AMC10 2002 B

AMC10 2002 B · Q19

AMC10 2002 B · Q19. It mainly tests Sequences & recursion (algebra), Averages (mean).

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 $a_2 - a_1$?

设 $\{a_n\}$ 是等差数列，且 $a_1 + a_2 + \cdots + a_{100} = 100$，$a_{101} + a_{102} + \cdots + a_{200} = 200$。求 $a_2 - a_1$ 的值。

(A) 0.0001 0.0001

(B) 0.001 0.001

(D) 0.1 0.1

(E) 1 1

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) Let $d=a_2-a_1$. Then $a_{k+100}=a_k+100d$, and $a_{101}+a_{102}+\cdots+a_{200}=(a_1+100d)+(a_2+100d)+\cdots+(a_{100}+100d)$ $=a_1+a_2+\cdots+a_{100}+10{,}000d$. Thus $200=100+10{,}000d$ and $d=\frac{100}{10{,}000}=0.01$.

（C）令 $d=a_2-a_1$。则 $a_{k+100}=a_k+100d$，并且 $a_{101}+a_{102}+\cdots+a_{200}=(a_1+100d)+(a_2+100d)+\cdots+(a_{100}+100d)$ $=a_1+a_2+\cdots+a_{100}+10{,}000d$。因此 $200=100+10{,}000d$，并且 $d=\frac{100}{10{,}000}=0.01$。

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