/

AMC8 1990

AMC8 1990 · Q16

AMC8 1990 · Q16. It mainly tests Arithmetic sequences basics, Arithmetic misc.

1990 - 1980 + 1970 - 1960 + \dots - 20 + 10 =
1990 - 1980 + 1970 - 1960 + \dots - 20 + 10 =
(A) -990 -990
(B) -10 -10
(C) 990 990
(D) 1000 1000
(E) 1990 1990
Answer
Correct choice: (D)
正确答案：（D）
Solution
By grouping as shown below, there are $\frac{199+1}{2} = 100$ groups of 10, for a sum of 1000: \[[1990 - 1980] + [1970 - 1960] + \cdots + [30 - 20] + 10.\]
By grouping as shown below, there are $\frac{199+1}{2} = 100$ groups of 10, for a sum of 1000: \[[1990 - 1980] + [1970 - 1960] + \cdots + [30 - 20] + 10.\]
Topics
AR.012 Arithmetic sequences basics AR.015 Arithmetic misc
Related Questions
AMC10 2010 A · Q8 AMC10 2022 B · Q4 AMC12 2021 A · Q16 AMC10 2023 B · Q9 AMC10 2014 A · Q3 AMC10 2021 A · Q6 AMC10 2013 B · Q13 AMC8 2013 · Q15
Practice full AMC exams on amcdrill.
Try full-length practice and diagnostics at www.amcdrill.com.