geometric sequence problem 2?

Ans:

For the Geometric Progression, the first term is a and the common ratio is r.

T(1) + T(2) = 36
a + ar = 36
a(1 + r) = 36 …… [1]

T(1) × T(3) = 9 × T(2)
a × ar² = 9 × ar
a²r² = 9ar
As a ≠ 0 and r ≠ 0 : ar = 9
a = 9/r …… [2]

Substitute [2] into [1] :
(9/r) (1 + r) = 36
9 + 9r = 36r
27r = 9
r = 1/3

Substitute r = 1/3 into [2] :
a = 9/(1/3)
a = 27

Sum of the first 8 terms, S(8)
= a [1 - rⁿ] / [1 - r]
= 27 × [1 - (1/3)⁸] / [1 - (1/3)]
= 3280/81

The answer: b. 3280/81

Let r be the common ratio.
a₂ = a₁r
a₃ = a₁r²

"The sum of the first two terms is 36"
a₁ + a₂ = 36
a₁ + a₁r = 36
a₁(1+r) = 36

"the product of the first and third terms is 9 times the second term"
a₁a₃ = 9a₂
a₁(a₁r²) = 9(a₁r)
a₁²r² = 9a₁r
a₁r = 9
r = 9/a₁

a₁(1+r) = 36
a₁(1+9/a₁) = 36
a₁ + 9 = 36
a₁ = 27 = 3³
r = 9/a₁ = 3²/3³= ⅓

a₈ = a₁r⁷ = 27(⅓)⁷ = 3³/3⁷ = 1/3⁴
∑ = 3³ + 3² + 3¹ + ... + 1/3⁴
 = 3⁷/3⁴ + 3⁶/3⁴ + ... + 3⁰/3⁴
 = 3280/3⁴

a+ar = 36
a^2r^2 = 9ar
ar = 9
a = 27
r = 1/3
S_8 = 27[1-(1/3)^8]/(2/3)
= 3280/81
Answer b)

a1 + a2 = 36
a1 + a1 * r = 36

a1 * a3 = 9a2
a1 * a1 * r^2 = 9a1 * r
a1 * r = 9

=>
a1 + a1 * r = 36
a1 * r = 9

a1 = 27, r = 1/3

S8 = a1 (1 - r^8)/(1 - r)
S8 = 27 (1 - (1/3)^8)/(1 - 1/3)

b. 3280/81

Let the first number be 'a' the common ratio be 'r'.
2nd term is 'ar' and 3rd term is 'ar^2' .
Given a + ar = 36 and a x ar^2 = 9 x ar or, ar = 9.
a +9 = 36 or a = 36 -9 = 27.
ar = 9 or, 27 x r = 9 or r = 9/27 = 1/3.
Sum of a GP series is given by the formula, S = a(1 - r^n)/(1 - r)
= 27[1 - (1/3)^8] / (1 - 1/3) = 27(1- 1/6561)/ 2/3 = (27 x 6560 x 3)/(2 x 6561)
= 531,360 / 13,122 = 3280/81. So, option (b) is the correct answer. (Ans.)

First term = k
Second term = kn
Third term = kn^2
k*kn^2 = 9kn
kn = 9
k + kn = 36
k = 27
n = 1/3
Terms 27, 9, 3, 1, 1/3, 1/9, 1/27, 1/81 = 40 40/81
B