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The sixth term of an arithmetic progression is 265 and the sum of the first 5 terms is 1445. Find the minimum value of n so that the sum of the first n terms is negative.

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    a+5d = 265
    5/2 (2a+4d) = 1445

    (a,d) = (305,-8)

    So, we want

    n/2 (2a+(n-1)d) < 0
    n/2 (2*305 + (n-1)(-8)) < 0
    n > 77.25

    so, you need 78 terms before the sum is negative

    makes sense, since the 39th term is negative, so you need another 39 terms to offset the first 39.

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