KEAM 2011 Math Paper

Given,
The expansion of (1+ax)n is,
On comparing the coefficient of like powers of x in Eqs. (i) and (ii),
na=6.....(iii)
‌

27

2

=‌

n(n−1)

2

⋅a2
⇒‌‌27=(n−1)(na)⋅a
27=(n−1)a6   [from Eq. (iii)]
(n−1)a=‌

9

2

.....(iv)
From Eqs. (iii) and (iv),
‌

(n−1)6

n

=‌

9

2

⇒‌

n−1

n

=‌

3

4

⇒‌‌4n−4=3n
⇒‌‌n=4
From Eq. (iii), a=‌

6

4

⇒‌‌a=3/2