Question

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

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 $a_2 - a_1?$

$0.0001$

$0.001$

$0.01$

$0.1$

$1$

答案：C

知识点：等差数列求和

难度评级：1460

解答：

设 $d = a_2 - a_1$ 。因为 $a_{k+100} = a_k + 100d$ ，第二组的一百项相对于第一组对应项的增量都相同，所以第二组的总和比第一组多 $100\cdot 100 d$ ： $\begin{aligned} &a_{101} + \cdots + a_{200} \\ &= (a_1 + \cdots + a_{100}) \\ &\quad {}+ 10000d. \end{aligned}$

因此 $200 = 100 + 10000d$ ，解得 $d = \dfrac{100}{10000} = 0.01$ 。

所以正确答案是 C。

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.

2002 AMC 10B 第 19 题

答案：C

难度评级：1460

解答：

其他年份的第 19 题