Question

What is the sum of the geometric series in which a1 = −4, r = 2, and an = −128?

Hint: r ≠ 1, where a1 is the first term and r is the common ratio.
Sn = 1,020
Sn = −252
Sn = 43,690
Sn = −1,020

Answer 1

Given:
The geometric series is defined by
a = a₁ = -4
r = 2

The n-th term is

`a_(n) = ar^(n-1)`
Because the n-th term is -128, therefore
-4(2ⁿ⁻¹) = -128
2ⁿ⁻¹ = 32 = 2⁵
n-1 = 5
n = 6

The sum of the first n terms is

`S_(n) = (a(1-r^(n)))/(1-r)`

Therefore

`S_(n) = (-4(1-2^(6)))/(1-2) =4(1-64) = -252`

Answer: -252