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 {an}\{a_n\} is an arithmetic sequence with a1+a2++a100=100a_1 + a_2 + \cdots + a_{100} = 100 and a101+a102++a200=200.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

Answer: C
Concepts:arithmetic sequencesummation
Difficulty rating: 1460
Small Hint:

Each term ak+100a_{k+100} exceeds aka_k by 100d,100d, where d=a2a1d = a_2 - a_1

Big Hint:

Subtract the two block sums to isolate 10000d10000d

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. \begin{aligned} &a_{101} + \cdots + a_{200} \\ &= (a_1 + \cdots + a_{100}) \\ &\quad {}+ 10000d. \end{aligned}

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 18#18
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