Question

If the $2^{nd},3^{rd}$ and $4^{th}$ terms in the expansion of $(a+b{)}^n$ are $135,30$ and $\frac{10}{3}$ respectively, then the value of $n$ is:

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is A $5$

It is given that
$T_2=n{a}^{n-1}b=135$
$T_3=\frac{n(n-1)a^{n-2}{b}^2}{2}=30$

$T_4=\frac{n.(n-1)(n-2)a^{n-3}{b}^3}{6}=\frac{10}{3}$
Therefore, by taking ratios, we get

$\Rightarrow \frac{n{a}^{n-1}b}{\frac{n(n-1)a^{n-2}{b}^2}{2}}=\frac{135}{30}$

$\Rightarrow \frac{2a}{(n-1)b}=\frac{9}{2}$

$\Rightarrow 4a=9b(n-1)$ ...(i)

$\Rightarrow \frac{\frac{n(n-1)a^{n-2}{b}^2}{2}}{\frac{n.(n-1)(n-2)a^{n-3}{b}^3}{6}}=\frac{3\left(30\right)}{10}$

$\Rightarrow \frac{3a}{b(n-2)}=9$

$\Rightarrow a=3b(n-2)$

$\Rightarrow \frac{a}{b}=3(n-2)$ ...(ii)

Substituting the value of $\frac{a}{b}$ in equation (i), we get

$\Rightarrow \frac{9(n-1)}{4}=3(n-2)$

$\Rightarrow 3n-3=4n-8$
$\Rightarrow n=5$