Question

$x_1,x_2,x_3,....$ are in A.P.
If $x_1+x_7+x_{10}=-6$ and $x_3+x_8+x_{12}=-11$, then $x_3+x_8+x_{22}=?$

• A. $-21$
• B. $-15$
• C. $-18$
• D. $-31$

Answer 1

The correct option is A $-21$

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

Since, $x_1+x_7+x_{10}=-6$

$\Rightarrow \left(a\right)+(a+6d)+(a+9d)=-6$

$\Rightarrow 3a+15d=-6\to \text{eqn}\left(1\right)$

Also, $x_3+x_8+x_{12}=-11$

$\Rightarrow (a+2d)+(a+7d)+(a+11d)=-11$

$\Rightarrow 3a+20d=-11\to \text{eqn}\left(2\right)$

Solving eqn (1) and (2), we get $a=3$ and $d=-1$

Now, $x_3+x_8+x_{22}=3a+30d=-21$