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Question

Let nϵN.If (1+x)n=a0x+a1x+a2x2+...+anxn, and an3,an2,an1, are in AP then

A
a1,a2,a3 are in AP
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B
a1,a2,a3 are in HP
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C
n=7
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D
n=14
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Solution

The correct options are
A a1,a2,a3 are in AP
C n=7
an3,an2,an1 are in A.P indicates that
a3,a2,a1 are in A.P.
Since they are binomial coefficients.
For the above terms to be in A.P, they must follow the relationship
(n2r)2=n+2
Here r will be 2.
Therefore
n28r+16=n+2
n29r+14=0
(n7)(n2)=0
n=7 since n=2 is not possible here.

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