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If the sum of 12 th and 22 nd terms of an AP is 100 , then the sum of the first 33 terms of an $\mathrm{AP}$ is
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Verified Answer
The correct answer is:
1650
Here, $T_{12}=a+11 d$ and $T_{22}=a+21 d$
Since, $100=T_{12}+T_{22}$
$\begin{aligned}
& \therefore \quad 100=a+11 d+a+21 d \\
& \Rightarrow a+16 d=50 \quad \ldots(\mathrm{i})
\end{aligned}$
Now,
$\begin{aligned}
S_{33} & =\frac{33}{2}[2 a+(33-1) d] \\
& =33(a+16 d)=33 \times 50 \quad [From Eq. (i)] \\
& \approx 1650
\end{aligned}$
Thus, required sum be 1650 .
Since, $100=T_{12}+T_{22}$
$\begin{aligned}
& \therefore \quad 100=a+11 d+a+21 d \\
& \Rightarrow a+16 d=50 \quad \ldots(\mathrm{i})
\end{aligned}$
Now,
$\begin{aligned}
S_{33} & =\frac{33}{2}[2 a+(33-1) d] \\
& =33(a+16 d)=33 \times 50 \quad [From Eq. (i)] \\
& \approx 1650
\end{aligned}$
Thus, required sum be 1650 .
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