AMC12 2005 A

AMC12 2005 A · Q13

AMC12 2005 A · Q13. It mainly tests Linear equations, Arithmetic sequences basics.

In the five-sided star shown, the letters $A$, $B$, $C$, $D$ and $E$ are replaced by the numbers $3$, $5$, $6$, $7$ and $9$, although not necessarily in that order. The sums of the numbers at the ends of the line segments $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$ and $\overline{EA}$ form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?

在所示的五角星中，字母 $A$, $B$, $C$, $D$ 和 $E$ 被数字 $3$, $5$, $6$, $7$ 和 $9$ 替换（不一定按此顺序）。线段 $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$ 和 $\overline{EA}$ 两端数字之和形成一个等差数列（不一定按此顺序）。这个等差数列的中间项是多少？

(A) 9 9

(B) 10 10

(D) 12 12

(E) 13 13

Answer

Correct choice: (D)

正确答案：（D）

Solution

$(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)$ (i.e., each number is counted twice). The sum $A + B + C + D + E$ will always be $3 + 5 + 6 + 7 + 9 = 30$, so the arithmetic sequence has a sum of $2 \cdot 30 = 60$. The middle term must be the average of the five numbers, which is $\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}$.

$(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)$（即每个数字被计算了两次）。$A + B + C + D + E$ 恒为 $3 + 5 + 6 + 7 + 9 = 30$，因此该等差数列的和为 $2 \cdot 30 = 60$。中间项必须是这五个数的平均数，即 $\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}$。

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