Question

How many terms are needed in the series $-15$, $-12$, $-9$, $\dots \dots$ so that their sum is 18?

Answer 1

Let $a$ be the first term and $d$ be their common difference of the AP.

Then, Sum to $n$ terms of an AP $=\frac{n}{2}(2a+(n-1\left)d\right)$

Here, in the given AP,. $a=-15;d=-12+15=3$

So,$S_n=18$

$=>\frac{n}{2}\left(2\right(-15)+(n-1\left)\right(3\left)\right)=18$

$=>\frac{n}{2}(-30+3n-3)=18$

$=>\frac{n}{2}(3n-33)=18$

$=>\frac{n}{2}(n-11)=6$

$=>n(n-11)=12\times 1$

$=>n=12$