Math
posted by azmeerahim .
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.

a+5d = 265
5/2 (2a+4d) = 1445
(a,d) = (305,8)
So, we want
n/2 (2a+(n1)d) < 0
n/2 (2*305 + (n1)(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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