If x is real , the numbers $ 5^{1+x} + 5^{1-x} $ , a/2 , $ 25^x + 25^{-x} $ form an A.P. then to which interval must ' a ' belong to? A) [ 1 , 5 ] 
B) [ 2 , 5 ] 
C) [ 5 , 12 ] 
D) [ 12 , $ \infty $ ] 
My progress so far : 
Let  $ 5^{x} = y $
Since they are in AP so , 
=> $ 5y $ + $ 5\over y $ + $ y^2 $ + $ 1\over (y^2)  $ = a
=> $ {5y^3 + 5y + y^4 + 1 }\over (y)^2 $ = a 
As you can see , It becomes a mess . How do I solve it with a better approach . 

As they are in AP, we have
$$
a = 5^{1+x} + 5^{1-x} + 25^{x} + 25^{-x}.
$$
So say $5^x = t$, then,
$$
a = 5t + \frac{5}{t} + t^2 + \frac{1}{t^2}.
$$
Now 
$$
5t+\frac{5}{t} \geq 10
\qquad\text{and}\qquad
t^2+\frac{1}{t^2} \geq 2
$$
by the AM-GM inequality.
So $a \ge 12$, are the values that $a$ can take.
Hope that helps.
