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

Answer 1

Answer: (A)

1020

Answer 2

Given, $x_{1}, x_{2}, \ldots, x_{n}$ 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.,
$a_{1}+a_{n}=a_{2}+a_{n-1}=a_{3}+a_{n-2}=\ldots$
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 E (i),
$x_{1}+x_{4}+x_{9}+x_{11}+x_{20}+x_{22} + x_{27}+x_{30}=272$
$\Rightarrow\left(x_{1}+x_{30}\right)+\left(x_{4}+x_{27}\right)+\left(x_{9}+x_{22}\right)+\left(x_{11}+x_{20}\right)=272$
$\Rightarrow 4\left(x_{1}+x_{30}\right)=272$
$\Rightarrow x_{1}+x_{30}=\frac{272}{4}=68$
$\therefore S_{30}=\frac{30}{2}\left(x_{1}+x_{30}\right)=15(68)=1020$