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If $n$ is odd, then in how many ways $3$ numbers that are in A.P. can be selected from the numbers $1,2,3,...............n$?

$\begin{array}{1 1} (A) \large\frac{(n+1)^2}{4} \\ (B) \large\frac{(n-1)^2}{2} \\ (C) \large\frac{(n-1)^2}{4} \\ (D) \large\frac{(n+1)^2}{2} \end{array} $

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Since $n$ is odd let $n=2m+1$
The common differences of the different $A.P.^s$ can be
$1,2,3,.............m$
The no. of $A.P.^s$ with common differences $1, 2,........m$
respectively are $(2m-1),(2m-3),(2m-5)........1$
$\therefore$ The total no. of $A.P.^s$ =$(2m-1)+(2m-3)+..........+1$
$=m^2=\large\bigg(\frac{n-1}{2}\bigg)^2$
answered Sep 3, 2013 by rvidyagovindarajan_1
 

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