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ASHA 777 [7]
3 years ago
14

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

Mathematics
1 answer:
Katen [24]3 years ago
3 0

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.

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Read more on Brainly.com - brainly.com/question/3151349#readmore

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