During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

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

m (Solution 3)
Line 25: Line 25:
 
Subtract each of the terms by <math>10</math> to make the sequence <math>3 , 6 , 9,..., 60, 63</math>. Then divide the each term in the sequence by <math>3</math> to get <math>1, 2, 3,..., 20, 21</math>. Now it is clear to see that there are <math>21</math> terms in the sequence.
 
Subtract each of the terms by <math>10</math> to make the sequence <math>3 , 6 , 9,..., 60, 63</math>. Then divide the each term in the sequence by <math>3</math> to get <math>1, 2, 3,..., 20, 21</math>. Now it is clear to see that there are <math>21</math> terms in the sequence.
 
<math>\boxed{\textbf{(B)}\ 21}</math>.
 
<math>\boxed{\textbf{(B)}\ 21}</math>.
 +
 +
==Video Solution==
 +
https://youtu.be/fcWPfgKeCmA
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}
 
{{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:34, 7 July 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. $\boxed{\textbf{(B)}\ 21}$.

Video Solution

https://youtu.be/fcWPfgKeCmA

~savannahsolver

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