how many term of the A.P -6,-11/5,-5........are needed to the sum -25

Question

This much information i have

Answer 1

This is not an AP (arithmetic progression) because the terms T_n are not covered by the general requirement: T_n=T₁+(n-1)d where d is the common constant difference between terms. Also, the terms -6, -11/5, -5 are not in ascending or descending order. The correct order would be -6, -5, -11/5 (becoming less negative) or -11/5, -5, -6 (becoming more negative). The series as given isn't a GP (geometric progression) either because there's no common ratio.

Here's what I think the sequence should have been: -6, -11/2, -5 (a true AP, common difference ½).

First term a=-6 common difference d=½. General term t_n=a+(n-1)d=-6+½(n-1)=-13/2+n/2.

The AP can be written: a+(a+d)+(a+2d)+...+(a+(n-3)d)+(a+(n-2)d)+(a+(n-1)d).

This can be written: (a+a+(n-1)d)+(a+d+a+(n-2)d)+(a+2d+a+(n-3)d)+...

Each of these pair sums=2a+(n-1)d and there are n/2 pairs so the sum S_n is:

(n/2)(2a+(n-1)d). Plug in a=-6 and d=½: (n/2)(-12+½n-½)=(n/4)(n-25).

S_n=(n/4)(n-25)=-25, n(n-25)=-100,

n²-25n+100=0=(n-5)(n-20), so n=5 or 20. If we check the sum of the first 5 terms:

-6-11/2-5-9/2-4=-15-10=-25✔️