The correct options are
A a1,a2,a3 are in AP
C n=7
an−3,an−2,an−1 are in A.P
This implies
nCn−3,nCn−2,nCn−1 are in A.P
This implies
nC3,nC2,nC1 are in A.P
For
nCr−1,nCr,nCr+1 to be in A.P
It must follow
(n−2r)2=n+2
For the above case, r=2.
Substituting in the above expression, we get
(n−4)2=n+2
n2−8n+16=n+2
n2−9n+14=0
(n−2)(n−7)=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.