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

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

Question 1.
Find : (i) the 7th term of 2, 4, 8, …..
(ii) the 9 th term of 1, \(\frac { 1 }{ 2 }\), \(\frac{1}{2^2}\), ….
(iii) the nth term of \(\frac { 15 }{ 8 }\), \(\frac { 3 }{ 8 }\), \(\frac { 3 }{ 40 }\), ….
Solution:
(i) Given sequence be 2, 4, 8,…
Here \(\frac{T_2}{T_1}\) = \(\frac{4}{2}\) = 2; \(\frac{T_3}{T_2}\) = \(\frac{8}{4}\) = 2 ….
Clearly given progression form G.P with first term a = 2 and common ratio r = 2
We know that T_n = ar^{n – 1}
∴ T₇ = ar⁶ = 2 × 2⁶ = 128

(ii) Given sequence be 1, \(\frac { 1 }{ 2 }\), \(\frac{1}{2^2}\), ……
Here \(\frac{T_2}{T_1}\) = \(\frac{1}{2}\); \(\frac{T_3}{T_2}\) = \(\frac{\frac{1}{2^2}}{\frac{1}{2}}\) = \(\frac { 1 }{ 2 }\) = …..
Clearly given series form G.P with first term a = 1 and common ratio r = \(\frac { 1 }{ 2 }\)< 1
We know that T_n = ar^{n – 1}
∴ T₉ = ar⁸ = 1 × \(\left(\frac{1}{2}\right)^8\) = \(\frac { 1 }{ 256 }\)

(iii) Given sequence be, \(\frac { 15 }{ 8 }\), \(\frac { 3 }{ 8 }\), \(\frac { 3 }{ 40 }\), ……
Here \(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\) = \(\frac{\frac{3}{8}}{\frac{15}{8}}\) = \(\frac{1}{5}\); \(\frac{\mathrm{T}_3}{\mathrm{~T}_2}\) = \(\frac{\frac{3}{40}}{\frac{3}{8}}\) = \(\frac { 1 }{ 5 }\) ….
Thus, the given progression forms G.P. with first term a = \(\frac { 15 }{ 8 }\)
and r = common ratio = \(\frac { 1 }{ 5 }\)
We know that T_n = ar^n-1
⇒ T_n = \(\frac { 15 }{ 8 }\)\(\left(\frac{1}{5}\right)^{n-1}\)

Question 2.
The second term of a G.P. is 18 and the fifth term is 486. Find :
(i) the first term,
(ii) the common ratio
Solution:
Let a be the first term and r be the common ratio of given G.P.
Given T₂ = 18 ⇒ ar = 18 …(1) [∵ T_n = ar^n-1]
and T₅ = 486 ⇒ ar⁴ = 486
On dividing (2) by (1) ; we have
r³ = \(\frac { 486 }{ 18 }\) = 27 = 3³ ⇒ r = 3
∴ from (1); a = \(\frac { 18 }{ 3 }\) = 6
Hence the required first term be 6 and common ratio be 3 .

Question 3.
Find the value of x for which x + 9, x – 6, 4 are the first three terms of a geometrical progression and calculate the fourth term of progression in each case.
Solution:
Since x + 9, x – 6 and 4 are the three consecutive terms of G.P.
∴ (x – 6)² = 4(x + 9)
[∵ if a, b, c are in G.P Then b² = ac]
⇒ x² – 12x + 36 = 4x + 36
⇒ x² – 16x = 0 ⇒ x = 0, 16
When x = 0, given three terms of G.P becomes ; 9, -6 and 4 i.e. r = \(\frac{-6}{9}\) = \(\frac{-2}{3}\)
∴ T₄ = ar³ = 9\(\left(-\frac{2}{3}\right)^3\) = –\(\frac{8}{3}\)
When x = 16, given three terms of G.P becomes ; 25, 10 and 4 with r = \(\frac{10}{25}\) = \(\frac{2}{5}\)
∴ T₄ = ar³ = 25\(\left(\frac{2}{5}\right)^3\) = \(\frac{8}{5}\)

Question 4.
If 5, x, y, z 405 are the first five terms of a geometric progression, find the values of x, y and z.
Solution:
Since 5, x, y, z 405 are first five terms of G.P.
Let r be the common ratio of G.P
∴ 405 = T₅ = ar⁴ = 5r⁴ ⇒ r⁴ = 81 = 3⁴
⇒ r = 3
∴ x = T₂ = ar = 5 × 3 = 15
y = T₃ = ar² = 5 × 3² = 45
z = T₄ = ar³ = 5 × 3³ = 135

Question 5.
Insert 3 geometric means between 16 and 256.
Solution:
Let G₁, G₂, G₃ are three GM between 16 and 256
Thus 16, G₁, G₂, G₃, 256 are in G.P
Let r be the common ratio of given G.P.
∴ 256 = T₅ = ar⁴ = 16 × r⁴
⇒ r⁴ = 16 = 2⁴ ⇒ r = 2
Thus G₁ = T₂ = ar = 16 × 2 = 32
G₂ = T₃ = ar² = 16 × 2² = 64
and G₃ = T₄ = ar³ = 16 × 2³ = 128
Hence the required three G.M between 16 and 256 are 32, 64, 128.

