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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.
Solution
\[\text{ In the expansion of } \left( x + a \right)^n , \text{ the 2nd, 3rd and 4th terms are } ^{n}{}{C}_1 x^{n - 1} a^1 , ^{n}{}{C}_2 x^{n - 2} a^2 \text{ and } ^{n}{}{C}_3 x^{n - 3} a^3 , \ \text{ respectively } . \]
\[\text{ According to the question } , \]
\[ ^{n}{}{C}_1 x^{n - 1} a^1 = 240 \]
\[ ^{n}{}{C}_2 x^{n - 2} a^2 = 720\]
\[^{n}{}{C}_3 x^{n - 3} a^3 = 1080\]
\[ \Rightarrow \frac{^{n}{}{C}_2 x^{n - 2} a^2}{^{n}{}{C}_1 x^{n - 1} a^1} = \frac{720}{240}\]
\[ \Rightarrow \frac{n - 1}{2x}a = 3\]
\[ \Rightarrow \frac{a}{x} = \frac{6}{n - 1} . . . \left( 1 \right)\]
\[\text{ Also } , \]
\[\frac{^{n}{}{C}_3 x^{n - 3} a^3}{^{n}{}{C}_2 x^{n - 2} a^2} = \frac{1080}{720}\]
\[ \Rightarrow \frac{n - 2}{3x}a = \frac{3}{2}\]
\[ \Rightarrow \frac{a}{x} = \frac{9}{2n - 4} . . . \left( 2 \right)\]
\[\text{ Using } \left( 1 \right) \text{ and } \left( 2 \right) \text{ we get } \]
\[\frac{6}{n - 1} = \frac{9}{2n - 4}\]
\[ \Rightarrow n = 5\]
\[\text{ Putting in eqn } \left( 1 \right) \text{ we get } \]
\[ \Rightarrow 2a = 3x\]
\[\text{ Now } , ^{5}{}{C}_1 x^{5 - 1} \left( \frac{3}{2}x \right) = 240\]
\[ \Rightarrow 15 x^5 = 480\]
\[ \Rightarrow x^5 = 32\]
\[ \Rightarrow x = 2\]
\[\text{ By putting the value of x and n in} \left( 1 \right) \text{ we get} \]
\[a = 3\]
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