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Question

If the coefficients of 5th , 6th and 7th terms in the expansion of (1x)n are in A.P then find the value of n.

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Solution

It can be (1+x)n only as (1x)n terms are +ve/ve alternatively hence can't be in A.P.,
General term, TrH=nCrxr
T5=nC4x4,T6=nC5x5,T7=nC6x6
Coefficients are in A.P.
2nC5=nC4+nC6
2n!5!(n5)!=n!4!(n4)!+n!6!(n6)!
25(n5)=1(n4)(n5)+16×5
2(n4)=5+16(n4)(n5)
12(n4)=30+n29n+20
12n48=n29n+50
nn21n+98=0
n214n7n+98=0
n(n14)7(n14)=0
(n7)(n14)=0
n=7 or n=14.

1165370_1245319_ans_ac60fbb598f74ed29e8977dbc2f0dff7.jpg

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