Question

What is the 6th term of the geometric sequence where a1 = 128 and a3 = 8?

0.03125

0.0625

0.125

0.15625

Answer 1

The common ratio is dividing by four; 128 divided by 4 gives 32, divided by 4 gives 8.

So, divide by 4 three more times and you get .125

Answer 2

You must use the formula \[a _{n}=a _{1}r ^{n-1}\] They give you a1 and they give you a3, so n=3, and you can find r. Then go back and substitute n=6 and the value you obtained for r to find \[a _{6}\]