If in the expansion of

Solution:

Suppose $r^{t h},(r+1)^{t h}$ and $(r+2)^{t h}$ terms are the three consecutive terms.

Their respective coefficients are ${ }^{n} C_{r-1},{ }^{n} C_{r}$ and ${ }^{n} C_{r+1}$.

We have:

${ }^{n} C_{r-1}={ }^{n} C_{r+1}=56$

$\Rightarrow r-1+r+1=n \quad\left[\right.$ If ${ }^{n} C_{r}={ }^{n} C_{s} \Rightarrow r=s$ or $\left.r+s=n\right]$

$\Rightarrow 2 r=n$

$\Rightarrow r=\frac{n}{2}$

Now,

${ }^{n} C_{\frac{n}{2}}=70$ and ${ }^{n} C_{\left(\frac{n}{2}-1\right)}=56$

$\Rightarrow \frac{{ }^{n} C_{\left(\frac{n}{2}-1\right)}}{{ }^{n} C_{\frac{n}{2}}}=\frac{56}{70}$

$\Rightarrow \frac{\frac{n}{2}}{\left(\frac{n}{2}+1\right)}=\frac{8}{10}$

$\Rightarrow 5 n=4 n+8$

$\Rightarrow n=8$

So, $r=\frac{n}{2}=4$

Thus, the required terms are 4 th, 5 th and 6 th.