1965 AHSME Problems/Problem 20


For every $n$ the sum of n terms of an arithmetic progression is $2n + 3n^2$. The $r$th term is:

$\textbf{(A)}\ 3r^2 \qquad \textbf{(B) }\ 3r^2 + 2r \qquad \textbf{(C) }\ 6r - 1 \qquad \textbf{(D) }\ 5r + 5 \qquad \textbf{(E) }\ 6r+2\qquad$


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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