Suppose $$x_{1},x_{2},x_{3},...,x_{100}$$ are in arithmetic progression such that $$x_{5}=-4$$ and $$2x_{6}+2x_{9}=x_{11}+x_{13}$$, Then,$$x_{100}$$ equals

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, 
$$2d=-4$$
$$d=-2$$
$$a=4$$

$$x_{100}=a+99d$$
$$x_{100}=4-198=-194$$