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2002 AMC 10B Problem 19

Problem 19 of 25HarderAlgebra

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 a2−a1?a_2 - a_1?

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Solution

Let d=a2−a1.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 100⋅100d: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.
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