Correct Answer: 6974

Solution : Given: The series is
$\begin{aligned} & S 1=2,9,16, \ldots \ldots, 632 \\ & S 2=7,11,15, \ldots \ldots, 743\end{aligned}$.
From series $S1$ and $S2$, first common term = 23, second common term = 51.
⇒ $d=51–23=28$
Use the formulas,
$a_n=a+(n–1)d$,
$S_n=\frac{n}{2}[2a+(n–1)d]$,
where $a_n$, $n$, $a$, $d$, and $S_n$ are the $n$th term in the sequence, the number of terms, the first term in the sequence, the common difference, and the sum.
⇒ $a_n=a+(n–1)d\leq 632$
⇒ $23+(n–1)28\leq 632$
⇒ $(n–1)28\leq 632–23$
⇒ $(n–1)28\leq 609$
⇒ $n–1\leq 21.75$
⇒ $n\leq 22.75$
So, the value of $n=22$.
The sum of all the common terms between the given series S1 and S2
= $\frac{22}{2}[2\times 23+(22–1)\times28]$.
= $11\times [46+21\times28]$
= $11\times [46+588]$
= $11\times 634$
= $6974$
Hence, the correct answer is 6974.