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If in an $\mathrm{AP}, S_n=q n^2$ and $S_m=q m^2$, where $S_r$ denotes the sum of $r$ terms of the $\mathrm{AP}$, then $S_q$ equals to
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Verified Answer
The correct answer is:
$q^3$
$q^3$
Since, $S_n=q n^2$ and $S_m=q m^2$
So, $S_1=q, S_2=4 q, S_3=9 q$
$\therefore T_1=S_1=q$
$T_2=S_2-S_1=4 q-q=3 q$
$T_3=S_3-S_2=9 q-4 q=5 q$
So, the series is $q, 3 q, 5 q \ldots$.
Hence, $a=q$ and $d=3 q-q=2 q$
$$
\begin{aligned}
&\therefore S_q=\frac{q}{2}[2 q+2 q(q-1)] \\
&=\frac{q}{2}\left[2 q+2 q^2-2 q\right]=q^3
\end{aligned}
$$
So, $S_1=q, S_2=4 q, S_3=9 q$
$\therefore T_1=S_1=q$
$T_2=S_2-S_1=4 q-q=3 q$
$T_3=S_3-S_2=9 q-4 q=5 q$
So, the series is $q, 3 q, 5 q \ldots$.
Hence, $a=q$ and $d=3 q-q=2 q$
$$
\begin{aligned}
&\therefore S_q=\frac{q}{2}[2 q+2 q(q-1)] \\
&=\frac{q}{2}\left[2 q+2 q^2-2 q\right]=q^3
\end{aligned}
$$
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