PROBLEMS ON GEOMETRIC PROGRESSION AND SERIES

Problem 1 :

t₁₂ of the series -128, 64, -32, ................ is

(a) -1/16    (b) 16     (c)  1/16     (d) None

Solution :

12th term :

t_n = ar^n-1

a = -128, r = -64/128 ==> -1/2

t₁₂ = (-128)(-1/2)^12-1

t₁₂ = (-128)(-1/2)¹¹

t₁₂ = 1/2⁴

t₁₂ = 1/16

Problem 2 :

The last term of the series

x², x, 1.................. to 31 terms is 

(a) x²⁸    (b) 1/x     (c)  1/x²⁸     (d) None

Solution :

It is a geometric series. It consists of 31 terms. So, 31^st term should be the last term.

a = x², r = 1/x

t₃₁ = (x²) (1/x)^31-1

t₃₁ = (x²) (1/x)³⁰

t₃₁ = 1/x²⁸

Problem 3 :

The sum of the series 

243, 81, 27, ............. to 8 terms.

(a) 36    (b) 36  13/30     (c)  36  1/9     (d) None

Solution :

a = 243, r = 81/243 ==>  1/3 and n = 8

s_n = a(rⁿ - 1)/(r - 1)

s₈ = 243((1/3)⁸ - 1)/((1/3) - 1)

s₈ = 243((1 - 3⁸)/3⁸)/(-2/3)

s₈ = (-2/9)(1 - 6561)

s₈ = (-2/9)(-6560)

s₈ = 13120/9

Converting into mixed fraction, we get

s₈ = 13120/9

s₈ = 1457  7/9

So, the answer is none of the these.

Problem 4 :

The second terms of a G.P is 24 and the fifth term is 81. The series is

(a) 16, 36, 24, 54,....         (b)  24, 36, 53, .......

(c)  16, 24, 36, 54, ........    (d)  None

Solution :

t₂ = 24 and t₅ = 81

ar = 24 ----(1)     ar⁴ = 81----(2)

(2) / (1)

ar⁴ / ar = 81/24

r³ = 27/8

r = 3/2

Applying the value of r in (1), we get

a(3/2) = 24

a = 24 (2/3)

a = 16

16, 16(3/2), 16(3/2)²,.....

16, 24, 36,...........

Problem 5 :

The sum of n terms of the series 4 + 44 + 444 + .......... is

(a) (4/9)[(10/9) (10ⁿ - 1) - n]         (b)  (10/9)(10ⁿ - 1) - n 

(c) 4/9(10ⁿ - 1) - n         (d)  None

Solution :

4 + 44 + 444 + ..........n terms

4(1 + 11 + 111 + ...............n terms)

= 4 x (9/9)(1 + 11 + 111 + ...............n terms)

= (4/9) (9 + 99 + 999 +...........n terms)

= (4/9) [(10 - 1) + (100 - 1) + (1000 - 1) +...............n terms]

= (4/9) [ (10 + 100 + 1000 + ...........n terms) - (1 + 1 + 1+ .........n terms)]

= (4/9)[10(10ⁿ - 1)/(10- 1) - n(1)]

= (4/9)[(10/9)(10ⁿ - 1) - n]

Problem 6 :

The sum of first 20 terms of a G.P is 244 times the sum of the first 10 terms. The common ratio is 

(a) ±√3         (b)  ±3       (c)  √3       (d)  None

Solution :

s_n = a(rⁿ - 1)/(r - 1)

s₂₀ = 244 s₁₀

a(r²⁰ - 1)/(r - 1) = 244 a(r¹⁰ - 1)/(r - 1)

(r²⁰ - 1) = 244(r¹⁰ - 1)

(r¹⁰ + 1) (r¹⁰ - 1) = 244 (r¹⁰ - 1)

r¹⁰ + 1 = 244

r¹⁰ = 243

r¹⁰ = 3⁵

r² = 3

r = √3

Problem 7 :

The product of 3 numbers in GP is 729 and the sum of squares is 819, the numbers are 

(a) 9, 3, 27         (b)  27, 3, 9       (c)  3, 9, 27       (d)  None

Solution :

Let a/r, a and ar be three terms.

Product of three terms = 729

(a/r) a (ar) = 729

a3 = 729

a³ = 9³

a = 9

Sum of the squares = 819

(a/r)² + a² + (ar)² = 819

a²(1/r² + 1 + r²) = 819

81(r⁴ + r² + 1)/r² = 819

(r⁴ + r² + 1)/r² = 819/81

(r⁴ + r² + 1)/r² = 91/9

9(r⁴ + r² + 1) = 91r²

Let r² = t

9(t² + t + 1) = 91t

9t² + 9t - 91t + 9 = 0

9t² - 82t + 9 = 0

(9t - 1)(t - 1) = 0

t = 1/9 and t = 1

r² = 1/9 and r = ±1/3

a = 9, ar = 9(1/3), ar² = 9(1/3)²

a = 9, ar = 3, ar² = 1

So, the sequence is 9, 3, 1.

Problem 8 :

Four geometric means between 4 and 972 are 

(a) 12, 36, 108, 324         (b)  12, 24, 108, 320

(c)  10, 36, 108, 320          (d)  None

Solution :

Four geometric means lies between

4, ___, ____, ____, ____, 972

a = 4, ar⁵ = 972

Applying the value of a, we get

r⁵ = 972/4

r5 = 243

r⁵ = 3⁵

r = 3

ar = 4(3) ==> 12

ar² = 4(3)² ==>  36

ar³ = 4(3)³ ==>  108

ar⁴ = 4(3)⁴ ==>  324

So, four geometric means are 12, 36, 108, 324.