If n is positive integer and three consecutive coefficients in the expansion of are in the ratio 6 : 33 : 110, then n =
4
6
12
16
Let the consecutive coefficient of are
By condition,
....(1)
...(2)
Solving (i) and (ii), we get n=12 and r=2.
Aliter : We take first n=4 [By alternate (a)] but (a) does not hold. Similarly (b).
So alternate (c), n =12 gives
So coefficient of II, III and IV terms are
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