The number of ways in which three numbers in A.P. can be selected from 1, 2, 3, …, n is

Given numbers are 1, 2, 3, … n.
Let the three selected numbers in A.P. be a, b, c, then

$\mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2} \text{or }\mathrm{a}+\mathrm{c}=2\mathrm{b}----\left(\mathrm{i}\right)$

From (1) it is clear that a + c should be an even integer. This 
is possible only when both a and c are odd or both are even.
Case I. When n is even. Let n = 2m
The number of odd numbers = m
and number of even numbers = m
∴ number of selections of a and c from m odd numbers = ${ }^{\mathrm{m}}{\mathrm{C}}_2$
Number of selections of a and c from m even numbers = ${ }^{\mathrm{m}}{\mathrm{C}}_2$
∴ Number of ways in this case = 2 ⋅ ${ }^{\mathrm{m}}{\mathrm{C}}_2$ = m (m – 1)

$=\frac{\mathrm{n}}{2}(\frac{\mathrm{n}}{2}-1)=\frac{\mathrm{n}(\mathrm{n}-2)}{4}$
Case II. When n is odd. Let n = 2m + 1
Then, number of odd numbers = m + 1
and number of even numbers = m
∴ Required number in this case ${=}^{\mathrm{m}+1}{\mathrm{C}}_2{+}^{\mathrm{m}}{\mathrm{C}}_2$

$\begin{array}{rl} & =\frac{(\mathrm{m}+1)\mathrm{m}}{2}+\frac{\mathrm{m}(\mathrm{m}-1)}{2}={\mathrm{m}}^2={\left(\frac{\mathrm{n}-1}{2}\right)}^2\\ & =\frac{1}{4}(\mathrm{n}-1{)}^2\end{array}$