Question 1: Find the 6th term of the G.P.: \(-6, -3, -\frac{3}{2}, \dots\)
Pehle \(r = \frac{-3}{-6} = \frac{1}{2}\) nikalo, phir \(a_6 = a_1 r^5\).
\[
a_6 = -6\left(\frac{1}{2}\right)^5 = -\frac{6}{32} = -\frac{3}{16}
\]
Answer: \(-\frac{3}{16}\)
Punjab Board Class 11 Mathematics
Exercise 6.5 Solutions — Sequences and Series
Complete worked solutions for Exercise 6.5 from Chapter 6 (Sequences and Series), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.
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Mathematics 11 (Punjab Board)
Exercise 6.5 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 6: Sequences and Series.
Question 2: Find the 8th term of the G.P.: \(3, 3^2, 3^3, \dots\)
\(a_1 = 3, r = 3\) — \(a_8 = 3 \cdot 3^7 = 3^8\).
\[
a_8 = 3^8 = 6561
\]
Answer: \(6561\)
Question 3: Find the value of \(n\) if the \(n^{th}\) terms of the two G.P.s \(1, 2, 4, 8, \dots\) and \(256, 128, 64, \dots\) are equal.
Dono ke \(n\)th terms likh kar barabar rakho — powers of 2 mein likhne se kaam asan hoga.
\[
2^{n-1} = 256\left(\frac{1}{2}\right)^{n-1} = 2^{9-n} \implies n-1 = 9-n \implies n = 5
\]
Answer: \(n = 5\)
Question 4: Write the first five terms of the G.P. with given \(a_1\) and \(r\):
Har agli term pichli ko \(r\) se multiply karke banao.
(i) \(a_1 = 243, r = \frac{1}{3}\)
\[
243, \quad 81, \quad 27, \quad 9, \quad 3
\]
(ii) \(a_1 = 579, r = -\frac{1}{2}\)
\[
579, \quad -\frac{579}{2}, \quad \frac{579}{4}, \quad -\frac{579}{8}, \quad \frac{579}{16}
\]
Answer: (i) \(243, 81, 27, 9, 3\), (ii) \(579, -\frac{579}{2}, \frac{579}{4}, -\frac{579}{8}, \frac{579}{16}\)
Question 5: Find the 12th term of the G.P.: \(1+i, 2i, \dots\)
\(r = \frac{2i}{1+i} = 1+i\). Phir \(a_{12} = (1+i)^{12}\) — \((1+i)^2 = 2i\) use karo.
\[
r = \frac{2i}{1+i} = 1+i, \qquad a_{12} = (1+i)^{12} = \big((1+i)^2\big)^6 = (2i)^6 = -64
\]
Answer: \(-64\)
Question 6: The 4th and 9th terms of a G.P. are 54 and 13122 respectively. Find the G.P. and its general term.
Do equations divide karke \(r^5\) nikalo, phir \(a_1\).
\[
\frac{a_1 r^8}{a_1 r^3} = \frac{13122}{54} \implies r^5 = 243 \implies r = 3
\]
\[
a_1 = \frac{54}{27} = 2 \implies \text{G.P.: } 2, 6, 18, \dots \quad a_n = 2 \cdot 3^{n-1}
\]
Answer: \(2, 6, 18, \dots\); \(a_n = 2 \cdot 3^{n-1}\)
Question 7: If \(a, b, c, d\) are in G.P., show that: (i) \(a-b, b-c, c-d\) are in G.P. (ii) \(a^2-b^2, b^2-c^2, c^2-d^2\) are in G.P.
Terms ko \(A, AR, AR^2, AR^3\) rakho — common factor nikaalne par G.P. zahir ho jayegi.
(i)
\[
A-AR, \; AR-AR^2, \; AR^2-AR^3 = A(1-R), \; AR(1-R), \; AR^2(1-R)
\]
Common ratio \(R\) — in G.P.
(ii)
\[
A^2(1-R^2), \; A^2R^2(1-R^2), \; A^2R^4(1-R^2)
\]
Common ratio \(R^2\) — in G.P.
Answer: Proved.
Question 8: If \(a_{p+q} = m\) and \(a_{p-q} = n\) of a G.P., show that \(a_p = \sqrt{mn}\).
Dono equations multiply karo — \(a_1^2 r^{2p-2} = (a_1 r^{p-1})^2\) banega.
\[
(a_1 r^{p+q-1})(a_1 r^{p-q-1}) = mn \implies (a_1 r^{p-1})^2 = mn \implies a_p = \sqrt{mn}
\]
Answer: Proved.
