Question

If a₁, a₂, a₃ …. are in G.P. a₁ + a_n = 66, a₂ a_n–1 = 128 and , then n will be-

• A. 5
• B. 6
• C. 7
• D. None of these

Answer 1

Answer: (B)

6

Answer 2

a₂a_n–1 = 128

Since a₁, a₂, a₃,……., a_n are in G.P.

̃ = = …….$\frac { a _ { n } } { a _ { n - 1 } }$

\ a₂ a_n–1 = a₁a_n = 128

Also a₁ + a₂ = 66

\ a_n – a₁ = $\sqrt { 66 ^ { 2 } - 4 \times 128 }$= 62

Solving a_n = 64, a₁ = 2

Also

\ $\frac { \mathrm { a } \left( 1 - \mathrm { r } ^ { \mathrm { n } } \right) } { 1 - \mathrm { r } }$= 126 (a_n = a₁r^n–1)

̃ 64 = 2r^n–1 \ r^n–1 = 32

= 63 ̃ 32r –1 = 63r – 63

̃ 31r = 62 ̃ r = 2

2^n–1 = 32 = 2⁵ ̃ n –1 = 5 ̃ n = 6