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

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==Solution 3==
 
==Solution 3==
Using the formula for arithmetic series' nth term, we see that <math>a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21</math> <math>\boxed{\textbf{(B)}}</math>.
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Using the formula for arithmetic sequence's nth term, we see that <math>a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21</math> <math>\boxed{\textbf{(B)}}</math>.
  
 
==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 00:49, 14 March 2015

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

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

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