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Question

Let nϵN If (1+x)n=a0+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
This implies
nCn3,nCn2,nCn1 are in A.P
This implies
nC3,nC2,nC1 are in A.P
For
nCr1,nCr,nCr+1 to be in A.P
It must follow
(n2r)2=n+2
For the above case, r=2.
Substituting in the above expression, we get
(n4)2=n+2
n28n+16=n+2
n29n+14=0
(n2)(n7)=0
n=2 and n=7
However for n=2 we will have only 3 coefficients, a0,a1,a2
Hence
n=7 is the required answer.

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