Question

Identify the 8th term of the geometric sequence in which a3 = 16 and a6 = −128.

Answer 1

Answer:A. - 512

Step-by-step explanation:

Here is how I solved this problem:

-First take -128 and divide it by 16

(You should get -8)

-Then take -8 and cube root it

(You should get 2)

You have found your multiplier, which is 2.

After this you can start from -128 and multiply by 2, 2 times to get to the 8th term which is -512.

me know if you have any further questions.