OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Ex 14(g)

The availability of Class 11 OP Malhotra Solutions Chapter 14 Sequence and Series Ex 14(g) encourages students to tackle difficult exercises.

S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Ex 14(g)

Question 1.
Find three numbers in G.P. whose sum is 19 and product is 216 .
Solution:
Let the three numbers in G.P are \(\frac { a }{ r }\), a, ar and their product = 216
⇒ \(\frac { a }{ r }\) × a × ar = 216 ⇒ a³ = 6³ ⇒ a = 6
Also given their sum is 19 .
∴ \(\frac { a }{ r }\) + a + ar = 19 ⇒ \(\frac { 6 }{ r }\) + 6 + 6r = 19
⇒ 6 + 6r + 6r² = 19r
⇒ 6 r² – 13r + 6 = 0
⇒ r = \(\frac{13 \pm \sqrt{169-144}}{12}\) = \(=\frac{13 \pm 5}{12}\)
⇒ r = \(\frac { 3 }{ 2 }\), \(\frac { 2 }{ 3 }\)
When a = 6 and r = \(\frac { 3 }{ 2 }\)
Then required numbers are \(\frac{6}{\frac{3}{2}}\), 6, 6 × \(\frac { 3 }{ 2 }\)
i.e. 4. 6. 9
when r = 6, r = \(\frac { 2 }{ 3 }\)
Then required numbers are \(\frac{6}{\frac{2}{3}}\), 6, 6 × \(\frac { 2 }{ 3 }\)
i.e. 9, 6, 4.

Question 2.
The sum of the first terms of a G.P. is \(\frac { 13 }{ 12 }\) and their product is -1. Find the G.P.
Solution:
Let the required G.P be, a, ar, ar² …….
it is given that a + ar + ar² = \(\frac { 13 }{ 12 }\) …(1)
Also (a) (ar) (ar²) = -1 ⇒ (ar)³ = -1
⇒ ar = – 1 ⇒ a = –\(\frac { 1 }{ r }\) …(2)
∴ from (1) and (2); we have
–\(\frac { 1 }{ r }\)–\(\frac { 1 }{ r }\) × r – \(\frac { 1 }{ r }\) × r² = \(\frac { 13 }{ 12 }\)
⇒ – \(\frac { 1 }{ r }\) – 1 – r = \(\frac { 13 }{ 12 }\)
⇒ – 1 – r – r² = \(\frac { 13r }{ 12 }\)
⇒ – 12 – 12r – 12r² = 13r
⇒ 12r² + 25r + 12 = 0
∴ r = \(\frac{-25 \pm \sqrt{625-576}}{24}\) = \(\frac{-25 \pm 7}{24}\)
⇒ r = –\(\frac { 32 }{ 24 }\), –\(\frac { 18 }{ 24 }\) i.e. –\(\frac { 4 }{ 3 }\), –\(\frac { 3 }{ 4 }\)
Thus required G.P be;
\(\frac { 3 }{ 4 }\), \(\frac { 3 }{ 4 }\) × \(\left(-\frac{4}{3}\right)\), \(\frac { 3 }{ 4 }\) \(\left(-\frac{4}{3}\right)^2\) …..
i.e. \(\frac { 3 }{ 4 }\), -1, \(\frac { 4 }{ 3 }\), …..
when r = –\(\frac { 3 }{ 4 }\)
∴ from (2); a = \(\frac { 4 }{ 3 }\)
Thus required G.P be;
\(\frac { 4 }{ 3 }\), \(\frac { 4 }{ 3 }\) × \(\left(-\frac{3}{4}\right)\), \(\frac{4}{3}\) \(\left(-\frac{3}{4}\right)^2\) ……
i.e. \(\frac { 4 }{ 3 }\), -1, \(\frac { 3 }{ 4 }\), …….

