If in the expansion of

Solution:

(b) 7, 14

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:

$2=\frac{{ }^{n} C_{4}}{{ }^{n} C_{5}}+\frac{{ }^{n} C_{6}}{{ }^{n} C_{5}}$

$\Rightarrow 2=\frac{5}{n-4}+\frac{n-5}{6}$

$\Rightarrow 12 n-48=30+n^{2}-4 n-5 n+20$

$\Rightarrow n^{2}-21 n+98=0$

$\Rightarrow(n-14)(n-7)=0$

$\Rightarrow n=7$ and 14