AMC10 2003 B

AMC10 2003 B · Q24

AMC10 2003 B · Q24. It mainly tests Sequences & recursion (algebra), Manipulating equations.

The first four terms in an arithmetic sequence are x + y, x − y, xy, and x/y, in that order. What is the fifth term?

一个等差数列的前四项依次为 x + y, x − y, xy, 和 x/y。第几项是多少？

(A) -15/8 -15/8

(B) -6/5 -6/5

(D) 27/20 27/20

(E) 123/40 123/40

Answer

Correct choice: (E)

正确答案：（E）

Solution

(E) Since the difference of the first two terms is $-2y$, the third and fourth terms of the sequence must be $x-3y$ and $x-5y$. Thus $$x-3y=xy \text{ and } x-5y=\frac{x}{y},$$ so $xy-5y^2=x$. Combining these equations we obtain $$(x-3y)-5y^2=x \text{ and, therefore, } -3y-5y^2=0.$$ Since $y$ cannot be $0$, we have $y=-3/5$, and it follows that $x=-9/8$. The fifth term in the sequence is $x-7y=123/40$.

（E）由于前两项之差为 $-2y$，该数列的第三项和第四项必为 $x-3y$ 与 $x-5y$。因此 $$x-3y=xy \text{ 且 } x-5y=\frac{x}{y},$$ 所以 $xy-5y^2=x$。联立这些方程可得 $$(x-3y)-5y^2=x \text{，因此 } -3y-5y^2=0。$$ 由于 $y$ 不能为 $0$，故 $y=-3/5$，从而 $x=-9/8$。数列的第五项为 $x-7y=123/40$。

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