let the 3 consecutive terms be r-1, r, r+1
then nc(r-1): ncr: nc(r+1)= 1:7: 42
Lets consider first 2 terms initially,
nc(r-1)/ncr= r/(n-r+1)= 1/7 => n-r+1 = 7r => 8r=n+1
Last two terms,
ncr/ nc(r+1)= 1/6 => r+1/ (n-r+2)=1/6 => 6r+6 = n-r+2 => 7r= n-4
this gives, r=5 and n=39
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