AMC8 2025
AMC8 2025 · Q4
AMC8 2025 · Q4. It mainly tests Sequences & recursion (algebra).
Lucius is counting backward by $7$s. His first three numbers are $100$, $93$, and $86$. What is his $10$th number?
Lucius 按 $7$ 倒数计数。他的前三个数是 $100$、$93$ 和 $86$。他的第 $10$ 个数是多少？
(A)
\ 30
\ 30
(B)
\ 37
\ 37
(C)
\ 42
\ 42
(D)
\ 44
\ 44
(E)
\ 47
\ 47
Answer
Correct choice: (B)
正确答案：（B）
Solution
We plug $a=100, d=-7$ and $n=10$ into the formula $a+d(n-1)$ for the $n$th term of an arithmetic sequence whose first term is $a$ and common difference is $d$ to get $100-7(10-1) = \boxed{\text{(B) 37}}$.
将 $a=100$、$d=-7$ 和 $n=10$ 代入等差数列第 $n$ 项公式 $a+d(n-1)$，得 $100-7(10-1) = \boxed{\text{(B)}\ 37}$。
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