If ^nC_4, ^nC_5, ^nC_6 are in A.P., then n can be:

Question

If ${}^nC_4, {}^nC_5, {}^nC_6$ are in A.P., then $n$ can be:

Accepted Answer

$2\binom{n}{5} = \binom{n}{4}+\binom{n}{6} \Rightarrow 2 = \dfrac{5}{n-4}+\dfrac{n-5}{5}.$

$$12(n-4) = 30+n^2-9n+20 \implies n^2-21n+98=0 \implies (n-7)(n-14)=0.$$

$n=7$ or $14$; matching the given options, $n=14$.

Solutions · 1