Question:
The coefficients of 5th, 6th and 7th terms in the expansion of (1 + x)n are in A.P., find n.
Solution:
Coefficients of the 5 th, 6 th and 7 th terms in the given expansion are ${ }^{n} C_{4},{ }^{n} C_{5}$ and ${ }^{n} C_{6}$ These coefficients are in $A P$.
Thus, we have
$2^{n} C_{5}={ }^{n} C_{4}+{ }^{n} C_{6}$
On dividing both sides by ${ }^{n} C_{5}$, we get:
$2=\frac{{ }^{n} C_{4}}{{ }^{n} C_{5}}+\frac{{ }^{n} C_{6}}{{ }^{n} C_{5}}$
$\Rightarrow 2=\frac{5}{n-4}+\frac{n-5}{6}$
$\Rightarrow 12 n-48=30+n^{2}-4 n-5 n+20$
$\Rightarrow n^{2}-21 n+98=0$
$\Rightarrow(n-14)(n-7)=0$
$\Rightarrow n=7$ or 14