Let S_k= 1 + 2 + 3+… +k k. If S_1^2+S_2^2+… +S_10^2= 5 12A, then A is equal to :

$${S}_{k}=\frac{1+2+3+\ldots +k}{k}=\frac{k(k+1)}{2k}=\frac{k+1}{2}.........(i)$$Sk​=k1+2+3+…+k​=2kk(k+1)​=2k+1​.........(i)

$\therefore {S}_{1}^{2}+{S}_{2}^{2}+....+{S}_{10}^{2}={(\frac{1+1}{2})}^{2}+{(\frac{2+1}{2})}^{2}+...+{(\frac{10+1}{2})}^{2}$∴S12​+S22​+....+S102​=(21+1​)2+(22+1​)2+...+(210+1​)2 (using equation $(i)$(i))

$$={(\frac{2}{2})}^{2}+{(\frac{3}{2})}^{2}+...+{(\frac{11}{2})}^{2}$$=(22​)2+(23​)2+...+(211​)2

$$=\frac{1}{{2}^{2}}[{2}^{2}+{3}^{2}+...+{11}^{2}]$$=221​[22+32+...+112]

$$=\frac{1}{4}[({1}^{2}+{2}^{2}+{3}^{2}+...+{11}^{2})-{1}^{2}]$$=41​[(12+22+32+...+112)−12]

$=\frac{1}{4}[\frac{11\times 12\times 23}{6}-1]$=41​[611×12×23​−1] (using the identity \displaystyle \sum {n}^{2}_{n=1}^{n=n}=\frac{(n)(n+1)(2n+1)}{6})

$$=\frac{1}{4}\times 505$$=41​×505

Comparing with the given condition in question, we get

$$\frac{5}{12}A=\frac{505}{4}$$125​A=4505​

$\Rightarrow A=\frac{505\times 3}{5}=303$⇒A=5505×3​=303.