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 $\boxed{\textbf{(B)}\ 21}$ .

Solution 2

$\dfrac{73 - 13}{3} + 1 = 21$ $\boxed{\textbf{(B)}}$ .

Solution 3

Using the formula for arithmetic series' nth term, we see that $a + (n-1)d \Longarrow 13 + (n-1)3 =73,$ (Error compiling LaTeX. Unknown error_msg) so $(n-1)3 = 60 \Longarrow n = 21$ (Error compiling LaTeX. Unknown error_msg) $\boxed{\textbf{(B)}}$ .

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 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.

Page

Toolbox

Search

2015 AMC 10A Problems/Problem 7

Contents

Problem

Solution

Solution 2

Solution 3

See Also