Question

Number of ways in which 3 numbers in AP can be selected from 1, 2, 3,…n is
A. ${\left( {\dfrac{{n - 1}}{2}} \right)^2}$, if n is even
B. $\dfrac{{n\left( {n - 2} \right)}}{4}$, if n is odd
C. $\dfrac{{{{\left( {n - 1} \right)}^2}}}{4}$, if n is odd
D. $\dfrac{{n\left( {n - 2} \right)}}{4}$, if n is even

Answer 1

Here in this question, we have a sequence and we have to determine the number of ways in which the three numbers can be selected from the sequence. We determine the possible ways of sequence with the help of arithmetic sequence definition and common difference.

Complete step by step solution:
In the sequence we have three kinds of sequence namely, arithmetic sequence, geometric sequence and harmonic sequence.
In arithmetic sequence we the common difference between the two terms, In geometric sequence we the common ratio between the two terms, In harmonic sequence it is a ratio of arithmetic sequence to geometric sequence.
The general arithmetic progression is of the form $a,a + d,a + 2d,...$ where a is first term nth d is the common difference.
Let the three numbers of the AP be $a,a + d,a + 2d$
the $a > 1$ and $a + 2d \leqslant n \Rightarrow a < n - 2d$
$\Rightarrow 1 < a < n - 2d$
On simplifying the above inequality

$$\Rightarrow 1 < n - 2d \\\ \Rightarrow 2d < n - 1 \\\ \Rightarrow d < \dfrac{{n - 1}}{2} \\\$$

If d = 1 we can assume the sequence as 1,2, ..n – 2
If d = 2 we can assume the sequence as 1,2, ..n – 4
If n is even $d = \dfrac{{n - 2}}{2}$, $1 < a < 2$
The number of AP is $(n - 2) + (n - 4) + ... + 2$

$$\Rightarrow \dfrac{{n - 2}}{4}(2 + n - 2) \\\ \Rightarrow \dfrac{{n(n - 2)}}{4} \\\$$

If n is odd $d = \dfrac{{n - 1}}{2}$,
The number of AP will be is $1 + 3 + ... + (n - 2)$
$\Rightarrow \dfrac{{{{(n - 1)}^2}}}{4}$

So, the correct answer is “Option C”.

Note: By considering the formula of arithmetic sequence we verify the common difference which we obtained. We have to check the common difference for all the terms. Suppose if we check for the first two terms not for other terms then we may go wrong. The above method is a general method.