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Q.
Given that n is odd, the number of ways in which three numbers in A.P. can be selected from $1, 2, 3, ..., n$ is
Probability
Solution:
Exhaustive number of cases $=^nC_3$
Three numbers in A.P. can be selected in the following mutually exclusive ways :
Numbers, with common difference $= 1, (1, 2, 3); (2, 3, 4); ........$
$(n -2, n - 1, n),$ hence, $(n - 2)$ in number
Numbers with common difference $= 2, (1,3,5); (2,4,6); ..........$
$( n - 4, n - 2, n)$ hence, $(n - 4)$ in number.
Number with common difference $= 3, (1,4,7); (2,5,8)........$ $(n - 6, n - 3, n),$ hence, $(n - 6)$ in number
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Numbers with common difference $=\frac{n-1}{2}\left(1, \frac{n+1}{2}, n\right),$
hence 1 in number [Note that n is odd, so $\frac{n-1}{2}$ is integer]
$\therefore $ Total number of favourable case $(n-2) +(n-4) +(n-6+) +.....+3+1$
$=\frac{1}{2}\left(\frac{n-1}{2}\right)\left[1+\left(n-2\right)\right]=\frac{\left(n-1\right)^{2}}{4}$
[$\therefore $ total number of terms in above A.P. is $\frac{n-1}{2}$ ]
Required probability $=\frac{\frac{\left(n-1\right)^{2}}{4}}{^{n}C_{3}}$
$=\frac{\frac{\left(n-1\right)^{2}}{4}}{\frac{n\left(n-1\right)\left(n-2\right)}{6}}=\frac{3\left(n-1\right)}{2n\left(n-2\right)}$ ALTERNATE : There are in the set $\left\{1, 2, 3, ....., n\right\}$ (n being odd), $\frac{n-1}{2}$ even number, $\frac{n+1}{2}$ odd numbers; and for an A.P., the sum of the extremes is always even and hence the choice is either (both) 2 even or (both) 2 odd and this may be done in
$^{\frac{n-1}{2}}C_{2}+ ^{\frac{n+1}{2}} C_{2}=\frac{\left(n-1\right)^{2}}{4}$ ways.
Hence, the required probability $=\frac{4}{^{n}C_{3}}=\frac{3\left(n-1\right)}{2n\left(n-2\right)}$