Answer: (3)

Solution

$\therefore 1 + 2 + 3 + \ldots + k = \frac{k(k+1)}{2}$

$\therefore S_k = \frac{k(k+1)}{2 \cdot 2k} = \frac{k+1}{2}$ Wait, actually $S_k = \frac{k(k+1)/2}{2k}$? No.

$S_k = \frac{1+2+3+\ldots+k}{2} = \frac{k(k+1)}{4}$? No, looking at the image again: $S_k = \frac{1+2+\ldots+k}{2}$.

Actually the image says $S_k = \frac{1+2+3+\ldots+k}{2}$.

$\Rightarrow \frac{5}{12}A = \frac{1}{4}[1^2 + 2^2 + \ldots + 11^2 - 1]$

$= \frac{1}{4}\left[\frac{11(11+1)(2 \times 11+1)}{6} - 1\right]$

$= \frac{1}{4}\left[\frac{11 \times 12 \times 23}{6} - 1\right] = \frac{1}{4}[505]$

$A = \frac{505 \times 12}{4 \times 5} = 303$