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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.
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Solution
\[\text{ The 6th, 7th and 8th terms in the expansion of } (x + a )^n \text{ are } ^{n}{}{C}_5 x^{n - 5} a^5 , ^{n}{}{C}_6 x^{n - 6} a^6 \text{ and } ^{n}{}{C}_7 x^{n - 7} a^7 .\]
According to the question,
\[^{n}{}{C}_5 x^{n - 5} a^5 = 112\]
\[ ^{n}{}{C}_6 x^{n - 6} a^6 = 7\]
\[ ^{n}{}{C}_7 x^{n - 7} a^7 = \frac{1}{4}\]
\[\text{ Now } , \]
\[\frac{^{n}{}{C}_6 x^{n - 6} a^6}{^{n}{}{C}_5 x^{n - 5} a^5} = \frac{7}{112}\]
\[ \Rightarrow \frac{n - 6 + 1}{6} x^{- 1} a = \frac{1}{16}\]
\[ \Rightarrow \frac{a}{x} = \frac{3}{8n - 40} . . . \left( 1 \right)\]
\[\text{ Also, } \]
\[\frac{^{n}{}{C}_7 x^{n - 7} a^7}{^{n}{}{C}_6 x^{n - 6} a^6} = \frac{1/4}{7}\]
\[ \Rightarrow \frac{n - 7 + 1}{7} x^{- 1} a = \frac{1}{28}\]
\[ \Rightarrow \frac{a}{x} = \frac{1}{4n - 24} . . . \left( 2 \right)\]
\[\text{ From } \left( 1 \right) \text{ and } \left( 2 \right), \text{ we get: } \]
\[\frac{3}{8n - 40} = \frac{1}{4n - 24}\]
\[ \Rightarrow \frac{3}{2n - 10} = \frac{1}{n - 6}\]
\[ \Rightarrow n = 8\]
\[\text{ Putting in eqn } \left( 1 \right) \text{ we get} \]
\[ \Rightarrow a = x\]
\[\text{ Now, } ^{8}{}{C}_5 x^{8 - 5} \left( \frac{x}{8} \right)^5 = 112\]
\[ \Rightarrow \frac{56 x^8}{8^5} = 112\]
\[ \Rightarrow x^8 = 4^8 \]
\[ \Rightarrow x = 4\]
\[\text{ By putting the value of x and n in } \left( 1 \right) \text{ we get} \]
\[a = \frac{1}{2}\]
\[a = 3 \text{ and } x = 2\]
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