Question

Let $x_1,x_2,\cdots ,x_n$ be in an $A.P$. If $x_1+x_4+x_9+x_{11}+x_{20}+x_{22}+x_{27}+x_{30}=272,$ then $x_1+x_2+x_3+\cdots +x_{30}$ is equal to

• A. 1020
• B. 1200
• C. 716
• D. 2720
• E. 2072

Answer 1

Answer: (A)

Given, $x_1,x_2,\dots ,x_n$ are in AP.
and $x_1+x_4+x_9+x_{11}+x_{20}+x_{22}+x_{27}+x_{30}=272$
We know that in an $AP$ , the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term i.e.,
$a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\dots$
If an $AP$ consists of 30 terms, then
$x_1+x_{30}=x_4+x_{27}=x_9+x_{22}=x_{11}+x_{20}$
From Eq. (i),
$x_1+x_4+x_9+x_{11}+x_{20}+x_{22}+x_{27}+x_{30}=272$
$\Rightarrow \left(x_1+x_{30}\right)+\left(x_4+x_{27}\right)+\left(x_9+x_{22}\right)+\left(x_{11}+x_{20}\right)=272$
$\Rightarrow 4\left(x_1+x_{30}\right)=272$
$\Rightarrow x_1+x_{30}=\frac{272}{4}=68$
$\therefore S_{30}=\frac{30}{2}\left(x_1+x_{30}\right)=15(68)=1020$