Question 9: Three numbers are in G.P. whose product is 216 and sum is 26. Find the numbers.
Numbers \(\frac{a}{r}, a, ar\) lo — product se \(a = 6\), phir sum se \(r\).
\[
a^3 = 216 \implies a = 6, \qquad \frac{6}{r} + 6 + 6r = 26 \implies 3r^2 - 10r + 3 = 0
\]
\[
(3r-1)(r-3) = 0 \implies r = 3 \text{ or } \frac{1}{3} \implies \text{numbers: } 2, 6, 18
\]
Answer: \(2, 6, 18\)
Question 10: If \(a_3 = (a_1)^2\) and \(a_2 = 9\) of a G.P., find the 6th term.
\(a_1 r^2 = a_1^2\) se \(r^2 = a_1\); \(a_1 r = 9\) se \(r^3 = 9\). Phir \(a_6 = a_1 r^5\).
\[
r^3 = 9 \implies r = \sqrt[3]{9}, \quad a_1 = r^2, \quad a_6 = r^2 \cdot r^5 = r^7 = 9^{7/3}
\]
Answer: \(9^{7/3}\)
Question 11: If \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in G.P., show that the common ratio is \(\pm\sqrt{\frac{a}{c}}\).
\(r = \frac{1/b}{1/a} = \frac{a}{b}\) aur \(r = \frac{1/c}{1/b} = \frac{b}{c}\) — dono ko multiply karo.
\[
r^2 = \frac{a}{b} \cdot \frac{b}{c} = \frac{a}{c} \implies r = \pm\sqrt{\frac{a}{c}}
\]
Answer: Proved.
Question 12: Three numbers are in A.P. whose sum is 21. If 1, 4, 3 are subtracted from them respectively, the resulting numbers are in G.P. Find the numbers.
A.P. terms \(a-d, a, a+d\) — sum se \(a = 7\). Naye numbers \(6-d, 3, 4+d\) par \(b^2 = ac\) lagao.
\[
a = 7, \qquad 9 = (6-d)(4+d) \implies d^2 - 2d - 15 = 0 \implies d = 5 \text{ or } -3
\]
\[
\text{Numbers: } 2, 7, 12 \quad (\text{or } 10, 7, 4)
\]
Answer: \(2, 7, 12\)
Question 13: Three numbers are in A.P. whose sum is 6. If 1, 4, 15 are added to them respectively, the resulting numbers are in G.P. Find the numbers.
Sum se \(a = 2\). Naye numbers \(3-d, 6, 17+d\) par G.P. ki shart lagao.
\[
36 = (3-d)(17+d) \implies d^2 + 14d - 15 = 0 \implies d = 1 \text{ or } -15
\]
\[
\text{Numbers: } 1, 2, 3 \quad (\text{or } 17, 2, -13)
\]
Answer: \(1, 2, 3\)
Question 14: If the \(p^{th}\) term of a G.P. is \(q\) and the \(q^{th}\) term is \(p\), show that the \((p+q)^{th}\) term is \(\left(\frac{q^p}{p^q}\right)^{\!1/(p-q)}\).
Equations divide karke \(r^{p-q} = \frac{q}{p}\) nikalo, phir \(a_{p+q} = a_1 r^{p+q-1}\) mein rakho.
\[
r = \left(\frac{q}{p}\right)^{\!1/(p-q)}, \qquad a_{p+q} = p \cdot r^p = p\left(\frac{q}{p}\right)^{\!p/(p-q)} = \left(\frac{q^p}{p^q}\right)^{\!1/(p-q)}
\]
Answer: Proved.
Question 15: Find \(a\) if the numbers \(a, 2a+2, 3a+3\) are in G.P. Also find the 5th term of the G.P.
G.P. ki shart \((2a+2)^2 = a(3a+3)\) — quadratic solve karo. \(a = -1\) reject hoga (zero term).
\[
4a^2 + 8a + 4 = 3a^2 + 3a \implies a^2 + 5a + 4 = 0 \implies a = -4 \text{ or } -1
\]
\[
a = -4: \text{ G.P. } -4, -6, -9, \dots \quad r = \frac{3}{2}, \quad a_5 = -4\left(\frac{3}{2}\right)^4 = -\frac{81}{4}
\]
Common mistake: \(a = -1\) gives \(-1, 0, 0, \dots\) — 0 cannot be a term of a genuine G.P., so reject it.
Answer: \(a = -4\), \(a_5 = -\frac{81}{4}\)