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CAT 2012 | Question: 3

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$\textbf{Concept:}\:\:\dfrac{14}{8} = 1 + 6$

Where $14 = \text{divident(d)},8 = \text{divisor(D)},1 = \text{Quotient(Q)},6 = \text{Remainder(R)}$

We can write like this $\text{divident(d) = divisor(D)} \times \text{Quotient(Q) + Remainder(R)}$

$\implies d = DQ + R$

$2$-digit numbers which leave a remainder of $6$ when divided by $8:14,22,30,\dots,T_{n}$

First term $a = 14,$ common difference $d = 8$

We know that $:T_{n} = a + (n-1)d$

Put $n = 11$ and we get $T_{11} = 14 + 10 \times 8 = 14 + 80 = 94$

$\therefore$ We got the AP series $14,22,30,\dots, 94$

Now, $S_{n}  = \dfrac{n}{2}(a + l)\:;\:\text{where}\:\: l = \text{last term}$

$\implies S_{11} = \dfrac{11}{2}(14 + 94)$

$\implies S_{11} = \dfrac{11}{2}\times 108 = 594.$

So, the correct answer is $(B).$
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