If the sum of a certain number of terms of the AP 25, 22, 19, ... is 116.

The given A.P. is 25,.22,.19.....

Here, $a=25, d=22-25=-3$

$S_{n}=116$

$\Rightarrow \frac{n}{2}[2 a+(n-1) d]=116$

$\Rightarrow n[2 \times 25+(n-1)(-3)]=232$

$\Rightarrow 50 n-3 n^{2}+3 n=232$

$\Rightarrow 3 n^{2}-53 n+232=0$

$\Rightarrow 3 n^{2}-29 n-24 n+232=0$

$\Rightarrow n(3 n-29)-8(3 n-29)=0$

$\Rightarrow(3 n-29)(n-8)=0$

$\Rightarrow n=\frac{29}{3}$ or 8

Since $n$ cannot be a fraction, $n=8$.

Thus, the last term :

$a_{n}=a+(n-1) d$

$\Rightarrow a_{8}=25+(8-1) \times(-3)$

$\Rightarrow a_{8}=4$