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(Q). If ‘p’ is the number of ways in which three numbers in A.P. can be selected from 1, 2, 3, 4, …….., n. Then find the value of p if
(i). n is odd
(ii). n is even
Solution:-
Given numbers are :-
1, 2, 3, 4, ………., n
Let the three selected numbers in A.P. be a, b, c then
a + c = 2b
Since 2b is even 
Hence a+c should also be even
& This is possible only when both a & c are even or both a & c are odd.

Case(i). when n is odd
Let n = 2m + 1
∴ Number of odd numbers in given series = \frac{{(2m + 1) + 1}}{2}
= m+1
& Number of even numbers in given series = \frac{{(2m + 1) - 1}}{2}
= m
∴ number of ways of selection of a & c from (m+1) odd integers = ᵐ⁺¹C₂
& number of ways of selection of a & c from m even integers = ᵐC₂
∴ required number of ways = ᵐ⁺¹C₂ × ᵐC₂
\frac{{(m + 1)m}}{2} \times \frac{{m(m - 1)}}{2} 
= m²
{\left( {\frac{{n - 1}}{2}} \right)^2}         (∵ n = 2m + 1)
\frac{1}{4}{(n - 1)^2}   Answer

case(ii). when n is even
we can assume n = 2m
∴ Number of even integers in given series = m
& Number of odd integers in given series = m
∴ Number of ways of selection of a & c from m odd integers = ᵐC₂
Hence total number of ways =  ᵐC₂ +  ᵐC₂
=  \frac{{m(m + 1)}}{2} + \frac{{m(m - 1)}}{2}
= m(m-1)
=  \frac{n}{2}\left( {\frac{n}{2} - 1} \right)   (∵ n = 2m)
\frac{{n(n - 2)}}{4}     Answer