Question

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

• A. $a=3$
• B. $b=\frac{1}{3}$
• C. $n=5$
• D. $n=7$

Answer 1

Answer: (A), (B), (C)

$a=3$
$n=5$
$b=\frac{1}{3}$

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

$n{a}^{n-1}.b=135$ ...(i)

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

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

Taking ratio, we get

$\frac{a}{b(n-1)}=\frac{9}{4}$

$\frac{a}{b}=\frac{9(n-1)}{4}$ ... (A)

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

Hence, $\frac{T_3}{T_4}$

$\frac{6\left(a\right)}{2(n-2).b}=9$

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

Comparing A and B, we get

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

$4n-8=3n-3$

$n=5$

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

$=9$

Hence, the answers are A,B,C.