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If the sum of a certain number of terms of the A.P. 25 , $22,19, \ldots \ldots \ldots$ is 116 . Find the last term.
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$a=25, d=22-25=-3$. let $n$ be the no. of terms
Sum $=116 ; \quad$ Sum $=\frac{n}{2}[2 a+(n-1) d]$
$116=\frac{n}{2}[50+(n-1)(-3)]$
or $232=n[50-3 n+3]=n[53-3 n]$
$=-3 n^2+53 n$
$\begin{aligned}
&\Rightarrow \quad 3 n^2-53+232=0 \\
&\Rightarrow \quad(n-8)(3 n-29)=0 \\
&\Rightarrow \quad n=8 \text { or } n=\frac{29}{3}, n \neq \frac{29}{3} \therefore n=8
\end{aligned}$
$\therefore \quad$ Now, $T_8=a+(8-1) d=25+7 \times(-3)$
= 25 $-$ 21
$\therefore$ Last term $=4$
Sum $=116 ; \quad$ Sum $=\frac{n}{2}[2 a+(n-1) d]$
$116=\frac{n}{2}[50+(n-1)(-3)]$
or $232=n[50-3 n+3]=n[53-3 n]$
$=-3 n^2+53 n$
$\begin{aligned}
&\Rightarrow \quad 3 n^2-53+232=0 \\
&\Rightarrow \quad(n-8)(3 n-29)=0 \\
&\Rightarrow \quad n=8 \text { or } n=\frac{29}{3}, n \neq \frac{29}{3} \therefore n=8
\end{aligned}$
$\therefore \quad$ Now, $T_8=a+(8-1) d=25+7 \times(-3)$
= 25 $-$ 21
$\therefore$ Last term $=4$
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