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Q. Let $a_1, a_2, a_3$, ... be an arithmetic progression with nonzero common difference. It is given that $\sum^{12}_{i = 4} a_i = 63$ and $a_k = 7 $ for some k . Then the value of k is

UPSEEUPSEE 2019

Solution:

Given, $a_1, a_2, a_3, ... $ be an arithmetic progression
$ a_{1} +a_{2} +a_{3} + ...a_{n} = \frac{n}{2}\left(a_{1}+a_{n}\right)$
$ \Sigma_{i=4}^{12} a_{i} = 63$
$ \Rightarrow a_{4} +a_{5} +a_{6} + ... +a_{12} = 63 $
$\Rightarrow \frac{9}{2}\left(a_{4} +a_{12}\right) = 63$
$\Rightarrow a_{4} +a_{12} =14 $
$ \Rightarrow a +3d +a +11d = 14 $
$\Rightarrow 2a +14 d = 14 $
$\Rightarrow a+7d = 7 $
$ \Rightarrow a_{8} = 7 $
$\Rightarrow k = 8$