Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2010 AMC 12A Problems/Problem 10: Difference between revisions

Stargroup (talk | contribs)
28 edits
Created page with '== Problem 10 == The first four terms of an arithmetic sequence are <math>p</math>, <math>9</math>, <math>3p-q</math>, and <math>3p+q</math>. What is the <math>2010^\text{th}</ma…'
 
Thestudyofeverything (talk | contribs)
editor
1,113 edits
No edit summary
 
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
== Problem 10 ==
== Problem ==
The first four terms of an arithmetic sequence are <math>p</math>, <math>9</math>, <math>3p-q</math>, and <math>3p+q</math>. What is the <math>2010^\text{th}</math> term of this sequence?
The first four terms of an arithmetic sequence are <math>p</math>, <math>9</math>, <math>3p-q</math>, and <math>3p+q</math>. What is the <math>2010^\text{th}</math> term of this sequence?


<math>\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049</math>
<math>\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049</math>


== Solution ==
== Solution 1 ==
<math>3p-q</math> and <math>3p+q</math> are consecutive terms, so the common difference is <math>(3p+q)-(3p-q) = 2q</math>.
<math>3p-q</math> and <math>3p+q</math> are consecutive terms, so the common difference is <math>(3p+q)-(3p-q) = 2q</math>.


<cmath>\begin{align*}p+2q &= 9\\
9+2q &= 3p-q\\
q&=2\\
p&=5\end{align*}</cmath>


<math>p+2q = 9</math>
The common difference is <math>4</math>. The first term is <math>5</math> and the <math>2010^\text{th}</math> term is


<math>9+2q = 3p-q</math>
<cmath>5+4(2009) = \boxed{\textbf{(A) }8041}</cmath>


<math>q=2</math>
== Solution 2 ==
Since all the answer choices are around <math>2010 \cdot 4 = 8040</math>, the common difference must be <math>4</math>. The first term is therefore <math>9 - 4 = 5</math>, so the <math>2010^\text{th}</math> term is <math>5 + 4 \cdot 2009 = \boxed{\textbf{(A) }8041}</math>.


<math>p=5</math>
==Video Solution 1==
https://youtu.be/3gsf_XbOhhY


~Education, the Study of Everything


The common difference is <math>4</math>. The first term is <math>5</math> and the <math>2010^\text{th}</math> term is
== See also ==
{{AMC12 box|year=2010|num-b=9|num-a=11|ab=A}}


<math>5+4(2009) = \boxed{8041\ \textbf{(A)}}</math>
[[Category:Introductory Algebra Problems]]
{{MAA Notice}}

Latest revision as of 19:47, 27 October 2022

Problem

The first four terms of an arithmetic sequence are $p$, $9$, $3p-q$, and $3p+q$. What is the $2010^\text{th}$ term of this sequence?

$\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049$

Solution 1

$3p-q$ and $3p+q$ are consecutive terms, so the common difference is $(3p+q)-(3p-q) = 2q$.

\begin{align*}p+2q &= 9\\ 9+2q &= 3p-q\\ q&=2\\ p&=5\end{align*}

The common difference is $4$. The first term is $5$ and the $2010^\text{th}$ term is

\[5+4(2009) = \boxed{\textbf{(A) }8041}\]

Solution 2

Since all the answer choices are around $2010 \cdot 4 = 8040$, the common difference must be $4$. The first term is therefore $9 - 4 = 5$, so the $2010^\text{th}$ term is $5 + 4 \cdot 2009 = \boxed{\textbf{(A) }8041}$.

Video Solution 1

https://youtu.be/3gsf_XbOhhY

~Education, the Study of Everything

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_10&oldid=179581"