AMC8 2026

AMC8 2026 · Q14

AMC8 2026 · Q14. It mainly tests Arithmetic sequences basics.

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在线上选择了三个等间距的整数。第一个和第二个数的和是40，而第二个和第三个数的和是60。三个数的总和是多少？

(A) 70 70

(B) 75 75

(D) 85 85

(E) 90 90

Answer

Correct choice: (B)

正确答案：（B）

Solution

Since the integers are equally spaced, they form an arithmetic sequence, so we can label the middle term $a$, the smallest one $a-d$ and the largest one $a+d$. We are given that $(a-d)+a=2a-d=40$ and $a+(a+d)=2a+d=60$. Solving, $2d=60-40=20$ so $d=10$, which implies that $a=\dfrac{60-d}{2}=25$. Finally, the sum of all 3 numbers is $(a-d)+a+(a+d)=3a=3 \cdot 25 = \boxed{\textbf{(B)} 75}$.

因为这三个整数等间距，所以它们构成等差数列，因此可以将中间的数标记为$a$，最小的数为$a-d$，最大的数为$a+d$。题中给出$(a-d)+a=2a-d=40$，以及$a+(a+d)=2a+d=60$。解这个方程组得$2d=60-40=20$，所以$d=10$，由此得到$a=\dfrac{60-d}{2}=25$。最后，三个数的和为$(a-d)+a+(a+d)=3a=3 \cdot 25 = \boxed{\textbf{(B)} 75}$。

Topics

AR.012 Arithmetic sequences basics