Question

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.

Answer 1

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.

Answer 2

I can’t really understand why need to be 78😓