Question 3.
The product of first three terms of a G.P. is 1000 . If we add 6 to its second term and 7 to its third term, the resulting three terms form an A.P. Find the terms of the G.P.
Solution:
Let the required G.P be a, ar, ar², ……
it is given that a, ar, ar² = 1000
⇒ (ar)³ =10³ ⇒ ar = 10
⇒ a = \(\frac { 10 }{ r }\) …(1)
it is given that by adding 6 to second term and 7 to third term then the resulting terms are in A.P.
i.e. a, ar + 6, ar² + 7 forms A.P.
⇒ 2(ar + 6) = a + ar² + 7
⇒ 2(10 + 6) = \(\frac { 10 }{ r }\) + \(\frac { 10 }{ r }\) . r² + 7
⇒ 32 = \(\frac { 10 }{ r }\) + 10r + 7
⇒ 10r² + 7r + 10 = 32r
⇒ 10r² – 25r + 10 = 0
⇒ 2r² – 5r + 2 = 0
⇒ (r – 2)(2r – 1) = 0 ⇒ r = 2, \(\frac { 1 }{ 2 }\)
when r = 2 ∴ from (1); a = \(\frac { 10 }{ 5 }\) = 5
and infinite G.P be, 5, 10, 20, …….
when r = \(\frac { 1 }{ 2 }\)
∴ from (1); a = \(\frac{10}{\frac{1}{2}}\) = 20
and infinite G.P be, 20, 20 × \(\frac { 1 }{ 2 }\), 20 × \(\left(\frac{1}{2}\right)^2\), …..
i.e. 20, 10, 5, …..

Question 4.
If a, b, c are in G.P show that the following are also in G.P
(i) \(\frac { 1 }{ a }\), \(\frac { 1 }{ b }\), \(\frac { 1 }{ c }\)
(ii) \(\frac{1}{a^2}\), \(\frac{1}{b^2}\), \(\frac{1}{c^2}\)
(iii) a², b², c²
(iv) b²c², c²a², a²b²
Solution:
(i) Given a, b, c are in G.P
∴ b² = ac …(1)
⇒ \(\frac { 1 }{ ac }\) = \(\frac{1}{b^2}\) ⇒ \(\left(\frac{1}{b}\right)^2\) = \(\frac { 1 }{ a }\) × \(\frac { 1 }{ c }\)
⇒ \(\frac { 1 }{ a }\), \(\frac { 1 }{ b }\), \(\frac { 1 }{ c }\) are in G.P

(ii) On squaring eqn. (1); we have b⁴ = a²c²
⇒ \(\frac{1}{a^2 c^2}\) = \(\frac{1}{b^4}\) ⇒ \(\frac{1}{a^2}\) × \(\frac{1}{c^2}\) = \(\left(\frac{1}{b^2}\right)^2\)
⇒ \(\frac{1}{a^2}\), \(\frac{1}{b^2}\), \(\frac{1}{c^2}\) are in G.P

(iii) On squaring eqn. (1) ; we have
b⁴ = (ac)² = a² c² ⇒ (b²)² = a² × c²
Thus, a², b² and c² are in G.P.

(iv) On squaring eqn. (1); we iave
b⁴ = a² c²
Again squaring ; we have
⇒ (b⁴)² = (a²c²)²
⇒ b⁴ . b⁴ = (a²c²)²
⇒ b⁴a²c² = (a² c²)²
⇒ (a²b²) (b²c²) = (a²c²)²
Thus b²c², a²c² and a² b² are in G.P.

Question 5.
If a, b, c, d are in G.P., show that the following are also in G.P.
(i) a + b, b + c, c + d
(ii) a² + b², b² + c², c² + d²
Solution:
Given a, b, c and d are in G.P.
∴ \(\frac{b}{a}\) = \(\frac{c}{b}\) = \(\frac{d}{c}\) = r (say)
⇒ b = ar ; c = br = ar²; d = cr = ar³

(i) Here, (b + c)² = (ar + ar²)²
= a²r²(1 + r)²
and (a + b)(c + d) = (a + ar) (ar² + ar³)
= a(1 + r) ar²(1 + r)
= a²r²(1 + r)²
∴ (b + c)² = (a + b)(c + d)
Thus a + b, b + c and c + d are in G.P.

