1980 AMC 12 第 11 题

Question

已知某等差数列前

10

项之和为

100

，前

100

项之和为

10

，则前

110

项之和为

If the sum of the first

10

terms and the sum of the first

100

terms of a given arithmetic progression are

100

and

10,

respectively, then the sum of the first

110

terms is

90

-90

110

-110

-100

写出

S_n=\frac n2(2a+(n-1)d)

Write

S_n=\frac n2(2a+(n-1)d)

利用两个已知的部分和求出表达式

2a+109d

Use the two known partial sums to find the expression

2a+109d

由两个已知的和可得

\begin{aligned} 2a+9d&=20,\\ 2a+99d&=\frac15 \end{aligned}\text{。}

两式相减得到

90d=-\frac{99}{5}

，所以

d=-\frac{11}{50}

，且

2a=\frac{1099}{50}

。因此

\begin{aligned} S_{110}&=55(2a+109d)\\ &=55(-2)\\ &=-110 \end{aligned}\text{。}

因此，正确答案是 D。

The two given sums yield

\begin{aligned} 2a+9d&=20,\\ 2a+99d&=\frac15. \end{aligned}

Subtracting gives

90d=-\frac{99}{5},

so

d=-\frac{11}{50}

and

2a=\frac{1099}{50}.

Therefore

\begin{aligned} S_{110}&=55(2a+109d)\\ &=55(-2)\\ &=-110. \end{aligned}

Therefore, the correct answer is D.

其他年份的第 11 题