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