If the coefficients of three consecutive terms in the expansion of (1 + x)n be 76, 95 and 76, find n.
Suppose $r,(r+1)$ and $(r+2)$ are three consecutive terms in the given expansion.
The coefficients of these terms are ${ }^{n} C_{r-1},{ }^{n} C_{r}$ and ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$.
According to the question,
${ }^{n} C_{r-1}=76$
${ }^{n} C_{r}=95$
${ }^{n} C_{r+1}=76$
$\Rightarrow{ }^{n} C_{r-1}={ }^{n} C_{r+1}$
$\Rightarrow r-1+r+1=n \quad\left[\right.$ If ${ }^{n} C_{r}={ }^{n} C_{s} \Rightarrow r=s$ or $\left.r+s=n\right]$
$\Rightarrow r=\frac{n}{2}$
$\therefore \frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{95}{76}$
$\Rightarrow \frac{n-r+1}{r}=\frac{95}{76}$
$\Rightarrow \frac{\frac{n}{2}+1}{\frac{n}{2}}=\frac{95}{76}$
$\Rightarrow 38 n+76=\frac{95 n}{2}$
$\Rightarrow \frac{19 n}{2}=76$
$\Rightarrow n=8$