If the 6th, 7th and 8th terms in the expansion of


If the 6th, 7th and 8th terms in the expansion of (x + a)n are respectively 112, 7 and 1/4, find xan.


The 6 th, 7 th and 8 th terms in the expansion of $(x+a)^{n}$ are ${ }^{n} C_{5} x^{n-5} a^{5},{ }^{n} C_{6} x^{n-6} a^{6}$ and ${ }^{n} C_{7} x^{n-7} a^{7}$.

According to the question,

${ }^{n} C_{5} x^{n-5} a^{5}=112$

${ }^{n} C_{6} x^{n-6} a^{6}=7$

${ }^{n} C_{7} x^{n-7} a^{7}=\frac{1}{4}$


$\frac{{ }^{n} C_{6} x^{n-6} a^{6}}{{ }^{n} C_{5} x^{n-5} a^{5}}=\frac{7}{112}$

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

$\Rightarrow \frac{a}{x}=\frac{3}{8 n-40}$   ....(1)


$\frac{{ }^{n} C_{7} x^{n-7} a^{7}}{{ }^{n} C_{6} x^{n-6} a^{6}}=\frac{1 / 4}{7}$

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

$\Rightarrow \frac{a}{x}=\frac{1}{4 n-24}$   ....(2)

From $(1)$ and $(2)$, we get :

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

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

$\Rightarrow n=8$

Putting in eqn (1) we get

$\Rightarrow a=x$

Now, ${ }^{8} C_{5} x^{8-5}\left(\frac{x}{8}\right)^{5}=112$

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

$\Rightarrow x^{8}=4^{8}$

$\Rightarrow x=4$

By putting the value of $x$ and $n$ in (1) we get


$a=3$ and $x=2$





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