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. $a\le 12$
• C. $a\ge 12$
• D. None of these

Answer 1

Answer: (D)

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