Given that n is odd, the number of ways in which three numbers in AP can be selected from 1,2,3,....n is
A
(n−1)22
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B
(n+1)24
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C
(n+1)22
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D
(n−1)24
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Solution
The correct option is D(n−1)24 Let the numbers in A.P. be a,b,c ⇒a+c=2b i.e a+c is an even number which is possible only when a and c both are even or both are odd. Given n is odd; Let n=2k+1 Number of odd numbers from 1 to n=k+1 Number of even numbers from 1 to n=k number of ways in which numbers are selected =k+1C2+kC2 k+1C2+kC2=(k+1)k2+k(k−1)2=k2=(n−12)2