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3 October, 01:11

Find the sum of the geometric series for which a1 = 39, r = 1/3, n = 8 to the nearest ten-thousandth

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  1. 3 October, 01:12
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    58.4911 to the nearest ten thousandth.

    Step-by-step explanation:

    Sn = a1 * (1 - r^n) / (1 - r)

    S8 = 39 * (1 - (1/3) ^8) / (1 - 1/3)

    = 39 * 1.49977138

    = 58.49108382.
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