Let S k = 1   +   2   +   3 + … + k k . If S 1 2 + S 2 2 + … + S 10 2 = 5 12…

$S_k=\frac{1+2+3+\dots +k}{k}=\frac{k\left(k+1\right)}{2k}=\frac{k+1}{2} .........\left(i\right)$

$\therefore S_1^2+S_2^2+....+S_{10}^2={\left(\frac{1+1}{2}\right)}^2+{\left(\frac{2+1}{2}\right)}^2+...+{\left(\frac{10+1}{2}\right)}^2$  (using equation $\left(i\right)$)

$={\left(\frac{2}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2+...+{\left(\frac{11}{2}\right)}^2$

$=\frac{1}{2^2}\left[2^2+3^2+...+{11}^2\right]$

$=\frac{1}{4}\left[\left(1^2+2^2+3^2+...+{11}^2\right)-1^2\right]$

$=\frac{1}{4}\left[\frac{11\times 12\times 23}{6}-1\right]$ (using the identity $\underset{n=1}{\overset{n=n}{\sum n^2}}=\frac{\left(n\right){\displaystyle \left(n+1\right)}{\displaystyle \left(2n+1\right)}}{6}$)

$=\frac{1}{4}\times 505$

Comparing with the given condition in question, we get

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

$\Rightarrow A=\frac{505\times 3}{5}=303$.