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algebra

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What number must be added to each of the numbers 0, 8, and 32 so that they form consecutive terms of a geometric sequence?

I don't understand what the question is asking first of all.

The answer is 4.

Help is much appreciated.

  • algebra -

    "A geometric sequence goes from one term to the next by always multiplying (or dividing) by the same value. So 1, 2, 4, 8, 16,... and 81, 27, 9, 3, 1, 1/3,... are geometric, since you multiply by 2 and divide by 3, respectively, at each step."

    -- http://www.purplemath.com/modules/series3.htm

  • algebra -

    A geometric sequence cannot start with 0, or all the terms will just stay 0.

    So, you want n such that each term is a constant multiple of the one before.

    (8+n)/(0+n) = (32+n)/(8+n)

    (8+n)^2 = n(32+n)
    64 + 16n + n^2 = 32n + n^2
    64 = 16n
    n=4

    So, the sequence starts out 4,12,36,... with each term 3x the previous one.

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