Question:
Let $S_{k}=\frac{1+2+3+\ldots .+k}{k}$. If
$\mathrm{S}_{1}^{2}+\mathrm{S}_{2}^{2}+\ldots .+\mathrm{S}_{10}^{2}=\frac{5}{12} \mathrm{~A}$, then $\mathrm{A}$ is equal to :
Correct Option: 1
Solution:
$\mathrm{S}_{\mathrm{K}}=\frac{\mathrm{K}+1}{2}$
$\Sigma \mathrm{S}_{\mathrm{k}}^{2}=\frac{5}{12} \mathrm{~A}$
$\sum_{\mathrm{K}=1}^{10}\left(\frac{\mathrm{K}+1}{2}\right)^{2}=\frac{2^{2}+3^{2}+--+11^{2}}{4}=\frac{5}{12} \mathrm{~A}$
$\frac{11 \times 12 \times 23}{6}-1=\frac{5}{3} \mathrm{~A}$
$505=\frac{5}{3} \mathrm{~A}, \quad \mathrm{~A}=303$
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