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Question

Given Cn−3,Cn−2,Cn−1 are in A.P., then

A
n=14
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B
n=7
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C
n=12
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D
C1,C2,C3 are in H.P
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Solution

The correct option is B n=7
Since nn3Cn2,nCnn1C are in AP, it means that n3C,n2C,n1C are in AP
Thus, n3C+n1C=2.n2Cn(n1)(n2)6+n=2.n(n1)2n23n+2+6=6n6n29n+14=0(n2)(n7)=0=>n=2,7
Since n>2n=7

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