Question

A set of three numbers are chosen from the set $S=\{1,2,3,\dots ,(2n+1\left)\right\}.$ If the probability that the numbers chosen are in arithmetic progression is $\frac{4}{21},$ then the value of $n$ is

Answer 1

Let three numbers in A.P. be $a,b,c$
$2b=a+c$
Either $a,c$ both will be odd or both will be even.

$1,3,5,\dots ,2n+1\to (n+1)$ numbers
$2,4,6,\dots ,2n\to n$ numbers

Let $E$ be the event of selecting three numbers in A.P.
$\Rightarrow n\left(E\right)={}^{n+1}{C}_2+{}^n{C}_2$
$\Rightarrow n\left(E\right)=\frac{(n+1)n}{2}+\frac{n(n-1)}{2}$
$\Rightarrow n\left(E\right)=n^2$

Three numbers can be selected in ${}^{2n+1}{C}_3$ ways.
$\therefore \text{Probability}=\frac{n^2}{{}^{2n+1}{C}_3}$
$\Rightarrow \frac{4}{21}=\frac{6n^2}{(2n+1)2n(2n-1)}\Rightarrow \frac{4}{21}=\frac{3n}{4n^2-1}\Rightarrow 16n^2-4=63n\Rightarrow 16n^2-63n-4=0\Rightarrow 16n^2-64n+n-4=0\Rightarrow 16n(n-4)+(n-4)=0\Rightarrow (n-4)(16n+1)=0$
$\Rightarrow n=4$ or $n=\frac{-1}{16}$ (rejected)