Question

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

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

Answer 1

The correct option is A $1020$

Given $AP:x_1,x_2,....x_n$

Let $a$ be first term and $d$ be common difference.

$x_1+x_4+x_9+x_{11}+x_{20}+x_{22}+x_{27}+x_{30}=272$

$\Rightarrow a+a+3d+a+8d+a+10d+a+19d+a+21d+a+26d+a+29d=272$

$\Rightarrow 8a+116d=272$

$\Rightarrow 2a+29d=68$

Now,

$S_n=x_1+x_2+x_3+......+x_{30}$

$=\frac{n}{2}(x_1+x_{30})$

$=\frac{30}{2}(a+a+29d)$

$=15(2a+29d)$

$=15\times 68$

$=1020$

Hence, A is the correct option.