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John wants to calculate the sum of a geometric series with 10 terms, where the 10th term is 5 859 37... ... series with 10 terms, where the 10th term is 5 859 375 and the common ration is 5/3. John solved the problem by considering another ... show more ... Using John's method, a = 5 859 375, r = 3/5.

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  1. well, since T10 = ar^9
    a(5/3)^9 = 5859375 = 3*5^9
    So, a = 3^10 = 59049

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