Question

Find a, b and n in the expansion of (a + b)ⁿ, if the first three terms in the expansion are 729, 7290 and 30375 respectively.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ {T}_1=729,T_2=7290\mathrm{and}{T}_3=30375 \\ \mathrm{Now}, \\ {}^{n}C_0{a}^n{b}^0=729 \\ \Rightarrow a^n=729 \\ \Rightarrow a^n=3^6 \\ {}^{n}C_1{a}^{n-1}{b}^1=7290 \\ {}^{n}C_2{a}^{n-2}{b}^2=30375 \\ \mathrm{Also}, \\ \frac{{}^{n}C_2{a}^{n-2}{b}^2}{{}^{n}C_1{a}^{n-1}{b}^1}=\frac{30375}{7290} \\ \Rightarrow \frac{n-1}{2}\times \frac{b}{a}=\frac{25}{6}...\left(\mathrm{i}\right) \\ \Rightarrow \frac{(n-1)b}{a}=\frac{25}{3} \\ \mathrm{And}, \\ \frac{{}^{n}C_1{a}^{n-1}{b}^1}{{}^{n}C_0{a}^n{b}^0}=\frac{7290}{729} \\ \Rightarrow \frac{nb}{a}=\frac{10}{1}...\left(\mathrm{ii}\right) \\ \mathrm{On}\mathrm{dividing}\left(\mathrm{ii}\right)\mathrm{by}\left(i\right),\mathrm{we}\mathrm{get} \\ \frac{{\displaystyle \frac{nb}{a}}}{{\displaystyle \frac{(n-1)b}{a}}}=\frac{10\times 3}{25} \\ \Rightarrow \frac{n}{n-1}=\frac{6}{5} \\ \Rightarrow n=6 \\ \mathrm{Since},a^6=3^6 \\ \mathrm{Hence},a=3 \\ \mathrm{Now},\frac{nb}{a}=10 \\ \Rightarrow b=5 \end{gathered}$$