Question

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

• A. 6
• B. 5
• C. 4
• D. 3
• E. 2

Answer 1

Answer: (D)

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