JEE Advance - Mathematics (2011 - Paper 1 Offline - No. 6)
Let $${{a_1}}$$, $${{a_2}}$$, $${{a_3}}$$........ $${{a_{100}}}$$ be an arithmetic progression with $${{a_1}}$$ = 3 and $${S_p} = \sum\limits_{i = 1}^p {{a_i},1 \le } \,p\, \le 100$$. For any integer n with $$1\,\, \le \,n\, \le 20$$, let m = 5n. If $${{{S_m}} \over {{S_n}}}$$ does not depend on n, then $${a_{2\,}}$$ is
Answer
9
Explanation
It is given that a1, a2, a3, ......, a100 is an A.P.
$${a_1} = 3,\,{S_p} = \sum\limits_{i = 1}^p {{a_1}} ,\,1 \le p \le 100$$
$${{{S_m}} \over {{S_n}}} = {{{S_{5n}}} \over {{S_n}}} = {{{{5n} \over 2}(6 + (5n - 1)d)} \over {{n \over 2}(6 - d + nd)}}$$
$${{{S_m}} \over {{S_n}}}$$ is independent of n of $$6 - d = 0 \Rightarrow d = 6$$.
Therefore, $${a_2} = {a_1} + d = 3 + 6 = 9$$.
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