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

Revision as of 18:54, 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 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS