RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.3

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.3

Question 1.
(i) Find the 7^th term of series 1 + 3 + 9 + 27 + …
(ii) Find the 10th term of

Solution:
(i) Given series = 1 + 3 + 9 + 27 + ….
Hedre, a = 1, r = 3/1 = 3
7^th term = T₁ (From T_n – ar^n-1)
= 1 × (3)^7-1
= (3)⁶ = 729

Question 2.
(i) Which term of series 64 + 32 + 16 + 8 + … is 1/64 ?
(ii) Which term of series 6 + 3 + 3/2 + 3/4 + … is 3/256 ?
Solution:
(i) Given series =64 + 32 + 16 + 8 +….

Question 3.
Find the common ratio and 12^th term of G.P. 5 + 10 + 20 + 40 + …
Solution:
Given
series = 5 + 10 + 20 + 40 + ….

n^th term = T_n = ar^{n – 1}
= 5 × (2)^{n – 1} = 5.2^{n – 1}

Question 4.
Find the fifth term from the last term of G.P. 2, 6, 18, 54, … 118098.
Solution:
Given
series = 2, 6, 18, 54,…..,118098
Common ratio r = 6/2 = 3
Last term = 118098
⇒ ar^{n – 1} = 118098
⇒ 2 × (3)^{n – 1} = 118098
⇒ (3)^{n – 1} = 59049 = (3)¹⁰
On comparing, n – 1 = 10
⇒  n = 11
5^th term from last =ar^{n – 5}
= 2 × (3)^{11 – 5} =2 × (3)⁶
= 2 × 72 = 1458

Question 5.
Third term of a G.P. is 32 and 7th term is 8192, then find 10th term of GP.
Solution:
Given, T₃= ar² = 32
T₇ = ar⁶ = 8192
Divide equation (ii) by (i), we get

⇒ r⁴ = 256
⇒ r⁴ = 4⁴
On comparing r = 4
Then, from equation (i)
ar² = 32
⇒ a × 4² = 32
⇒ a = 2
10th term = T₁₀ = ar⁹
= 2 × (4)⁹ = 524288

Question 6.
Find G.P., whose 3^rd term is 1 and 7^th term is 16.
Solution:
Let a is first term of G.P. and r is common ratio, then
Third term T₃ = 1
⇒ ar²   = 1
Seventh term T₇ = 16
ar⁶ = 16
Divide equation (ii) from (i), we get

⇒ r⁴ = 16 = 2⁴
Hence, r = 2
ar² = 1
a × (2)² = 1

Question 7.
(i) Find 3 G.M. between 3 and 48.
(ii) Find 6 G.M. between 2 and 256.
Solution:
(i) Let G₁, G₂, G₃ be three GM. between 3 and 48, then
3, G₁, G₂, G₃, 48 will be in G.P.
Here a = 3,
Last term (5th term) = 48 and n = 5
3.r⁴ = 48  [∵ l = ar^{n – 1}]
⇒ r⁴ = 16 = 2⁴
⇒ r = 2
G₁ = 3 × 2 = 6
G₂ = 3 × 2² = 12
G₃ = 3 × 2³ = 24
Hence, three G.M. are 6, 12, 24.

(ii) Let G₁, G₂, G₃, G₄, G₅, G₆ be six G.M. between 2 and 256, then
2, G₁, G₂, G₃, G₄, G₅, G₆, 256 will be in G.P.
Here a = 6, last term (8th term) = 256 and n = 8
Hence, 2.r⁷ = 256 [∵ l = ar^{n – 1}]
⇒ r⁷= 128 = 2⁷
⇒ r = 2
G₁ = 2 × 2 = 4
G₂ = 4 × 2 = 8
G₃ = 8 × 2 = 16
G₄ = 16 × 2 = 32
G₅ = 32 × 2 = 64
G₆ = 64 × 2 = 128
Hence six G.M. are 4, 8, 16, 32, 64, 128.

Question 8.
For which value of x, numbers x, x + 3, x + 9 are in G.P. ?
Solution:
x, x + 3, x + 9 are in G.P.
Then, (x + 3)² = x × (x + 9)
⇒ x² + 9 + 6x = x² + 9x
⇒ x² + 6x + 9 – x² – 9x = 0
⇒ – 3x + 9 = 0
⇒ \(\frac { -9 }{ -3 }\) = 3
Hence, x = 3

Question 9.
Find four terms which are in G.P., whose third term is 4 more than first term and second term is 36 more than 4^th term.
Solution:
Let first four terms of G.P. are a, ar, ar². ar³ where a is first term and r is common ratio.
According to question,
Third term = First term + 4
⇒ ar² = a + 4
⇒ a(r² – 1) = 4
According to question,
Second term = Fourth term + 36
⇒ ar = ar³ + 36
⇒ ar – ar³ = 36
⇒ ar(1 – r²) = 36
Divide equation (ii) by (i), we get

⇒ – r = 9
⇒ r = – 9
Hence, common ratio of G.P.,
r = – 9
Putting r = – 9 in equation (i),
a[(-9)² – 1] = 4
⇒ a (81 – 1) = 4

Question 10.
If 4^th term of any G.P. is p, 7^th term is q and 10^th term is r, then prove that q² – pr.
Solution:
Let a is first term and R is common ratio of G.P., then according to question
T₄ = aR^{4 – 1}
p = aR³           ….. (i)
T₇ = aR^{7 – 1}
q = aR⁶           ….. (ii)
T₁₀ = aR^{10 – 1}
r = aR⁹            ….. (iii)
Multiplying equation (i) and (iii), we get
a² R¹² = pr    ….. (iv)
Squaring of equation (ii),
q² = a² R¹² Hence, from equation (iv) and (v)
q² = pr         ….. (v)
Hence Proved.

Question 11.
If (p + q)^th term in GP. is x and (p – q)^th term is y, then find p^th term.
Solution:
Let a is first term and r is common ratio of G.P.
So, (p + q)^th term = ar^p + q – 1 = x   … (i)
and (p – q)^th term = ar^p – q – 1 = y   … (ii)
Multiplying equation (i) and (ii), we get
a² (r^p + q – 1 + p – q) = x × y
or a² r^2p-2 = xy
ora² r^{2(p – 1)} = xy
or (ar^{p – 1})² = xy
or ar^{p – 1} = √xy
Hence, pth term = ar^{p – 1}
= √xy

Question 12.
If a, b, c are in G.P. and  a^x = b^y = c^z , then prove that

Solution:
Let a^x= b^y = C^z = k

and a, b, c are in G.P.
b² = ac
Putting the value of a, b, c in equation (i),

Question 13.
If n G.M. inserted between a and b then prove that product of all geometric means will be
\({ \left( \sqrt { xy } \right) }^{ n }\)
Solution:
Let G₁, G₂, G₃, G₄, ……G_n   are in G.M. in between x and y then

Question 14.
If x, y, z are in G.P., arithmetic mean of x, y is A₁ and A.M. of y, z is A₂ then prove that :

Solution:
x, y, z are in G.P.
∴ y² = xz     …… (i)
Arithmetic mean of x and y is A₁

Arithmetic mean of y and z is A₂

RBSE Solutions for Class 11 Maths