If in the expansion of (1+x)n, a, b, c are three consecutive coefficients, then n =
ac+ab+bcb2+ac
2ac+ab+bcb2−ac
ab+acb2−ac
2ac+ab+bcb2+ac
Here a=nCr-1,b=nCr and c=nCr+1
ncrncr-1=ba,⇒n-r+1r=ba⇒an-ar+a=br⇒an-ar=br-a............(I)
⇒an-a+br=-a
ncr+1ncr=cb⇒n-rr+1=cb⇒bn-br=cr+c...........(Ii)
solve (I) and (Ii)
n=2ac+ab+bcb2−ac