AMC 8 2026 Problem 14
AMC8 2026 年第 14 题
数列与规律★★★☆☆
Jami picked three equally spaced integer numbers on the number line. The sum of the first and the second numbers is 40, while the sum of the second and third numbers is 60. What is the sum of all three numbers?
Jami 在数轴上选取了三个等间距的整数。第一个数与第二个数之和为 40，第二个数与第三个数之和为 60。这三个数的总和是多少？
- A.70
- B.75
- C.80
- D.85
- E.90
正确答案 · Correct Answer
B
解析 · Solution
解法1
设中间数为 a，三数为 a−d, a, a+d。已知 (a−d)+a=2a−d=40，a+(a+d)=2a+d=60。两式相加：4a=100，a=25。三数之和 = 3a = 75。
两和之差 60−40=20 是公差的两倍，故公差 d=10。由 2a+d=60 得 a=25。三数为 15, 25, 35，总和 75。
Solution 1
Let the middle number be a, the three numbers are a−d, a, a+d. Given: (a−d)+a=2a−d=40 and a+(a+d)=2a+d=60. Adding: 4a=100, so a=25. Sum of all three = 3a = 75.
The difference between the two given sums is 60−40=20, which is twice the common difference. So d=10. From 2a+d=60 or 2a−d=40, we get a=25. The three numbers are 15, 25, 35. Sum = 75.
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