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If the middle term of the AP is 300 , then the sum of its first 51 terms is
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Verified Answer
The correct answer is:
15300
Given, number of terms, $n=51$
$\because n$ is odd.
$\therefore$ Middle term will be $\left(\frac{n+1}{2}\right)$ th term.
$\begin{aligned}
&=\left(\frac{51+1}{2}\right) \text { th term }=26 \text { th term } \\
\therefore \quad & T_{26}=300
\end{aligned}$
$a+25 d=300 \quad\left[\because T_{n}=a+(n-1) d\right]$
$\Rightarrow \quad T_{1}+25 d=300 \quad\left[\because T_{1}=a\right]$
$T_{1}=300-25 d$
and $T_{51}=a+50 d=T_{1}+50 d=300-25 d+50 d$
$=300+25 d$
$\therefore S_{51}=\frac{51}{2}[300-25 d+300+25 d]$
$\quad=\frac{51}{2}[600]=15300$
$\because n$ is odd.
$\therefore$ Middle term will be $\left(\frac{n+1}{2}\right)$ th term.
$\begin{aligned}
&=\left(\frac{51+1}{2}\right) \text { th term }=26 \text { th term } \\
\therefore \quad & T_{26}=300
\end{aligned}$
$a+25 d=300 \quad\left[\because T_{n}=a+(n-1) d\right]$
$\Rightarrow \quad T_{1}+25 d=300 \quad\left[\because T_{1}=a\right]$
$T_{1}=300-25 d$
and $T_{51}=a+50 d=T_{1}+50 d=300-25 d+50 d$
$=300+25 d$
$\therefore S_{51}=\frac{51}{2}[300-25 d+300+25 d]$
$\quad=\frac{51}{2}[600]=15300$
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