2008 AMC 10B Problem 13 — Solution | LIVE by Po-Shen Loh

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2008 AMC 10B Problem 13

Attempt Problem 13 of the 2008 AMC 10B 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 10B solutions, or check the answer key.

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13.

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?

$2008$

$4015$

$4016$

$4{,}030{,}056$

$4{,}032{,}064$

Answer: B

Concepts:mean summation algebraic manipulation

Difficulty rating: 1170

Solution:

Since the mean of the first $n$ terms is $n,$ their sum is $n^2.$

The $n$ th term is $n^2-(n-1)^2=2n-1,$ so the $2008$ th term is $2\cdot 2008-1=4015.$

Thus, the correct answer is B.

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