Suppose the three consecutive terms in the expansion of (1+x)n are (r−1) th, r th and (r+1) th terms.
Now, Tr−1=‌nCr−2(x)r−2 Tr=‌nCr−1(x)r−1 Tr+1=‌nCr(x)r ∴ Their coefficients are ‌nCr−2,‌nCr−1,‌nCr Since, the coefficients are in the ratio 1:7:42, so we have,
‌nCr−2
‌nCr−1
=
1
7
⇒
r−1
n−r+2
=
1
7
⇒7r−7=n−r+2 ⇒n−8r+9=0 .......(i) and
‌nCr−1
‌nCr
=
7
42
=
1
6
⇒
r
n−r+1
=
1
6
⇒6r=n−r+1 ⇒n−7r+1=0 .........(ii) On solving Eqs. (i) and (ii), we get r=8and n=55