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

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(Problem and Solution 7)
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<math> \textbf{(a)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 </math>
 
<math> \textbf{(a)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 </math>
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==Solution==
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73-13 is 60, so the amount of terms in the sequence 13, 16, 19,..., 70, 73 is the same as in the sequence 0, 3, 6,..., 57, 60.
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In this sequence, the terms are the multiples of 3 going up to 60, and there are 20 multiples of 3 in 60.
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However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B) }21}</math>.

Revision as of 15:47, 4 February 2015

Problem

How many terms are in the arithmetic sequence 13, 16, 19,..., 70, 73?

$\textbf{(a)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$


Solution

73-13 is 60, so the amount of terms in the sequence 13, 16, 19,..., 70, 73 is the same as in the sequence 0, 3, 6,..., 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 $\boxed{\textbf{(B) }21}$.

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