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. If $a_1$ and $a_5$ are positive integers and $a_5\le 162$, then $a_5$ is equal to

• A. $162$
• B. $64$
• C. $32$
• D. $2$

Answer 1

The correct option is A $162$

$\text{Solve:}$

$\text{Given,}$

$a_1=2,a_n=\frac{a_{n=1}^2}{a_{n-2}},n⩾3$

$\text{on taking}n=3\text{we get,}$

$a_3=\frac{a_2^2}{a_1}$

$=>a_1\cdot a_3=a_2^2$

So, $a_1,a_2,a_3$ are in GP

taking $n=4$ we get,

$a_3^2=a_2\cdot a_4$

so, $a_2,a_3,a_4$ are in GP

Similarly all other next terms are in GP

$=>a_1,a_2,a_3,a_4,a_5$ are in $GP$

and $a_1=2,$ lelt $a_2=x$

but $a_3=\frac{a_2^2}{a_1}=\frac{x^2}{2}$

so, eerms of GP will be $2,x,\frac{x^2}{2},\frac{x^3}{4},\frac{x^4}{8}$

=> $a_5=\frac{x^4}{8}$ ( positive integer)

$\because$ 8 is in denominator so, for

$a_5$ being an integer x must be even

number.

takin $x=2$

so, $\frac{x^4}{8}=2$

but $a_1=2$

so, $x\ne 2$

at $x=4$

$a_5=\frac{4^4}{8}=32$

$\text{and at}x=5$

$a_5=\frac{6^4}{8}=162$ So, $a_5$ will be $32$ and $162.$