Permutations and Combinations 2 Question 5
5. If ${ }^{n} C _4,{ }^{n} C _5$ and ${ }^{n} C _6$ are in $AP$, then $n$ can be
(2019 Main, 12 Jan II)
(a) 9
(b) 11
(c) 14
(d) 12
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Answer:
Correct Answer: 5. (c)
Solution:
- If ${ }^{n} C _4,{ }^{n} C _5$ and ${ }^{n} C _6$ are in AP, then
$$ 2{ }^{n} C _5={ }^{n} C _4+{ }^{n} C _6 $$
[If $a, b, c$ are in AP, then $2 b=a+c$ ]
$\Rightarrow 2 \frac{n !}{5 !(n-5) !}=\frac{n !}{4 !(n-4) !}+\frac{n !}{6 !(n-6) !}$
$\Rightarrow \frac{2}{5 \cdot 4 !(n-5)(n-6) !}$
$$ \because{ }^{n} C _r=\frac{n !}{r !(n-r) !} $$
$=\frac{1}{4 !(n-4)(n-5)(n-6) !}+\frac{1}{6 \cdot 5 \cdot 4 !(n-6) !}$
$\Rightarrow \quad \frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{30}$
$\Rightarrow \quad \frac{2}{5(n-5)}=\frac{30+(n-4)(n-5)}{30(n-4)(n-5)}$
$\Rightarrow \quad 12(n-4)=30+n^{2}-9 n+20$
$\Rightarrow \quad n^{2}-21 n+98=0$
$\Rightarrow \quad n^{2}-14 n-7 n+98=0$
$\Rightarrow \quad n(n-14)-7(n-14)=0$
$\Rightarrow \quad(n-7)(n-14)=0$
$$ \Rightarrow \quad n=7 \text { or } 14 $$
