Picky arithmetic progression

What is the total number of ways in which three distinct numbers in an arithmetic progression can be selected from the numbers 1 to 24 inclusive?


The answer is 132.

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1 solution

Otto Bretscher
May 7, 2015

Clever problem!

The lowest and the highest term of the A.P. will both be even or they will both be odd. To build an A.P with even end terms we choose 2 numbers out of the 12 even numbers in our range, giving us ( 12 2 ) \dbinom{12}{2} options; the middle term is then given as the average of the end terms. We can build just as many sequences with odd end terms, for a total of 2 ( 12 2 ) = 132 2\dbinom{12}{2}=132 .

But it can be ascending or can be descending as the common difference will be different so I think the answer should be 132 x 2=264 and that I had got try to shot this confusion........

shubham gupta - 6 years, 1 month ago

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The question is carefully worded: They ask for "the number of ways in which three distinct numbers in an A.P. can be selected." That's why I'm talking about the lowest and the highest term, not the first and the last.

Otto Bretscher - 6 years, 1 month ago

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thumbs up once again for you sir. :)

Karan Arora - 6 years, 1 month ago

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