A) \[\frac{{{(n-1)}^{2}}}{2}\]
B) \[\frac{{{(n+1)}^{2}}}{2}\]
C) \[\frac{{{n}^{2}}-1}{4}\]
D) \[\frac{{{(n-1)}^{2}}}{4}\]
Correct Answer: D
Solution :
| Let \[n=2m+1\] |
| For the three numbers in A. P., we have the following pattern. Common Numbers |
| Ways difference |
| 1 \[\left( 1,2,3 \right),\text{ }\left( 2,3,4 \right),\ldots .\]\[\left( n-2,n-1,n \right)\]\[\left( n-2 \right)\] |
| 2 \[\left( 1,3,5 \right),\]\[\left( 2,4,6 \right),....\]\[\left( n-4,n-2,n \right)\]\[\left( n-4 \right)\] |
| 3 \[\left( 1,4,7 \right),\]\[\left( 2,5,8 \right),....\]\[\left( n-6,n-9,n \right)\]\[\left( n-6 \right)\] |
| 4 \[.....................................................\] |
| 5 \[.....................................................\] |
| . \[.....................................................\] |
| . \[.....................................................\] |
| . \[.....................................................\] |
| \[m\] \[\left( 1,+m+1,2m+1 \right)1\] |
| \[\therefore \] Favourable number of ways \[=\left( n-2 \right)+\left( n-4 \right)+\left( n-6 \right)+.......+3+1\] |
| \[m\] terms \[=\frac{m}{2}\left( n-2+1 \right)\]\[=\frac{n-1}{2}.\frac{n-1}{2}=\frac{{{\left( n-1 \right)}^{2}}}{2}\] |
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