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Question

If the 5th, 6th, and 7th term in the expansion of (1+y)n are in A.P. then find the value of n.

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Solution

The coefficients of the 5th,6th and 7th terms are nC4,nC5,nC6

Since the coefficients of the 5th,6th and 7th terms are in AP. we have,

nC5nC4=nC6nC5

n!5!(n5)!n!4!(n4)!=n!6!(n6)!n!5!(n5)!

1120(n5)!124(n4)(n5)!=1720(n6)!1120(n5)(n6)!

124(n5)![151n4]=1120(n6)![161n5]

1(n5)[n455(n4)]=15[n566(n5)]

n9n4=n116

n221n+98=0

n=7,14

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