The numbers $$-16, 2^{x+3} -2^{2x -1}-16, 2^{2x-1} + 16$$ are in an arithmetic progression. Then x equals _________.

We know, if a,b,c are in A.P. then 2b=a+c

So, $$2\cdot(2^{x+3}-2^{2x-1}-16)=2^{2x-1}+16+(-16)$$

or, $$2\cdot(2^{x+3}-2^{2x-1}-16)=2^{2x-1}$$

Multiplying both sides with 2,

$$2^2\cdot(2^{x+3}-2^{2x-1}-16)=2^{2x}$$

or, $$2^{x+5}-2^{2x+1}-64=2^{2x}$$

or, $$2^{x+5}-2^{2x}\cdot2-64=2^{2x}$$

or, $$2^{x+5}-64=2^{2x}\cdot3$$

or, $$2^{2x}\cdot3-2^{x+5}+64=0$$

or, $$\left(2^{2x}\right)\cdot3-\left(2^x\right)\cdot32+64=0$$

let, $$2^x=z$$

So, $$3z^2-32z+64=0$$

or, $$3z^2-24z-8z+64=0$$

or, $$3z\left(z-8\right)-8\left(z-8\right)=0$$

or, $$\left(z-8\right)\left(3z-8\right)=0$$

or, $$z=8$$ or $$z=\dfrac{8}{3}$$

Considering only integral values of $$z$$, $$z=8$$

So, $$2^x=8$$

or, $$x=3$$.