## Interesting problem involving arithmetic progression

I just came up with a problem I hope you will find interesting, but I can't seem it solve it myself. I thought of induction as some guide, but am not sure how to proceed.

There are N terms in some finite arithmetic progression. Two of those terms are equal to 3. Prove that all terms in this progression also equal 3.

BiP
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 Recognitions: Homework Help You are making life too hard for yourself :) An arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. If term xi=a and xi+n=b then the difference between successive terms Δx must be Δx=(b-a)/n. If a=b then Δx must be ...

 Quote by Simon Bridge You are making life too hard for yourself :) An arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. If term xi=a and xi+n=b then the difference between successive terms Δx must be Δx=|a-b|/n. If a=b then Δx must be ...
Thanks!

So 2=2+n*d
0 = n*d
d = 0, since n is not 0.

Since d=0, every term must equal the ith term, which happens to be 2!

Genius!

BiP

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## Interesting problem involving arithmetic progression

 Quote by Bipolarity There are N terms in some finite arithmetic progression. Two of those terms are equal to 3. Prove that all terms in this progression also equal 3. BiP
 Quote by Bipolarity Thanks! So 2=2+n*d 0 = n*d d = 0, since n is not 0. Since d=0, every term must equal the ith term, which happens to be 2!
I'm assuming that the work above is related to the problem in the original post. The i-th term can't be 2 if you're proving that all the terms in the progression are 3.

 Quote by Mark44 I'm assuming that the work above is related to the problem in the original post. The i-th term can't be 2 if you're proving that all the terms in the progression are 3.
Sorry my mistake, I meant to write 3.

BiP

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