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Suppose the sequence <$a_n$> is an Arithmetic Progression if its $n^{th}$ term is a linear expression in $n$ then show that common difference is equal to the coefficient of $n$.

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  • $\begingroup$ I can’t proceed on any result except ... Isn't that precisely what a "linear expression in $n$" means? $\endgroup$ – dxiv Jan 16 '17 at 3:27
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Let $$a_n=An+B$$then$$a_{n+1}=A(n+1)+B$$now$$a_{n+1}-a_n=An+A+B-An-B$$ $$=A$$ Hope it helps!!!

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