How many terms of the AP – 15,

Let n number of terms are needed to make the sum – 55.

Here, first term (a) = -15, common difference (d) = -13+15 = 2

$\because \quad$ Sum of $n$ terms of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow \quad-55=\frac{n}{2}[2(-15)+(n-1) 2] \quad\left[\because S_{n}=-55\right.$ (given)]

$\Rightarrow \quad-55=-15 n+n(n-1)$

$\Rightarrow \quad n^{2}-16 n+55=0$

$\Rightarrow \quad n^{2}-11 n-5 n+55=0$ [by factorisation method]

$\Rightarrow \quad n(n-11)-5(n-11)=0$

$\Rightarrow \quad(n-11)(n-5)=0$

$\therefore \quad n=5,11$

Hence, either 5 and 11 terms are needed to make the sum – 55.