Question

Let $S_k=\frac{1+2+3+....+k}{k}.$ If
$S_1^2+S_2^2+...+S_{10}^2=\frac{5}{12}A$, then $A$ is equal to:

• A. $156$
• B. $283$
• C. $301$
• D. $303$

Answer 1

The correct option is D $303$

$S_k=\frac{1+2+3+\cdots +k}{k}$
$\Rightarrow S_k=\frac{\frac{k(k+1)}{2}}{k}=\frac{k+1}{2}$
Now, $S_1^2+S_2^2+S_3^2+\cdots +S_{10}^2=\frac{5}{12}A$
$\Rightarrow \sum _{k=1}^{10}{S}_k^2=\frac{5}{12}A$
$\Rightarrow \sum _{k=1}^{10}{\left(\frac{k+1}{2}\right)}^2=\frac{5}{12}A$
$\Rightarrow \frac{2^2+3^2+\cdots +{11}^2}{4}=\frac{5}{12}A$
$\Rightarrow 1^2+2^2+3^2+\cdots +{11}^2-1^2=\frac{5}{3}A$
$\Rightarrow \frac{\left(11\right)\cdot (11+1)\cdot (2\cdot 11+1)}{6}-1=\frac{5}{3}A$
$\Rightarrow 506-1=\frac{5}{3}A\Rightarrow A=303$