2008 AMC 12B Problem 12

Attempt Problem 12 of the 2008 AMC 12B 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 2008 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

For each positive integer

n,

the mean of the first

n

terms of a sequence is

n.

What is the

2008

th term of the sequence?

2008

4015

4016

4{,}030{,}056

4{,}032{,}064

If the mean of the first

n

terms is

n,

then their sum is

n^2

The

n

th term equals the sum of the first

n

minus the sum of the first

n - 1

Since the mean of the first

n

terms is

n,

their sum is

n \cdot n = n^2.

The

n

th term is the difference of consecutive sums,

n^2 - (n-1)^2 = 2n - 1.

For

n = 2008,

the term is

2 \cdot 2008 - 1 = 4015.

Thus, the correct answer is B.

Problem 12 in other years