Correct Answer - Option 4 : 303
From question
\({S_k} = \frac{{1 + 2 + 3 + \ldots + R}}{k}\)
\(\Rightarrow {S_k} = \frac{{k\left( {k + 1} \right)}}{{2k}}\)
\(\therefore {S_k} = \frac{{k + 1}}{2}\)
So,
\(S_k^2 = {\left( {\frac{{k + 1}}{2}} \right)^2} = \frac{1}{4}{\left( {k + 1} \right)^2}\) ----(1)
Now,
\(\Rightarrow \frac{5}{{12}}A = S_1^2 + S_2^2 + S_3^2 + \ldots S_{10}^2 = \mathop \sum \limits_{k = 1}^{10} 1S_k^2\)
\(\Rightarrow \frac{5}{{12}}A = \frac{1}{4}\mathop \sum \limits_{k = 1}^{10} {\left( {k + 1} \right)^2} = \frac{1}{4}\left[ {{2^2} + {3^2} + {4^2} + \ldots {{11}^2}} \right]\)
\(\because \left[ \sum {{n}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right]\)
\(\Rightarrow \frac{5}{{12}}A = \frac{1}{4}\left[ {\frac{{11 \times \left( {11 + 1} \right)\left( {2 \times 11 + 1} \right)}}{6} - {1^2}} \right]\)
\(\Rightarrow \frac{5}{{12}}A = \frac{1}{4}\left[ {\frac{{11 \times 12 \times 23}}{6} - 1} \right]\)
\(\Rightarrow \frac{5}{{12}}A = \frac{1}{4}\left[ {\left( {22 \times 23} \right) - 1} \right]\)
\(\Rightarrow \frac{5}{{12}}A = \frac{1}{4}\left[ {506 - 1} \right]\)
\(\Rightarrow \frac{5}{{12}}A = \frac{1}{4}\left[ {505} \right]\)
\(\Rightarrow \frac{A}{3} = 101\)
∴ A = 303
