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Question

In the binomial expansion of (1+y)n, where n is a natural number, the coefficients of the 5th,6th and 7th terms are in A.P, find n

A
n=7
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B
n=14
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C
n=8
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D
n=16
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Solution

The correct options are
A n=14
D n=7
The coefficient of the fifth term of the binomial expansion is nC4 , sixth term of the binomial expansion is nC5 , seventh term of the binomial expansion is nC6
Given that the 5th,6th and the 7th coefficient are in arithmetic progression,
Therefore
nC4+nC6 = 2nC5
n!(n4)!4!+n!(n6)!6!=2n!(n5)!5!
n(n1)(n2)(n3)4!+n(n1)(n2)(n3)(n4)(n5)6.5.4!2
n(n1)(n2)(n3)(n4)5.4!=0
Taking out n(n1)(n2)(n3)4! common, we get,
(n(n1)(n2)(n3)4!)(1+(n4)(n5)302(n4)5)=0
Therefore,
(1+(n4)(n5)302(n4)5)=0
30+n29n+20+3012n+48=0
n221n+98=0
(n7)(n14)=0
So, we get n=7 or 14

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