(ii) Here, (b² + c²)² = (a²r² + a²r⁴)²
= (a²r²)²[1 + r²]²
= a⁴r⁴(1 + r²)²
and (a² + b²) (c² + d²)
= (a² + a²r²) (a²r⁴ + a²r⁶)
= a²(1 + r²) a²r⁴(1 + r²)
= a⁴r⁴(1 + r²)²
∴ (b² + c²)² = (a² + b²) (c² + d²)
Thus a² + b², b² + c² and c² + d² are in G.P.

Question 6.
If a, b, c, d are in G.P., prove that
(i) (b + c)(b + d) = (c + a)(c + d)
(ii) (a – d)² = (b – c)² + (c – a)² + (d – b)²
Solution:
Given a, b, c and d are in G.P
∴ \(\frac { b }{ a }\) = \(\frac { c }{ b }\) = \(\frac { d }{ c }\) = r ≠ 0 (say)
∴ b = ar; c = br = ar²; d = cr = ar³

(i) L.H.S = (b + c)(b + d)
= (ar + ar²) (ar + ar³)
= ar(1 + r) ar(1 + r²)
= a²r²(1 + r) (1 + r²)
R.H.S = (c + a)(c + d)
= (ar² + a) (ar² + ar³)
= a(r² + 1) ar²(1 + r)
= a²r²(1 + r) (1 + r²)
∴ L.H.S = R.H.S

(ii) L.H.S = (a – d)² = (a – ar³)²
= a² (1 – r³)²
R.H.S = (b – c)² + (c – a)² + (d – b)²
= (ar – ar²)² + (ar² – a)² + (ar³ – ar)²
= (ar)²(1 – r)² + a²(r² – 1)² + (ar)² (r² – 1)²
= a²[r²(1 – r)² + (r² – 1)² + r²(r² – 1)²]
= a²[r²(1 – r)² + (r² – 1)² (1 + r²)]
= a²(1 – r)² [r² + (r + 1)²(1 + r²)]
= a²(1 – r)²[r² + (r² + 2r + 1) (1 + r²)]
= a²(1 – r)²[r² + (1 + r²)² + 2r (1 + r²)]
= a²(1 – r)² (r + 1 + r²)²
= a²[(1 – r) (1 + r + r²)]²
= a²(1 – r³)²
∴ L.H.S = R.H.S

Question 7.
If the pth, rth and rth terms of an A.P. are in G.P., prove that the common ratio of the G.P. is \(\frac{q-r}{p-q}\).
Solution:
Let a be the first term and d be the common difference of A.P respectively.
Since T_p, T_{q/sub> and Tr are in G.P.
∴ Common ratio = \(\frac{\mathrm{T}_q}{\mathrm{~T}_p}\) = \(\frac{\mathrm{T}_r}{\mathrm{~T}_q}\)
= \(\frac{a+(q-1) d}{a+(p-1) d}\) = \(\frac{a+(r-1) d}{a+(q-1) d}\)
= \(\frac{(q-1-r+1) d}{(p-1-q+1) d}\) = \(\frac{q-r}{p-q}\)}

Question 8.
If \(\frac{1}{x+y}\), \(\frac{1}{2 y}\), \(\frac{1}{y+z}\) are the three consecutive terms of an A.P. prove that x, y, z are the three consecutive terms of a G.P.
Solution:
Since \(\frac{1}{x+y}\), \(\frac{1}{2 y}\) and \(\frac{1}{y+z}\) are three consecutive terms of A.P.
⇒ \(\frac{1}{2 y}\) – \(\frac{1}{x+y}\) = \(\frac{1}{y+z}\) – \(\frac{1}{2 y}\)
⇒ \(\frac{x+y-2 y}{2 y(x+y)}\) = \(\frac{2 y-y-z}{2 y(y+z)}\)
⇒ \(\frac{x-y}{2 y(x+y)}\) = \(\frac{y-z}{2 y(y+z)}\)
⇒ (x – y)(y + z) = (x + y)(y – z)
⇒ xy + xz – y² – yz = xy – xz + y² – yz
⇒ 2y² = 2xz ⇒ y² = xz
Thus x, y and z are in G.P.