Question

If the second, third and fourth terms in the expansion of $(b+a{)}^n$ are $135,30$, and $\frac{10}{3}$ respectively, then:

• A. $n=3$
• B. $n=2$
• C. $n=7$
• D. $n=5$

Answer 1

The correct option is D $n=5$

Given

$(b+a{)}^n=(a+b{)}^n$

The second term is

$T_2={}^n{C}_1{a}^{n-1}b=135$

$T_2=n{a}^{n-1}b=135$

The third term is

$T_3={}^n{C}_2{a}^{n-2}{b}^2=30$

$T_3=\frac{n(n-1)a^{n-2}{b}^2}{2}=30$

The fourth term is

$T_4={}^n{C}_3{a}^{n-3}{b}^3=\frac{10}{3}$

$T_4=\frac{n(n-1)(n-2)a^{n-3}{b}^3}{6}=\frac{10}{3}$

On taking ratio of $T_2,T_3$

$\Rightarrow \frac{T_2}{T_3}=\frac{135}{30}$

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

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

$\Rightarrow 4a=9b(n-1)------\left(1\right)$

On taking ratio of $T_3,T_4$

$\Rightarrow \frac{T_3}{T_4}=\frac{3\left(30\right)}{10}$

$\Rightarrow \frac{\frac{n(n-1)a^{n-2}{b}^2}{2}}{\frac{n(n-1)(n-2)a^{n-3}{b}^3}{6}}=9$

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

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

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

Substituting $\frac{a}{b}$ in eq (1), we get

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

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

$\Rightarrow n=5$