2002 AMC 10B Problem 19

Below is the professionally curated solution for Problem 19 of the 2002 AMC 10B, 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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Concepts:arithmetic sequencesummation

Difficulty rating: 1460

19.

Suppose that {an}\{a_n\} is an arithmetic sequence with a1+a2++a100=100anda101+a102++a200=200.a_1 + a_2 + \cdots + a_{100} = 100 \quad\text{and}\quad a_{101} + a_{102} + \cdots + a_{200} = 200. What is the value of a2a1?a_2 - a_1?

0.00010.0001

0.0010.001

0.010.01

0.10.1

11

Solution:

Let d=a2a1.d = a_2 - a_1. Then ak+100=ak+100d,a_{k+100} = a_k + 100d, so the second block sum is the first plus 100100d:100\cdot 100 d: a101++a200=(a1++a100)+10000d.a_{101} + \cdots + a_{200} = (a_1 + \cdots + a_{100}) + 10000d.

Therefore 200=100+10000d,200 = 100 + 10000d, giving d=10010000=0.01.d = \dfrac{100}{10000} = 0.01.

Thus, the correct answer is C.

Problem 19 in Other Years