Question

Consider a sequence $\left\{a_n\right\}$ with $a_1=2$ and $a_n=\frac{a_{n-1}^2}{a_{n-2}}$for all $n\ge 3$, terms of the sequence being distinct. Given that $a_2$ and $a_5$ are positive integers and $a_5\le 162$ then the possible value(s) of $a_5$ can be

• A. $164$
• B. $62$
• C. $32$
• D. $2$

Answer 1

Answer: (A), (C)

$32$

Given:
$a_1=2\frac{a_n}{a_{n-1}}=\frac{a_{n-1}}{a_{n-2}}$
So the sequence is in $G.P.$
Let $a_2=x$,
Then the common ratio will be,
$r=\frac{x}{2}$
So,
$a_5\le 162\Rightarrow 2{\left(\frac{x}{2}\right)}^{5-1}\le 162\Rightarrow \frac{x^4}{8}\le 162\Rightarrow x\le 6$
As $a_2$ and $a_5$ are integers ,
For $x$ and $\frac{x^4}{8}$ to be integers,
$x=4,6[\because x\ne 2\text{as terms are distinct}]$
$a_5=\frac{4^4}{8},\frac{6^4}{8}\Rightarrow a_5=32,162$