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

Small hint:

Each term

a_{k+100}

exceeds

a_k

by

100d,

where

d = a_2 - a_1

Big hint:

Subtract the two block sums to isolate

10000d

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.