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 $n \geq 3$, terms of the sequence being distinct. Given that $a _2$ and $a _5$ are positive integers and $a _5 \leq 162$ then the possible value(s) of $a _5$ can be

• A. 2
• B. 32
• C. 64
• D. 162

Answer 1

Answer: (B), (D)

Given $a_1=2 ; \frac{a_n}{a_{n-1}}=\frac{a_{n-1}}{a_{n-2}} \Rightarrow a_1, a_2, a_3, a_4, a_5, \ldots \ldots$ in G.P.
Let $a _2= x$ then for $n =3$ we have
$\frac{a_3}{a_2}=\frac{a_2}{a_1} \Rightarrow a_1^2=a_1 a_3 \Rightarrow a_3=\frac{x^2}{2}$
i.e. $ 2, x, \frac{x^2}{2}, \frac{x^3}{4}, \frac{x^4}{8}, \ldots \ldots .$. with common ratio $r=\frac{x}{2}$
given $\frac{x^4}{8} \leq 162 \Rightarrow x^4 \leq 1296 \Rightarrow x \leq 6$
Also $x$ and $\frac{x^4}{8}$ are integers $\Rightarrow x$ must be even then only $\frac{x^4}{8}$ will be an integer. hence possible values of $x$ is 4 and 6. ( $x \neq 2$ as terms are distinct)
hence possible value of $a_5=\frac{x^4}{8}$ is $\frac{4^4}{8}, \frac{6^4}{8}$
32,162 $\Rightarrow B , D ]$