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

Revision as of 00:21, 14 February 2017

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 $a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21$

Solution 4

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

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.

@@ Line 29: / Line 29: @@
 ==Solution 4==
 Minus each of the terms by 12 to make the the sequence 1,4,7,....,61.
--1/3=20
+-1/3=20 20+1=21
-+1=21
 ==See Also==
 {{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}
 {{MAA Notice}}

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