2010 AMC 12A Problems/Problem 10

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2010 AMC 12A Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

Video Solution 1

See also