A JEE Advanced Progression!

Here a solution If $p=0$ then $2^{-x}=5$ means $-x=log_{2}5$ which is unique.

if $p\neq0$ , put $2^x=t$ , $t>0$ . implies

$pt + \frac{1}{t} =5$

implies $pt^2-5t+1=0$

Therefore for unique solution , discriminant $=0$ and

$\frac{-b}{2a}>0$

implies $25-4p=0$

implies $p= \frac{25}{4}$ and $\frac{5}{2p}>0$ means $p>0$

hence $p$ belongs to { $0$ , $\frac{25}{4}$ } implies $2$ values

Since $2,\alpha_1,\alpha_2,...,\alpha_{20},6$ are in H.P.

Therefore $\frac{1}{2}$ , $\frac{1}{\alpha_1}$ ,..., $\frac{1}{\alpha_{20}}$ , $\frac{1}{6}$ are in A.P.

Let common of this A.P be $d$ . Therefore $\frac{1}{6}$ = $\frac{1}{2}$ + $21d$

implies $d$ = - $\frac{1}{63}$ .

on solving we get $\alpha_{18}$ = $\frac{14}{3}$ .

Similarly $2, \beta_1, \beta_2,..., \beta_{20},6$ are in A.P.

$6=2+21d$

$d$ = $\frac{4}{21}$

on solving $\beta_3$ = $\frac{18}{7}$

Therefore $\alpha_{18}\beta_3$ = $12$ .

So the answer is $12$ ...