Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 10A Problems/Problem 7"

m (→‎Solution 4)
m (→‎Solution 3)
Line 18: Line 18:
  
 
==Solution 3==
 
==Solution 3==
−
Minus each of the terms by 12 to make the the sequence <math>1 , 4 , 7,..., 61</math>.
+
Minus each of the terms by <math>12</math> to make the the sequence <math>1 , 4 , 7,..., 61</math>. <math>61-1/3=20, 20 + 1 = 21</math>
−
 
 
−
<math>61-1/3=20, 20 + 1 = 21</math>
 
  
 
<math>\boxed{\textbf{(B)}\ 21}</math>.
 
<math>\boxed{\textbf{(B)}\ 21}</math>.

Revision as of 20:20, 10 March 2020

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

Using the formula for arithmetic sequence's nth term, we see that $a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21$ $\boxed{\textbf{(B)}\ 21}$.

Solution 3

Minus each of the terms by $12$ to make the the sequence $1 , 4 , 7,..., 61$. $61-1/3=20, 20 + 1 = 21$

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

Solution 4

Subtract each of the terms by $10$ to make the sequence $3 , 6 , 9,..., 60, 63$. Then divide the each term in the sequence by $3$ to get $1, 2, 3,..., 20, 21$. Now it is clear to see that there are $21$ terms in the sequence. $\boxed{\textbf{(B)}\ 21}$.

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_7&oldid=119037"