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If the sum of a certain number of terms of the A.P. $25,22,19, \ldots$ is 116 then the last term is
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The correct answer is:
4
$a=25, d=22-25=-3 \text {. }$
Let $n$ be the no. of terms
$\begin{array}{l}
\text { Sum }=116 ; \\
\text { Sum }=\frac{n}{2}[2 a+(n-1) d] \\
116=\frac{n}{2}[50+(n-1)(-3)] \\
\text { or } 232=n[50-3 n+3]=n[53-3 n]=-3 n^{2}+53 n \\
\Rightarrow 3 n^{2}-53+232=0 \\
\Rightarrow(n-8)(3 n-29)=0 \\
\Rightarrow n=8 \text { or } n=\frac{29}{3}, \\
n \neq \frac{29}{3} \\
\therefore n=8 \\
\therefore \text { Now, } \mathrm{T}_{8}=a+(8-1) d=25+7 \times(-3)=25-21=4 \\
\therefore \text { Last term }=4
\end{array}$
Let $n$ be the no. of terms
$\begin{array}{l}
\text { Sum }=116 ; \\
\text { Sum }=\frac{n}{2}[2 a+(n-1) d] \\
116=\frac{n}{2}[50+(n-1)(-3)] \\
\text { or } 232=n[50-3 n+3]=n[53-3 n]=-3 n^{2}+53 n \\
\Rightarrow 3 n^{2}-53+232=0 \\
\Rightarrow(n-8)(3 n-29)=0 \\
\Rightarrow n=8 \text { or } n=\frac{29}{3}, \\
n \neq \frac{29}{3} \\
\therefore n=8 \\
\therefore \text { Now, } \mathrm{T}_{8}=a+(8-1) d=25+7 \times(-3)=25-21=4 \\
\therefore \text { Last term }=4
\end{array}$
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