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 $n$.
Prove that the coefficient of $n$ is the common difference.
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0`I can’t proceed on any result except ...` Isn't that precisely what a "*linear expression in $n$*" means? – 2017-01-16
1 Answers
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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!!!