Question

In the expansion of ${(\sqrt{a}+\frac{1}{\sqrt{\left(3a\right)}})}^n$, if the ratio of the binomial coefficient of the $4^{th}$ term to the binomial coefficient of the $3^{rd}$ term is $\frac{10}{3}$, the $5^{th}$ term is

• A. $\frac{88a}{\sqrt{3}}$
• B. $\frac{88a}{3}$
• C. $50a^2$
• D. $55a^2$

Answer 1

The correct option is D $55a^2$

The ratio of the binomial coefficient of $T_4$ to that of $T_3$ will be: $\frac{{}^n{C}_3}{{}^n{C}_2}$
$=\frac{\frac{\left(n\right)(n-1)(n-2)}{6}}{\frac{n(n-1)}{2}}$
$=\frac{n-2}{3}$
$=\frac{10}{3}$ ...(given)
Hence, $n=12$.
Therefore,

$T_5$ $={}^{12}{C}_4{a}^4\frac{1}{9a^2}$
$=\frac{\mathrm{12.11.10.9}}{4!}{a}^4\frac{1}{9a^2}$
$=55a^2$