Posted by **azmeerahim** on Saturday, February 9, 2013 at 12:28pm.

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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**Steve**, Saturday, February 9, 2013 at 2:28pm
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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