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

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Solution

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

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