If the sum of a certain number of terms of the

To Find: Last term of the AP.

Let the number of terms be n.

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow \frac{n}{2}[54+(n-1)(-3)]=-30$

$\Rightarrow \mathrm{n}[54-3 \mathrm{n}+3]=-60$

$\Rightarrow 3 n^{2}-57 n-60=0$

$\Rightarrow n=\frac{57 \pm 63}{6}$

Either n = 20 or n = - 1 (n cannot be negative)

Therefore n = 20

Also,

$S=\frac{n}{2}(a+l)$ where l is the last term.

$\Rightarrow-30=\frac{20}{2}(27+l)$

$\Rightarrow-30=270+101$

$\Rightarrow-\frac{300}{10}=l$

$\Rightarrow 1=-30$

The last term is - 30.