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^{2}_{k} = \frac{5}{12}A$
$ \sum^{10}_{K=1} \left(\frac{K +1}{2}\right)^{2} = \frac{2^{2}+3^{2} +--+11^{2}}{4} = \frac{5}{12} A $
$ \frac{11\times12\times23}{6} - 1 = \frac{5}{3} A $
$ 505 = \frac{5}{3}A, A =303 $