1965 AHSME Problems/Problem 20

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In order to calculate the $r$th term of this arithmetic sequence, we can subtract the sum of the first $r-1$ terms from the sum of the first $r$ terms of the sequence. Plugging in $r$ and $r-1$ as values of $n$ in the given expression and subtracting yields \[(3r^2+2r)-(3r^2-4r+1)\]. Simplifying gives us the final answer of $\boxed{6r-1}$.

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