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

\[\text{ The 6th, 7th and 8th terms in the expansion of }  (x + a )^n \text{ are }  ^{n}{}{C}_5 x^{n - 5} a^5 , ^{n}{}{C}_6 x^{n - 6} a^6 \text{ and }  ^{n}{}{C}_7 x^{n - 7} a^7 .\]

According to the question,

\[^{n}{}{C}_5 x^{n - 5} a^5 = 112\]

\[ ^{n}{}{C}_6 x^{n - 6} a^6 = 7\]

\[ ^{n}{}{C}_7 x^{n - 7} a^7 = \frac{1}{4}\]

\[\text{ Now } , \]

\[\frac{^{n}{}{C}_6 x^{n - 6} a^6}{^{n}{}{C}_5 x^{n - 5} a^5} = \frac{7}{112}\]

\[ \Rightarrow \frac{n - 6 + 1}{6} x^{- 1} a = \frac{1}{16}\]

\[ \Rightarrow \frac{a}{x} = \frac{3}{8n - 40} . . . \left( 1 \right)\]

\[\text{ Also, } \]

\[\frac{^{n}{}{C}_7 x^{n - 7} a^7}{^{n}{}{C}_6 x^{n - 6} a^6} = \frac{1/4}{7}\]

\[ \Rightarrow \frac{n - 7 + 1}{7} x^{- 1} a = \frac{1}{28}\]

\[ \Rightarrow \frac{a}{x} = \frac{1}{4n - 24} . . . \left( 2 \right)\]

\[\text{ From }  \left( 1 \right) \text{ and } \left( 2 \right), \text{ we get: } \]

\[\frac{3}{8n - 40} = \frac{1}{4n - 24}\]

\[ \Rightarrow \frac{3}{2n - 10} = \frac{1}{n - 6}\]

\[ \Rightarrow n = 8\]

\[\text{ Putting in eqn } \left( 1 \right) \text{ we get} \]

\[ \Rightarrow a = x\]

\[\text{ Now, } ^{8}{}{C}_5 x^{8 - 5} \left( \frac{x}{8} \right)^5 = 112\]

\[ \Rightarrow \frac{56 x^8}{8^5} = 112\]

\[ \Rightarrow x^8 = 4^8 \]

\[ \Rightarrow x = 4\]

\[\text{ By putting the value of x and n in } \left( 1 \right) \text{ we get} \]

\[a = \frac{1}{2}\]

\[a = 3 \text{ and } x = 2\]