2008 AMC 12B Problem 12 — Solution | LIVE by Po-Shen Loh

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2008 AMC 12B Problem 12

Below is the professionally curated solution for Problem 12 of the 2008 AMC 12B, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:mean summation

Difficulty rating: 1500

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?

$2008$

$4015$

$4016$

$4{,}030{,}056$

$4{,}032{,}064$

Solution:

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.

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