સમાંતર શ્રેણી $25,22,19, \ldots \ldots .$ નાં નિશ્ચિત સંખ્યાના શરૂઆતના પદનો સરવાળો $116$ હોય તો છેલ્લું પદ શોધો. 

Let the sum of $n$ terms of the given $A.P.$ be $116$

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

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

$\therefore S_{n}=\frac{n}{2}[2 \times 25+(n-1)(-3)]$

$\Rightarrow 116=\frac{n}{2}[50-3 n+3]$

$\Rightarrow 232=n(53-3 n)=53 n-3 n^{2}$

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

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

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

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

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

Howerer, $n$ cannot be equal to $\frac{29}{3}$ therefore, $n=8$

$\therefore a_{8}=$ Last term $=a+(n-1) d=25+(8-1)(-3)$

$=25+(7)(-3)=25-21$

$=4$

Thus, the last term of the $A.P.$ is $4.$