Question

If the 6th, 7th and 8th terms in the expansion of (x + a)ⁿ are respectively 112, 7 and 1/4, find x, a, n.

Answer 1

$\mathrm{The}6\mathrm{th},7\mathrm{th}\mathrm{and}8\mathrm{th}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}(x+a{)}^n\mathrm{are}{}^{n}C_5{x}^{n-5}{a}^5,{}^{n}C_6{x}^{n-6}{a}^6\mathrm{and}{}^{n}C_7{x}^{n-7}{a}^7.$
According to the question,
$$\begin{gathered} {}^{n}C_5{x}^{n-5}{a}^5=112 \\ {}^{n}C_6{x}^{n-6}{a}^6=7 \\ {}^{n}C_7{x}^{n-7}{a}^7=\frac{1}{4} \\ \mathrm{Now}, \\ \frac{{}^{n}C_6{x}^{n-6}{a}^6}{{}^{n}C_5{x}^{n-5}{a}^5}=\frac{7}{112} \\ \Rightarrow \frac{n-6+1}{6}{x}^{-1}a=\frac{1}{16} \\ \Rightarrow \frac{a}{x}=\frac{3}{8n-40}...\left(1\right) \\ \mathrm{Also}, \\ \frac{{}^{n}C_7{x}^{n-7}{a}^7}{{}^{n}C_6{x}^{n-6}{a}^6}=\frac{1/4}{7} \\ \Rightarrow \frac{n-7+1}{7}{x}^{-1}a=\frac{1}{28} \\ \Rightarrow \frac{a}{x}=\frac{1}{4n-24}...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ \frac{3}{8n-40}=\frac{1}{4n-24} \\ \Rightarrow \frac{3}{2n-10}=\frac{1}{n-6} \\ \Rightarrow n=8 \end{gathered}$$


$$\begin{gathered} \mathrm{Putting}\mathrm{in}\mathrm{eqn}\left(1\right)\mathrm{we}\mathrm{get} \\ \Rightarrow a=x \\ \mathrm{Now},{}^{8}C_5{x}^{8-5}{\left(\frac{x}{8}\right)}^5=112 \\ \Rightarrow \frac{56x^8}{8^5}=112 \\ \Rightarrow x^8=4^8 \\ \Rightarrow x=4 \\ \mathrm{By}\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{and}n\mathrm{in}\left(1\right)\mathrm{we}\mathrm{get} \\ a=\frac{1}{2} \end{gathered}$$

$a=3\mathrm{and}x=2$