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Given that n is the odd, the number of ways in which three numbers in AP can be selected from 1,2,3,4, - - - , n is

Option: 1

\frac{(n-1)^2}{2}


Option: 2

\frac{(n+1)^2}{4}


Option: 3

\frac{(n+1)^2}{2}

 


Option: 4

\frac{(n-1)^2}{4}


Answers (1)

best_answer

Let\: n=2 m+1\: If
\\a, b, c \: are\: in\: \mathrm{AP}\:\\ then\: 2 b=a+c
ie, sum of two numbers is even.
Here, m+1 is odd and m is even number.

\because  Sum of two numbers is even, then both numbers are even or odd.
\therefore \text { Required number of ways }={ }^m C_2+{ }^{m+1} C_2
                                                                \begin{aligned} & =\frac{m(m-1)}{2}+\frac{(m+1) m}{2} \\ & =m^2 \\ & =\left(\frac{n-1}{2}\right)^2 \\ & =\frac{(n-1)^2}{4} \end{aligned}

 

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