If the coefficients of 5th,6th and 7th terms in the expansion of (1+x)n are in A.P., then the value n is equal to
A
7 only
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B
14 only
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C
7 or 14
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D
None of these
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Solution
The correct option is C7 or 14 Here, coefficient of T5 will be nC4, that of T6 will be nC5 and that of T7 will be nC6.
Since, it is given they are in A.P. ∴2nC5=nC4+nC6
Hence, ⇒2[n!(n−5)!5!]=[n!(n−4)!4!+n!(n−6)!6!] ⇒2[1(n−5)5]=[1(n−4)(n−5)+16×5] ⇒12n−48=30+n2−9n+20⇒n2−21n+98=0 ⇒n=7 or n=14