AMC10 2022 A

AMC10 2022 A · Q20

AMC10 2022 A · Q20. It mainly tests Sequences & recursion (algebra), Word problems (algebra).

A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are $57$, $60$, and $91$. What is the fourth term of this sequence?

一个四项等差正整数数列的每项与一个四项等比正整数数列对应项相加形成一个四项数列。结果数列的前三项为 $57$、$60$ 和 $91$。该数列的第四项是多少？

(A) 190 190

(B) 194 194

(D) 202 202

(E) 206 206

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let the arithmetic sequence be $a,a+d,a+2d,a+3d$ and the geometric sequence be $b,br,br^2,br^3.$ We are given that \begin{align*} a+b&=57, \\ a+d+br&=60, \\ a+2d+br^2&=91, \end{align*} and we wish to find $a+3d+br^3.$ Subtracting the first equation from the second and the second equation from the third, we get \begin{align*} d+b(r-1)&=3, \\ d+br(r-1)&=31. \end{align*} Subtract these results, we get \[b(r-1)^2=28.\] Note that either $(r-1)^2=1$ or $(r-1)^2=4.$ We proceed with casework: - If $(r-1)^2=1,$ then $r=2,b=28,a=29,$ and $d=-25.$ The arithmetic sequence is $29,4,-21,-46,$ arriving at a contradiction. - If $(r-1)^2=4,$ then $r=3,b=7,a=50,$ and $d=-11.$ The arithmetic sequence is $50,39,28,17,$ and the geometric sequence is $7,21,63,189.$ This case is valid. Therefore, The answer is $a+3d+br^3=17+189=\boxed{\textbf{(E) } 206}.$

设等差数列为 $a,a+d,a+2d,a+3d$，等比数列为 $b,br,br^2,br^3$。已知 \begin{align*} a+b&=57, \\ a+d+br&=60, \\ a+2d+br^2&=91, \end{align*} 我们求 $a+3d+br^3$。将第一式从第二式减去，第二式从第三式减去，得到 \begin{align*} d+b(r-1)&=3, \\ d+br(r-1)&=31. \end{align*} 将这两结果相减，得到 \[b(r-1)^2=28.\] 注意 $(r-1)^2=1$ 或 $(r-1)^2=4$。我们分情况讨论： - 若 $(r-1)^2=1$，则 $r=2,b=28,a=29,d=-25$。等差数列为 $29,4,-21,-46$，矛盾。 - 若 $(r-1)^2=4$，则 $r=3,b=7,a=50,d=-11$。等差数列为 $50,39,28,17$，等比数列为 $7,21,63,189$。此情况有效。因此，答案是 $a+3d+br^3=17+189=\boxed{\textbf{(E) } 206}$。

Topics