Question

What is the 6th term of the geometric sequence where a1 = 128 and a3 = 8? (1 point) 0.03125 0.0625 0.125 0.15625

Answer 1

I am pretty sure it's 0.125

Answer 2

6th term is 0.125

Explanation:

a1 = 128 and a3 = 8

To find nth term of geometric sequence we use formula

`a_n = a_1(r)^(n-1)`

a1 is the first term

r is the common ratio

a1= 128

Given a3 = 8, n=3, a1= 128

`a_n = a_1(r)^(n-1)`

`a_3 = 128(r)^(3-1)`

`8 = 128(r)^(2)`

Divide by 128 on both sides

1/16 = r^2

take square root on both sides

So r= 1/4

To find 6th term, replace r= 1/4

`a_6 = 128((1)/(4))^(6-1)`

a_6= 0.125