AMU Engineering 2017 solved paper Solved Paper

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=8 and n=55