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Question

If the 5th,6th&7th term in the expansion of (1+y)n are in AP then find n.

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Solution

In the expansion of (1+y)n we get, 5th, 6th and 7th terms as, nC4,nC5,nC6

Given,

nC4,nC5,nC6 are in A.P.

2×nC5=nC4+nC6

2×n!5!(n5)!=n!4!(n4)!+n!6!(n6)!

2×15(n5)=1(n4)(n5)+16×5

25(n5)=30+(n4)(n5)30(n4)(n5)

2×6(n4)=30+n29n+20

n221n+98=0

(n7)(n14)=0

n=7,14

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