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Question

For what values of the parameter a are there values of x such that  $$\displaystyle 5^{1+x}+5^{1-x}, a/2, 25^{x}+25^{-x}$$ are three consecutive terms of an A.P.?


Solution

Since $$\displaystyle 5^{1+x}+5^{1-x}, a/2, 25^{x}+25^{-x}$$ are in A. P., we have
$$\displaystyle 2a/2= 5^{1+x}+5^{1-x}+25^{x}+25^{-x},$$
Now put $$\displaystyle 5^{x}= t$$ so that $$\displaystyle t> 0,$$ we then have
$$\displaystyle a= 5t+5/t^{2}+1/t^{2}= +ive$$ ...(1)
$$\displaystyle = \left ( t^{2}+1/t^{2} \right )+5\left ( t+1/t \right )\geq 2+5.2= 12$$
because $$\displaystyle A.M.\geq G.M.$$
Also when $$\displaystyle a\geq 12$$ then (1) becomes
$$\displaystyle a= \left ( t+\frac{1}{t} \right )^{2}-2+5\left ( t+\frac{1}{t} \right )$$
$$\Rightarrow\displaystyle y^{2}+5y-\left ( a+2 \right )= 0$$
where $$\displaystyle y= t+\frac{1}{t}\geq 2\:as\:A.M.\geq G.M.$$
$$\displaystyle \therefore y= \frac{-5\pm \sqrt{25+4a+8}}{2}$$
$$\displaystyle y= \frac{-5\pm \sqrt{33+4a}}{2}\geq \frac{-5\pm \sqrt{33+48}}{2}$$
$$\displaystyle = \frac{-5\pm 9}{2}= \frac{9-5}{2}= 2 \therefore y\geq 2$$
$$\displaystyle \therefore y= t+\frac{1}{t}= 5^{x}+5^{-x}\geq 2$$
which we know is true and will give a solution for $$x.$$

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