Question 6.
Insert 5 geometric means between \(\frac { 1 }{ 3 }\) and 243.
Solution:
Let G₁, G₂, G₃, G₄ and G₅ are five G.M’s between \(\frac { 1 }{ 3 }\) and 243 .
Then \(\frac { 1 }{ 3 }\), G₁, G₂, G₃, G₄ and G₅, 243 are in G.P.
Let r be its common ratio.
∴243 = T₇ = ar⁶ = \(\frac { 1 }{ 3 }\)r⁶ ⇒ r⁶ = 243 × 3 = 3⁶
∴ r = 3
∴ G₁ = T₂ = ar = \(\frac { 1 }{ 3 }\) × 3 = 1
G₂ = T₃ = ar² = \(\frac { 1 }{ 3 }\) × 3² = 3
G₃ = T₄ = ar³ = \(\frac { 1 }{ 3 }\) × 3³ = 9
G₄ = T₅ = ar⁴ = \(\frac { 1 }{ 3 }\) × 3⁴ = 27
and G₅ = T₆ = ar⁵ = \(\frac { 1 }{ 3 }\) × 3⁵ = 81
Thus the required five G.M’s between \(\frac { 1 }{ 3 }\) and 243 are 1, 3, 9, 27 and 81 .

Question 7.
If the A.M. and G.M. between two numbers are respectively 17 and 8, find the numbers.
Solution:
Let the required two numbers are a and b
Then \(\frac { a + b }{ 2 }\) = 17 ⇒ a + b = 34 …(1)
Also G.M between a and b be 8
∴ \(\sqrt{a b}\) = 8 ⇒ ab = 64 …(2)
Now, (a – b)² = (a + b)² – 4ab
(a – b)² = 34² – 4 × 64 = 1156 – 256 = 900 = 30²
⇒ a – b = ± 30

Case-I. When a – b = 30
On adding (1) and (3); we have
2a = 64 ⇒ a = 32 ∴ from (1); b = 2

Case-II. When a – b = – 30 …(4)
On adding eqn. (1) and eqn. (4); we have
2a = 4 ⇒ a = 2 ∴ from (1); b = 32
Hence the required numbers are 2 and 32 .

Question 8.
The second, third and sixth terms of an A.P. are consecutive terms of a geometric progression. Find the common ratio of the geometric progression.
Solution:
Let a be the first term and d be the common difference of given A.P.
Given T₂, T₃ and T₆ of A.P are consecutive terms of G.P.
∴ \(\mathrm{T}_3^2\) = T₂ T₆
[if a, b and c are in G.P. Then b² = ac]
⇒ (a + 2d)² = (a + d)(a + 5d)
[∵ T_n = a + (n – 1) d]
⇒ a² + 4ad + 4d² = a² + 5d² + 6ad
⇒ 2ad = – d² ⇒ 2a = – d
⇒ d = – 2a …(1)
∴ required common ratio of G.P = \(\frac{\mathrm{T}_3}{\mathrm{~T}_2}\)
= \(\frac{a+2 d}{a+d}\) = \(\frac{a+2(-2 a)}{a-2 a}\) = \(\frac{-3 a}{-a}\) = 3

Question 9.
The 5th, 8th and 11 th terms of a G.P. are P, Q and S respectively. Show that Q² = PS.
Solution:
Let a be the first term and r be the common ratio of G.P.
We know that T_n = ar^{n – 1}
Given T₅ = P ⇒ ar⁴ = P …(1)
T₈ = Q ⇒ ar⁷ = Q …(2)
T₁₁ = S ⇒ ar¹⁰ = S …(3)
∴ PS = (ar⁴) (ar¹⁰) [using (1) and eqn. (2)]
= a²r¹⁴ = (ar⁷)² = Q² [using eqn. (2)]

Question 10.
The (p + q)th term and (p – q)th terms of a G.P. are a and b respectively. Find the pth term.
Solution:
Let A be the first term and R be the common ratio of given G.P.
given T_p+q = a ⇒ AR^p+q-1 = a …(1) [∵ T_n = ar^n-1]
T_p-q = b ⇒ AR^p-q-1 = b …(2)
On dividing eqn. (1) by (2); we have
(R)^p+q-1-p+q+1 = \(\frac{a}{b}\)

