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Combinations — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsCombinations4 Marks
Question
If ${^\text{n}}\text{C}_{\text{4}},{^\text{n}}\text{C}_{\text{5}}$ and ${^\text{n}}\text{C}_{\text{6}}$ are in A.P. then find n.
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Answer

${^\text{n}}\text{C}_{\text{4}},{^\text{n}}\text{C}_{\text{5}}$ and ${^\text{n}}\text{C}_{\text{6}}$ are in A.P.
$\therefore​​​​{^\text{n}}\text{C}_{\text{5}}-{^\text{n}}\text{C}_{\text{4}}={^\text{n}}\text{C}_{\text{6}}-{^\text{n}}\text{C}_{\text{5}}$
$\Rightarrow \frac{\text{n}!}{5!(\text{n}-5)!}-\frac{\text{n}!}{4(\text{n}-4)!}=\frac{\text{n}!}{6!(\text{n}-6)!}-\frac{\text{n}!}{6(\text{n}-5)!}$
$\Rightarrow \frac{\text{n}!}{5!(\text{n}-5)!}\Big[\frac{1}{5}-\frac{1}{\text{n}-4}\Big]=\frac{\text{n}!}{5!(\text{n}-6!)}\Big[\frac{1}{6}-\frac{1}{\text{n}-5}\Big]$
$\Rightarrow \frac{1}{\text{n}-5}\Big[\frac{\text{n}-4-5}{5(\text{n}-4)}\Big]=\frac{1}{5}\Big[\frac{\text{n}-5-6}{6(\text{n}-5)}\Big]$
$\Rightarrow \frac{\text{n}-9}{\text{n}-4}=\frac{\text{n}-11}{6}$
$\Rightarrow 6\text{n}-54=\text{n}^{2}-15\text{n}+44$
$\Rightarrow \text{n}^{2}-21\text{n}+98=0$
$\text{n}=7,14$

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