2002 AMC 10B Problem 19

Attempt Problem 19 of the 2002 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 2002 AMC 10B solutions, or check the answer key.

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19.

Suppose that $\{a_n\}$ is an arithmetic sequence with $a_1 + a_2 + \cdots + a_{100} = 100$ and $a_{101} + a_{102} + \cdots + a_{200} = 200.$ What is the value of $a_2 - a_1?$

$0.0001$

$0.001$

$0.01$

$0.1$

$1$

Answer: C

Concepts:arithmetic sequence summation

Difficulty rating: 1460

Solution:

Let $d = a_2 - a_1.$ Then $a_{k+100} = a_k + 100d,$ so the second block sum is the first plus $100\cdot 100 d:$ $\begin{aligned} &a_{101} + \cdots + a_{200} \\ &= (a_1 + \cdots + a_{100}) \\ &\quad {}+ 10000d. \end{aligned}$

Therefore $200 = 100 + 10000d,$ giving $d = \dfrac{100}{10000} = 0.01.$

Thus, the correct answer is C.

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