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For what value of $k$ are $2k-7$, $k+5$ and $3k+2$ consecutive terms of an Arithmetic Progression?

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Well, $3k + 2 = (k + 5) + d, 3k + 2 = (2k -7) + 2d$, etc. Can you take it from here? –  The Chaz 2.0 Oct 30 '12 at 16:08
    
You’ve been working with arithmetic progressions for a while now; do you have any ideas on how to proceed? –  Brian M. Scott Oct 30 '12 at 16:09
    
Yes, It's my one of the favourite topic in Mathematics. And now I know that how can we proceed but, I am feeling difficulty in finding the relation between the given variable terms to get an AP from it. –  Alpha Oct 30 '12 at 16:12
    
Another useful thing to remember is that the sum of three consecutive terms of an arithmetic progression is equal to three times the middle term. If the terms can be taken out of order (i.e. they are three consecutive terms of a progression, but not in order) there is a second value of $k$ to be found. –  Mark Bennet Oct 30 '12 at 18:06

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up vote 3 down vote accepted

Hint: The numbers $a,b,c$ are consecutive terms of an AP if and only if $c-b=b-a$.

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Yah! That's the point to be noted. Thanks @Andre Nicolas!! –  Alpha Oct 30 '12 at 16:19

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