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Q. An $A.P.$ consists of $23$ terms. If the sum of the $3$ terms in the middle is $141$ and the sum of the last $3$ terms is $261$, then the firm is

KEAMKEAM 2015Sequences and Series

Solution:

Let the first term and common difference of AP be $a$ and $d$, respectively. Then, $a_{11}+a_{12}+a_{13}=141$
$\Rightarrow 3 a+33 d=141$
$\Rightarrow a+11 d=47\,...(i)$
Also, $ a_{21}+a_{22}+a_{23}=261$
$\Rightarrow 3 a+63 d=261$
$\Rightarrow a+21 d=87 \,......(ii) $
From Eqs. (i) and (ii), $a=3$