A) 303
B) 283
C) 301
D) 156
Correct Answer: A
Solution :
\[{{S}_{k}}=\frac{k(k+1)}{2k}=\frac{k+1}{2}\] Now,\[\sum\limits_{k=1}^{10}{{{({{S}_{k}})}^{2}}=\frac{5}{12}A\Rightarrow \sum\limits_{k=1}^{10}{{{\left( \frac{k+1}{2} \right)}^{2}}}=\frac{5}{12}A}\] \[\Rightarrow \]\[\frac{1}{4}({{2}^{2}}+{{3}^{2}}+....+{{11}^{2}})=\frac{5}{12}A\] \[\Rightarrow \]\[\frac{1}{4}\left( \frac{11\times 12\times 23}{6}-1 \right)=\frac{5}{12}A\] \[\Rightarrow \]\[\frac{505}{4}=\frac{5}{12}A\]\[\Rightarrow \]\[A=303\]
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