Axiom Learning·AP PrecalculusChapter 6 · Sequences
数列 — Arithmetic, Geometric & Connections to Linear/Exponential Functions
- 区分等差数列与等比数列 · Distinguish arithmetic from geometric sequences
- 推导并使用等差/等比数列的通项公式 · Derive and apply general term formulas
- 比较数列与线性/指数函数的联系与区别 · Connect sequences to linear and exponential functions
- 理解数列的定义域为整数，图像为离散点 · Understand discrete domain of sequences
- 由两个值写出线性/指数函数，处理间隔不为 1 的表格 · Build functions from two values; handle equal intervals of any length
- 在情境中建立并解释数列 · Model and interpret sequences in context
| Term 术语 | Definition 定义 | Chinese 中文 |
|---|---|---|
| Sequence | Ordered list of numbers following a rule | 数列 |
| Term ($a_n$) | Each number in the sequence | 项 |
| Common Difference ($d$) | Constant added between consecutive terms (arithmetic) | 公差 |
| Common Ratio ($r$) | Constant multiplied between consecutive terms (geometric) | 公比 |
| Finite Sequence | Has a definite first and last term | 有限数列 |
| Infinite Sequence | Continues indefinitely (indicated by …) | 无限数列 |
| Explicit Formula | Formula that gives the $n$th term directly from $n$ | 通项公式 |
| Recursive Formula | Each term defined using the previous term | 递推公式 |
6.1 What Is a Sequence? 数列是什么？
数列是按特定顺序排列的一组数字，每个数字称为一项。
- Domain 定义域: positive integers $\{1, 2, 3, \ldots\}$ or whole numbers $\{0, 1, 2, \ldots\}$ — always integers, always discrete
- Range 值域: set of real numbers (the terms themselves)
- Graph: isolated dots, not a continuous curve 图像是离散点，不是连续曲线
数列的定义域只包含整数，所以图像是一个个分离的点（离散）。
对应的线性/指数函数定义域是全体实数，图像是连续的直线或曲线。
因此数列和对应函数的定义域不同，但描述的变化模式相同。
Sequence Generator · 数列生成器
Build each term one step at a time, then connect the pattern to explicit and recursive rules. 逐项生成数列，并同时观察通项公式与递推公式。
| n | an | Difference 差 |
|---|
6.2 Arithmetic Sequences 等差数列
$d$: common difference 公差
$n$: term number 项数
Matches linear function $f(x)=b+mx$ form
If you know the $k$th term $a_k$: $$a_n = a_k + (n-k)d$$ 不知道首项时，用已知的第 $k$ 项代入。
An arithmetic sequence has common difference $d = -3.5$ and 11th term $a_{11} = 7$. Find the 50th term.
We know $a_{11}=7$, $d=-3.5$, want $a_{50}$.
$a_{50} = a_{11} + (50-11)\cdot(-3.5)$
$= 7 + 39\times(-3.5)$
$= 7 - 136.5 = \boxed{-129.5}$
An arithmetic sequence (increasing) includes $-2$ and $13$, with exactly 4 terms between them (so they are separated by 5 steps). Find the terms in between.
If $-2$ and $13$ are separated by 4 terms, there are $4+1=5$ steps between them.
$d = \dfrac{13-(-2)}{5} = \dfrac{15}{5} = 3$
Starting from $-2$, add $d=3$ each time:
$-2,\; 1,\; 4,\; 7,\; 10,\; 13$
The 4 terms between $-2$ and $13$ are: $\boxed{1,\; 4,\; 7,\; 10}$
Given the arithmetic sequence $\{5, 8, 11, 14, 17, \ldots\}$:
(a) Find $d$ and write the explicit formula for $a_n$.
(b) Find $a_{20}$.
(c) Is 200 a term of this sequence?
$d = 8-5 = 3$; first term $a_1 = 5$
$a_n = 5 + (n-1)\cdot3 = 5+3n-3 = 3n+2$
$a_{20} = 3(20)+2 = 60+2 = \boxed{62}$
Set $a_n = 200$: $3n+2 = 200 \Rightarrow 3n=198 \Rightarrow n=66$
$n=66$ is a positive integer ✓ → Yes, $200 = a_{66}$.
An arithmetic sequence has $a_3 = 11$ and $a_7 = 27$.
(a) Find the common difference $d$.
(b) Write the explicit formula $a_n$.
(c) Find $a_{15}$.
