Question

If the 2nd, 3rd and 4th terms in the expansion of (x + a)ⁿ are 240, 720 and 1080 respectively, find x, a, n.

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{the}\mathrm{expansion}\mathrm{of}{\left(x+a\right)}^n,\mathrm{the}2\mathrm{nd},3\mathrm{rd}\mathrm{and}4\mathrm{th}\mathrm{terms}\mathrm{are}{}^n\mathrm{C}_1{x}^{\mathrm{n}-1}{a}^1,{}^{\mathrm{n}}\mathrm{C}_2{\mathrm{x}}^{\mathrm{n}-2}{\mathrm{a}}^2\mathrm{and}\mathit{}{}^{n}C_3{x}^{\mathrm{n}-3}{a}^3,\mathrm{respectively}. \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ {}^{n}C_1{x}^{n-1}{a}^1=240 \\ {}^{n}C_2{x}^{n-2}{a}^2=720 \\ {}^{n}C_3{x}^{n-3}{a}^3=1080 \\ \Rightarrow \frac{{}^{n}C_2{x}^{n-2}{a}^2}{{}^{n}C_1{x}^{n-1}{a}^1}=\frac{720}{240} \\ \Rightarrow \frac{n-1}{2x}a=3 \\ \Rightarrow \frac{a}{x}=\frac{6}{n-1}...\left(1\right) \\ \mathrm{Also}, \\ \frac{{}^{n}C_3{x}^{n-3}{a}^3}{{}^{n}C_2{x}^{n-2}{a}^2}=\frac{1080}{720} \\ \Rightarrow \frac{n-2}{3x}a=\frac{3}{2} \\ \Rightarrow \frac{a}{x}=\frac{9}{2n-4}...\left(2\right) \\ \mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get} \\ \frac{6}{n-1}=\frac{9}{\mathit{2}n-4} \\ \Rightarrow n=5 \\ \mathrm{Putting}\mathrm{in}\mathrm{eqn}\left(1\right)\mathrm{we}\mathrm{get} \\ \Rightarrow 2a=3x \\ \mathrm{Now},{}^{5}C_1{x}^{5-1}\left(\frac{3}{2}x\right)=240 \\ \Rightarrow 15x^5=480 \\ \Rightarrow x^5=32 \\ \Rightarrow x=2 \\ \mathrm{By}\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{and}n\mathrm{in}\left(1\right)\mathrm{we}\mathrm{get} \\ a=3 \end{gathered}$$