JEE Main 12 Jan 2019 Morning Question 10
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 [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 303
B) 283
C) 301
D) 156
Show Answer
Answer:
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}{{{( \frac{k+1}{2} )}^{2}}}=\frac{5}{12}A} $
$ \Rightarrow $ $ \frac{1}{4}(2^{2}+3^{2}+….+11^{2})=\frac{5}{12}A $
$ \Rightarrow $ $ \frac{1}{4}( \frac{11\times 12\times 23}{6}-1 )=\frac{5}{12}A $
$ \Rightarrow $ $ \frac{505}{4}=\frac{5}{12}A $
$ \Rightarrow $ $ A=303 $
