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. $162$
• B. $64$
• C. $32$
• D. $2$

Answer 1

Answer: (A), (C)

The correct options are
A $162$
C $32$

$a_1=2$ and $\frac{a_n}{a_{n-1}}=\frac{a_{n-1}}{a_{n-2}}$
Hence, $a_1,a_2,a_3,a_4,a_5,...$ are in G.P.

Let $a_2=x$. Then for $n=3$, we have
$\Rightarrow \frac{a_3}{a_2}=\frac{a_2}{a_1}$
$\Rightarrow a_2^2=a_1{a}_3$
$\Rightarrow a_3=\frac{x^2}{2}(\because a_1=2)$

i.e., the sequence is $2,x,\frac{x^2}{2},\frac{x^3}{4},\frac{x^4}{8},...$ with common ratio, $r=\frac{x}{2}$

Given, $a_5=\frac{x^4}{8}\le 162$
$\Rightarrow x^4\le 1296$
$\Rightarrow x\le 6$
Also, $x$ and $\frac{x^4}{8}$ are integers.
So, if $x$ is even, then only $\frac{x^4}{8}$ will be an integer.

Hence, possible values of $x$ are $4$ and $6$. $(x\ne 2,$ as terms are distinct)

Hence, possible value of $a_5=\frac{4^4}{8},\frac{6^4}{8}$
$\Rightarrow a_5=32,162$