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. 303
• B. 283
• C. 156
• D. 301

Answer 1

Answer: (A)

$S_K=\frac{K+1}{2}$
$\sum S_k^2=\frac{5}{12}A$
${\sum }_{K=1}^{10}{\left(\frac{K+1}{2}\right)}^2=\frac{2^2+3^2+--+11^2}{4}=\frac{5}{12}A$
$\frac{11\times 12\times 23}{6}-1=\frac{5}{3}A$
$505=\frac{5}{3}A,A=303$