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8 October, 07:02

What is the 8th term in a geometric sequence where the first term is equal to 4 and the common ration is 2?

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  1. 8 October, 07:25
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    512

    Step-by-step explanation:

    1st term = a

    common ratio = r

    a*r with the exponent n-1

    a=4 r=2 n=8

    8th term 4 * (2 at the seventh power) = 512

    OR

    1st term : 4

    2nd term = 4*2=8

    3rd term 8*2=16

    4th term 12*2=32 ... 8th term 256*2=512
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