2010 AMC 12A Problem 10

Attempt Problem 10 of the 2010 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12A solutions, or check the answer key.

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10.

The first four terms of an arithmetic sequence are

p,

9,

3p-q,

and

3p+q.

What is the

2010

th term of this sequence?

8041

8043

8045

8047

8049

The common difference equals

(3p+q)-(3p-q)=2q

Also

9-p=2q

and

(3p-q)-9=2q;

solve for

p

and

q

Consecutive terms differ by a common difference

d.

From the last two terms,

d=(3p+q)-(3p-q)=2q.

From the first two terms,

9-p=d=2q,

and from the second and third,

(3p-q)-9=d=2q.

Solving this system gives

p=5,

q=2,

and

d=4.

The

2010

th term is

\begin{aligned} p+2009d &= 5+2009\cdot4 \\ &= 8041. \end{aligned}

Thus, A is the correct answer.

Problem 10 in other years