# 2015 AMC 10A Problems/Problem 7

## Problem

How many terms are in the arithmetic sequence $13$, $16$, $19$, $\dotsc$, $70$, $73$? $\textbf{(a)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$

## Solution $73-13 = 60$, so the amount of terms in the sequence $13$, $16$, $19$, $\dotsc$, $70$, $73$ is the same as in the sequence $0$, $3$, $6$, $\dotsc$, $57$, $60$.

In this sequence, the terms are the multiples of $3$ going up to $60$, and there are $20$ multiples of $3$ in $60$.

However, one more must be added to include the first term. So, the answer is $\boxed{\textbf{(B)}\ 21}$.

## See Also

 2015 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS