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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.
Solution
\[\text{ We have: } \]
\[ T_1 = 729, T_2 = 7290 \text{ and } T_3 = 30375\]
\[\text{ Now,} \]
\[ ^{n}{}{C}_0 a^n b^0 = 729\]
\[ \Rightarrow a^n = 729\]
\[ \Rightarrow a^n = 3^6 \]
\[ ^{n}{}{C}_1 a^{n - 1} b^1 = 7290\]
\[^{n}{}{C}_2 a^{n - 2} b^2 = 30375\]
\[\text{ Also, } \]
\[\frac{^{n}{}{C}_2 a^{n - 2} b^2}{^{n}{}{C}_1 a^{n - 1} b^1} = \frac{30375}{7290}\]
\[ \Rightarrow \frac{n - 1}{2} \times \frac{b}{a} = \frac{25}{6} . . . (i)\]
\[ \Rightarrow \frac{(n - 1)b}{a} = \frac{25}{3}\]
\[\text{ And } , \]
\[\frac{^{n}{}{C}_1 a^{n - 1} b^1}{^{n}{}{C}_0 a^n b^0} = \frac{7290}{729}\]
\[ \Rightarrow \frac{nb}{a} = \frac{10}{1} . . . (ii)\]
\[\text{ On dividing (ii) by (i), we get } \]
\[\frac{\frac{nb}{a}}{\frac{(n - 1)b}{a}} = \frac{10 \times 3}{25}\]
\[ \Rightarrow \frac{n}{n - 1} = \frac{6}{5}\]
\[ \Rightarrow n = 6\]
\[\text{ Since } , a^6 = 3^6 \]
\[\text{ Hence,} a = 3\]
\[\text{ Now } , \frac{nb}{a} = 10\]
\[ \Rightarrow b = 5\]
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