AMC12 2009 A

AMC12 2009 A · Q17

AMC12 2009 A · Q17. It mainly tests Sequences & recursion (algebra), Algebra misc.

Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$, and the sum of the second series is $r_2$. What is $r_1 + r_2$?

设 $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ 和 $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ 是两个不同的正数无限等比级数，且首项相同。第一个级数的和为 $r_1$，第二个级数的和为 $r_2$。求 $r_1 + r_2$。

(A) 0 0

(B) $\frac{1}{2}$ $\frac{1}{2}$

(D) $\frac{1+\sqrt{5}}{2}$ $\frac{1+\sqrt{5}}{2}$

(E) 2 2

Answer

Correct choice: (C)

正确答案：（C）

Solution

Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$. Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$. This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$. As we are given that $r_1$ and $r_2$ are distinct, these must be precisely the two roots of the equation $x^2 - x + a = 0$. Using Vieta's formulas we get that the sum of these two roots is $\boxed{1}$.

利用等比级数求和公式可得，这两个级数的和分别为 $\frac a{1-r_1}$ 和 $\frac a{1-r_2}$。因此有 $\frac a{1-r_1}=r_1$ 且 $\frac a{1-r_2}=r_2$。这可改写为 $r_1(1-r_1)=r_2(1-r_2)=a$。由于已知 $r_1$ 与 $r_2$ 不同，它们必为方程 $x^2-x+a=0$ 的两个根。由韦达定理，这两个根之和为 $\boxed{1}$。

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