Question 11.
If the pth, qth, rth terms of a G.P. are x, y, z respectively, prove that x^q-r, y^r-p, z^p-q = 1.
Solution:
Let A be the first term and R be the common ratio of given G.P.
Given T_p = x ⇒ AR^p-1 = x …(1)
T_q = y ⇒ AR^q-1 = y …(2)
T_r = z ⇒ AR^r-1 = z …(3)
L.H.S = x^q-r y^r-p z^p-q
= [AR^p-1]^p-r [AR^q-1]^r-p [AR^r-1]^p-r
= \(\mathrm{A}^{q-r+r-p+p-q} \mathrm{R}^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}\)
= \(\mathrm{A}^0 \mathrm{R}^{p q-p r-q+r+r q-p q-r+p+r p-r q-p+q}\)
= A⁰R⁰ = 1 = R.H.S.

Question 12.
In a set of four numbers, the first three are in G.P. and the last three are in A.P. with difference 6 . If the first number is the same as the fourth, find the four numbers.
Solution:
Since it is given that, out of four numbers the first three are in G.P and last three are in A.P with common difference 6 .
Let the numbers are x, a – 6, a, a + 6. Since first three are in G.P.
∴ (a – 6)² = ax …(1)
Also x = a + 6 ∴ from (1); we have
(a – 6)² = a(a + 6)
⇒ a² – 12a + 36 = a² + 6a
⇒ 18a = 36 ⇒ a =2
∴ x = 2 + 6 = 8
Hence the required four numbers are
8, 2 -6 , 2, 2 + 6 i.e. 8, -4, 2, 8.

Question 13.
If \(a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}\) and a, b, c, are in G.P., prove that x, y, z are in A.P.
Solution:
Given \(a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}\) = k ≠ 0 (say)
∴ a = k^x; b = k^y; c = k^z
Since a, b and c are in G.P.
∴ b² = ac ⇒ (k^y)² = k^x . k^z ⇒ k^2y = k^x+z
⇒ 2y = x + z ∴ x, y, z are in A.P.

Question 14.
If one G.M., G and two A.M’s p and q be inserted between two given numbers, prove that G² =(2p – q)(2q – p).
Solution:
Let a and b are two given numbers given G be the G.M between a and b
∴ G = \(\sqrt{a b}\) ⇒ G² = ab
Also given p, q are two A.M’s between a and b
Then a, p, q, b are in A.P

Question 15.
Construct a quadratic equation in x such that the A.M. of its roots is A and G.M. is G.
Solution:
Let α and β are the roots of quadratic eqn.
s.t \(\frac{\alpha+\beta}{2}\) = A ⇒ α + β = 2A
and \(\sqrt{\alpha \beta}\) = G ⇒ αβ = G²
∴ required quadratic eqn. having roots α and β be given by
x² – (α + β) x +αβ = 0
⇒ x² – 2Ax + G² = 0

Question 16.
The fourth term of a G.P. is greater than the first term, which is positive, by 372 . The third term is greater than the second by 60 . Calculate the common ratio and the first term of the progression.
Solution:
Let a be the first term and r be the common ratio of G.P.
Given T₄ = T₁ + 372
⇒ ar³ – a = 372 …(1)
T₃ = T₂ + 60
⇒ ar² – ar = 60 …(2)
On dividing eqn. (1) and eqn. (2); we have
\(\frac{r^3-1}{r(r-1)}\) = \(\frac{372}{60}\) = \(\frac{62}{10}\) = \(\frac{31}{5}\)
⇒ \(\frac{r^2+r+1}{r}\) = \(\frac{31}{5}\)
⇒ 5r² – 26r + 5 = 0
⇒ (r – 5)(5r – 1) = 0 ⇒ r = 5, \(\frac{1}{5}\)
When r = 5 ∴ from (1); we have
a[125 – 1] = 372 ⇒ a = \(\frac{372}{124}\) = 3
When r = \(\frac{1}{5}\) ∴ from (1); we have
a\(\left[\frac{1}{125}-1\right]\) = 372 ⇒ a \(\left(\frac{-124}{125}\right)\) = 372
⇒ a = – 375, which is not possible.
Thus required first term be 3 and common ratio be 5.

Question 17.
The first, english and twenty-second terms of an A.P. are three consecutive terms of a G.P. Find the common ratio of the G.P. Given also that the sum of the first twenty-two terms of the A.P. is 275 , find its first term.
Solution:
Let a be the first term and d be the common difference of an A.P.
It is given that T₁, T₈ and T₂₂ terms of an A.P. are three consecutive terms of G.P.

Hence the required first term be 5.