$\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 $A.P.^s$ =$(2m-1)+(2m-3)+..........+1$

$=m^2=\large\bigg(\frac{n-1}{2}\bigg)^2$

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