Question 14.

Using the arithmetic progression formula for the nth term, where

$x_n=a+\left(n-1\right)d$

Substituting the value for n and using that in the equation that is given, 

$2x_{6}+2x_{9}=x_{11}+x_{13}$, Then,$x_{100}$ equals

We get, $2\left(a+5d\right)+2\left(a+8d\right)=a+10d+a+12d$

$4a+26d=2a+22d$

$2a=-4d$

$a=-2d$

We are given, $x_5=-4$

$a+4d=-4$

Substituting the value for a in terms of d, 

$2d=-4$

$d=-2$

$a=4$

$x_{100}=a+99d$

$x_{100}=4-198=-194$