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. 2
• B. 32
• C. 64
• D. 162

Answer 1

Answer: (B), (D)

Given $a_1=2;\frac{a_n}{a_{n-1}}=\frac{a_{n-1}}{a_{n-2}}\Rightarrow a_1,a_2,a_3,a_4,a_5,\dots \dots$ 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. $2,x,\frac{x^2}{2},\frac{x^3}{4},\frac{x^4}{8},\dots \dots .$. with common ratio $r=\frac{x}{2}$
given $\frac{x^4}{8}\le 162\Rightarrow x^4\le 1296\Rightarrow x\le 6$
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. ( $x\ne 2$ 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 $\Rightarrow B,D]$