Question

If the second, third and fourth terms in the expansion of ${(x+a)}^n$ are $240,720$ and $1080$ respectively, then the value of $n$ is

• A. $15$
• B. $20$
• C. $10$
• D. $5$

Answer 1

Answer: (D)

$5$

Given, $T_2=n{\left(x\right)}^{n-1}{a}^1=240$ ....(i)

$T_3=\frac{n(n-1)}{1\cdot 2}{x}^{n-2}{a}^2=720$ ....(ii)

and $T_4=\frac{n(n-1)(n-2)}{1\cdot 2\cdot 3}{x}^{n-3}{a}^3=1080$ ....(iii)

To eliminate $x$,

$\frac{T_2\cdot T_4}{T_3^2}=\frac{240\cdot 1080}{720\cdot 720}=\frac{1}{2}$

$\Rightarrow \frac{T_2}{T_3}\cdot \frac{T_4}{T_3}=\frac{1}{2}$

Now, $\frac{T_{r+1}}{T_r}=\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}$

Putting $r=3$ and $2$ in above expression, we get

$\frac{n-2}{3}\cdot \frac{2}{n-1}=\frac{1}{2}\Rightarrow n=5$