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 

Let n = 2m + 1

 For the three numbers in A. P., we have the following pattern, Common Numbers Ways difference 

$\begin{array}{r}\begin{array}{ll}1. (1,2,3),(2,3,4),\dots \dots \dots \dots (n-2,n-1,n)& (n-2)\\ 2. (1,3,5),(2,4,6),\dots \dots \dots \dots (n-4,n-2,n)& (n-4)\\ 3. (1,4,7),(2,5,8),\dots \dots \dots \dots (n-6,n-3,n)& (n-6)\\ 4. .........................................................& \\ 5.& \\ .& \end{array}\\ \end{array}$

   $m(1,m+1,2m+1)$

 Favourable number of ways 

$=(n-2)+(n-4)+(n-6)+\dots \dots \dots +3+1$

m terms  $=\frac{m}{2}(n-2+1)=\frac{n-1}{2},\frac{n-1}{2}=\frac{(n-1{)}^2}{4}$

Alternatively, if a, b, care in A. P., then a+ c = 2b

 $\therefore$   Sum of terminal digits is even

$\therefore$     Terminal digits must be either both even or both odd .

$\therefore$    Required number of selections = number of ways of selecting 2 odd numbers from 

 $\frac{n+1}{2}$   odd numbers+   number of ways of selecting 2 even numbers from  $\frac{n-1}{2}$

even number [ ·: n is odd]  

$\begin{array}{l}=\frac{\frac{n+1}{2}{C}_2+\frac{n-1}{2}{C}_2=\frac{\frac{n+1}{2}\times \frac{n-1}{2}}{2}+\frac{\frac{n-1}{2}\times \frac{n-3}{2}}{2}}{ }\\ =\frac{(n-1{)}^2}{4}.\end{array}$