Question

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

• A. $\frac{3\left(n-1\right)}{n\left(n-2\right)}$
• B. $\frac{3{\left(n+1\right)}^2}{2n\left(n-1\right)\left(n-2\right)}$
• C. $\frac{\left(n-2\right)}{n\left(n-1\right)}$
• D. $\frac{3\left(n-1\right)}{2n\left(n-2\right)}$

Answer 1

Answer: (D)

Exhaustive number of cases ${=}^n{C}_3$
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 $=\frac{4}{{}^n{C}_3}=\frac{3\left(n-1\right)}{2n\left(n-2\right)}$