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Question

If in the expansion of (1+y)n, the coefficient of 5th,6th and 7th terms are in A.P., then n is equal to:

A
7,11
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B
7,14
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C
8,16
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D
none of the above
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Solution

The correct option is B 7,14
Given
(1+y)n=nC0+nC1y+nC2y2+......+nCnyn

Here Coefficient of 5th,6th,7th are nC4,nC5,nC6 respectively

Since nC4,nC5,nC6 are in A.P

nC5=nC4+nC62

2=nC4+nC6nC5

2=nC4nC5+nC6nC5

We know that
nCrnCr1=nr+1r

Hence

2=5n5+1+n6+16

2=5(6)+(n5)(n4)6(n4)

12(n4)=30+n24n5n+20

12n48=50+n29n

n221n+98=0

(n7)(n14)=0

n=7,14

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