \implies t = 2, 8
\]
Answer: \(2\) and \(8\)
Question 10: If \(\frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}\) are in A.P., prove that \(a, b, c\) are in H.P.
Har term mein 2 jama karo — \((a+b+c)\) common nikalne par reciprocals ki A.P. zahir hogi.
\[
\frac{b+c-a}{a} + 2 = \frac{a+b+c}{a}, \quad \text{similarly } \frac{a+b+c}{b}, \; \frac{a+b+c}{c}
\]
\[
\implies \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in A.P.} \implies a, b, c \text{ are in H.P.}
\]
Answer: Proved.
Question 11: If \(a, b, c, d\) are in H.P., show that \(3(a-b)(c-d) = (b-c)(a-d)\).
Reciprocals A.P. mein hain — unhein \(x-3d, x-d, x+d, x+3d\) rakho (symmetric form), phir \(a, b, c, d\) nikaal kar verify karo.
\[
\frac{1}{a} = x-3d, \; \frac{1}{b} = x-d, \; \frac{1}{c} = x+d, \; \frac{1}{d} = x+3d
\]
\[
a - b = \frac{2d}{(x-3d)(x-d)}, \quad c - d = \frac{2d}{(x+d)(x+3d)}
\]
\[
b - c = \frac{2d}{(x-d)(x+d)}, \quad a - d = \frac{6d}{(x-3d)(x+3d)}
\]
\[
3(a-b)(c-d) = \frac{12d^2}{(x-3d)(x-d)(x+d)(x+3d)} = (b-c)(a-d)
\]
Answer: Proved.
Question 12: If two A.M.s \(A_1, A_2\), two G.M.s \(G_1, G_2\) and two H.M.s \(H_1, H_2\) are inserted between two positive numbers \(a\) and \(b\), prove that \(\frac{G_1G_2}{H_1H_2} = \frac{A_1+A_2}{H_1+H_2}\).
Teenon progressions ki properties likho: \(A_1+A_2 = a+b\), \(G_1G_2 = ab\), \(\frac{1}{H_1}+\frac{1}{H_2} = \frac{1}{a}+\frac{1}{b}\).
\[
\frac{G_1G_2}{H_1H_2} = \frac{ab}{H_1H_2}, \qquad \frac{A_1+A_2}{H_1+H_2} = \frac{a+b}{H_1+H_2}
\]
\[
\frac{H_1+H_2}{H_1H_2} = \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} \implies \frac{ab}{H_1H_2} = \frac{a+b}{H_1+H_2}
\]
Answer: Proved.
Question 13: The H.M. between two numbers is 4 and \(2A + G^2 = 27\), where \(A\) and \(G\) are the A.M. and G.M. between them. Find the numbers.
\(s = a+b, p = ab\) rakho: \(s + p = 27\) aur \(\frac{2p}{s} = 4\) — solve karke quadratic banao.
\[
s + p = 27, \quad p = 2s \implies 3s = 27 \implies s = 9, \; p = 18
\]
\[
t^2 - 9t + 18 = 0 \implies t = 3, 6
\]
Answer: \(3\) and \(6\)
Question 14: If \(a, b, c\) are in A.P. and \(b, c, d\) are in H.P., prove that \(ad = bc\).
Pehli shart se \(c = 2b - a\); doosri se \(\frac{2}{c} = \frac{1}{b} + \frac{1}{d}\) — substitute karke simplify karo.
\[
2bd = c(b+d), \qquad c = 2b - a
\]
\[
2bd = (2b-a)(b+d) = 2b^2 + 2bd - ab - ad \implies ad = b(2b-a) = bc
\]
Answer: Proved.
Question 15: If \(a, b, c\) are in G.P., prove that \(\log_a x, \log_b x, \log_c x\) are in H.P.
H.P. ke liye reciprocals ka A.P. hona chahiye — \(\frac{1}{\log_a x} = \log_x a\) (base change).
\[
b^2 = ac \implies \log_x b^2 = \log_x (ac) \implies 2\log_x b = \log_x a + \log_x c
\]
So \(\log_x a, \log_x b, \log_x c\) are in A.P., hence \(\log_a x, \log_b x, \log_c x\) are in H.P.
Answer: Proved.
Question 16: If \(a, b, c\) are in H.P., show that: (i) \(\frac{a-b}{b-c} = \frac{a}{c}\) (ii) \((a-c)^2 = (a+c)(a+c-2b)\).
\(b = \frac{2ac}{a+c}\) rakho (H.P. se), phir dono identities verify karo.
(i)
\[
a - b = \frac{a(a-c)}{a+c}, \quad b - c = \frac{c(a-c)}{a+c} \implies \frac{a-b}{b-c} = \frac{a}{c}
\]
(ii)
\[
(a+c)(a+c-2b) = (a+c)^2 - 4ac = (a-c)^2
\]
Answer: Proved.
Question 17: If \(2+x, 5+x, 9+x\) are in H.P., find \(x\).
Reciprocals A.P. mein honge — darmiyani ka double sum ke barabar.
\[
\frac{2}{5+x} = \frac{1}{2+x} + \frac{1}{9+x} = \frac{11+2x}{(2+x)(9+x)}
\]
\[
2(x^2+11x+18) = (5+x)(11+2x) \implies x = 19
\]
Answer: \(x = 19\)
Question 18: If the roots of \(a(b-c)x^2 + b(c-a)x + c(a-b) = 0\) are equal, prove that \(a, b, c\) are in H.P.
Pehle dekho \(x = 1\) root hai (coefficients ka sum zero) — equal roots hon to dono roots 1, phir product of roots use karo.
\[
x = 1 \text{ satisfies the equation, so both roots are } 1
\]
\[
\text{Product of roots} = \frac{c(a-b)}{a(b-c)} = 1 \implies ac - bc = ab - ac
\]
\[
2ac = b(a+c) \implies \frac{2}{b} = \frac{1}{a} + \frac{1}{c}
\]
So \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P., hence \(a, b, c\) are in H.P.
Answer: Proved.
