Question

$5^{1+x}+5^{1-x}, \frac{a}{2}, 5^{2x}+5^{-2x}$ are in A.P., then the value of $a$ is:

• A. $a <\,12$
• B. $a \,\leq \,12$
• C. $a \,\geq \,12$
• D. None of these

Answer 1

Answer: (D)

Given $: 5^{1+ x }+5^{1- x }, \frac{ a }{2}, 5^{2 x }+5^{-2 x }$ are in A.P.
We know that if $a , b , c$ are in A.P. then $2 b = a + c$
$\therefore 2 \cdot \frac{ a }{2}=5^{1+ x }+5^{1- x }+5^{2 x }+5^{-2 x }$
$\Rightarrow a=5.5^{x}+5\left(5^{x}\right)^{-1}+\left(5^{x}\right)^{2}+\left(5^{x}\right)^{-2}$
Let $5^{x}=t $
$\therefore a=5 t+\frac{5}{t}+t^{2}+\frac{1}{t^{2}}$
$\Rightarrow a=t^{2}+\frac{1}{t^{2}}+5\left(t+\frac{1}{t}\right)$
$\Rightarrow a=\left(t+\frac{1}{t}\right)^{2}-2+5\left(t+\frac{1}{t}\right)$
Put $t+\frac{1}{t}=A$
$\therefore a = A ^{2}+5 A -2$ [ add & subtract $\left(\frac{b}{2 a}\right)^{2}]$
$\Rightarrow a =\left[ A ^{2}+5 A -\left(\frac{5}{2}\right)^{2}\right]+\left(\frac{5}{2}\right)^{2}-2$
$\Rightarrow a=\left(A-\frac{5}{2}\right)^{2}+\frac{17}{4}$
$ \Rightarrow a \geq \frac{17}{4}$