Given,
28−p,p,70−α,α are the coefficient of four consecutive terms in the expansion of (1+x)n
So, Crn=28-p,Cr+1n=p,Cr+2n=70-\alpha &Cr+3n=\alpha
\Rightarrow Crn+Cr+1n=28&Cr+2n+Cr+3n=70
\Rightarrow Cr+1n+1=28&Cr+3n+1=70
We know that, C28=28orC128=28,
Now on comparing with Cr+1n+1=28 we get, (n,r)=(7,1)or(n,r)=(27,0)
Now, checking with Cr+3n+1=70 we get, (n,r)=(7,1)
So, p=C27=21 and α=C47=35
Hence, the value of 2α−3p=70−63=7