$d = \dfrac{27-11}{4} = 4$
(b) $a_n = a_3 + (n-3)\cdot4 = 11+4(n-3) = 4n-1$
(c) $a_{15} = 4(15)-1 = \boxed{59}$
6.3 Geometric Sequences 等比数列
$r$: common ratio 公比
$n$: term number
The first term of a geometric sequence is 36 and the fifth term is $\dfrac{64}{9}$. What could be the 6th term?
$a_5 = a_1\cdot r^4 \Rightarrow \dfrac{64}{9} = 36\cdot r^4$
$r^4 = \dfrac{64}{9\times36} = \dfrac{64}{324} = \dfrac{16}{81}$
$r^2 = \dfrac{4}{9} \Rightarrow r = \pm\dfrac{2}{3}$
两个可能的公比：$r=\dfrac{2}{3}$ 或 $r=-\dfrac{2}{3}$
If $r=+\dfrac{2}{3}$: $a_6 = \dfrac{64}{9}\cdot\dfrac{2}{3} = \dfrac{128}{27}$
If $r=-\dfrac{2}{3}$: $a_6 = \dfrac{64}{9}\cdot\left(-\dfrac{2}{3}\right) = -\dfrac{128}{27}$
Arithmetic: $\{-1, 4, 9, 14, 19, 24, 29,\ldots\}$ Geometric: $\{1, 5, 25, 125, 625,\ldots\}$
Find the common difference/ratio. Then find the average rate of change (AROC) between terms 1–2 and terms 2–3 for each.
AROC between $a_1$ and $a_2$: $\dfrac{4-(-1)}{2-1}=5$
AROC between $a_2$ and $a_3$: $\dfrac{9-4}{3-2}=5$
→ AROC is constant = $d = 5$ 等差数列的AROC恒定（与线性函数斜率类比）
AROC between $a_1$ and $a_2$: $\dfrac{5-1}{2-1}=4$
AROC between $a_2$ and $a_3$: $\dfrac{25-5}{3-2}=20$
→ AROC is not constant — it increases (proportional growth) 等比数列的AROC不恒定，在增大
A geometric sequence has $a_2 = 6$ and $a_5 = 162$.
(a) Find the common ratio $r$.
(b) Write the explicit formula $a_n$ (index starts at 1).
(c) Find $a_8$.
(b) $a_1 = a_2/r = 6/3 = 2$; $a_n = 2\cdot3^{n-1}$
(c) $a_8 = 2\cdot3^7 = 2\times2187 = \boxed{4374}$
Identify whether each is arithmetic or geometric. Then write the explicit formula.
(a) $\{3, 7, 11, 15, 19,\ldots\}$
(b) $\left\{100, 50, 25, 12.5, \ldots\right\}$
(c) $\{2, -6, 18, -54,\ldots\}$
$a_n = 3+(n-1)\cdot4 = 4n-1$
(b) Ratios: $50/100=0.5$ (constant) → Geometric, $r=0.5$
$a_n = 100\cdot(0.5)^{n-1}$
(c) Ratios: $-6/2=-3$ (constant) → Geometric, $r=-3$
$a_n = 2\cdot(-3)^{n-1}$
6.4 Average Rate of Change & Sequence Types 变化率与数列类型
- Compute first differences (subtract consecutive terms): constant → Arithmetic
- Compute consecutive ratios (divide consecutive terms): constant → Geometric
对于等比数列，AROC 不恒定，在增大（或减小），与指数函数的变化率类比。
For each equation, describe the sequence it represents (arithmetic or geometric) and state the relevant parameter ($d$ or $r$):
- $f(n) = 3n + 4$
- $f(n) = -\left(\dfrac{1}{2}\right)^n$
- $f(n) = 4(n-1) - 3.5$
- $f(n) = -3 + 6(n-4)$
$d=3$ (coefficient of $n$); e.g., $f(1)=7, f(2)=10, f(3)=13,\ldots$
$r=\frac{1}{2}$ (constant ratio); $a_0=-1, a_1=-\frac{1}{2}, a_2=-\frac{1}{4},\ldots$
$d=4$ and $a_1=-3.5$
$d=6$; this is point-slope form with $a_4=-3$
6.5 Sequences vs. Functions — Domain & Graphs 数列与函数的区别
Function: domain = all real numbers — graph is a continuous line or curve
数列定义域为整数 → 图像是离散点；函数定义域为实数 → 图像是连续曲线
Given the sequence $\{-4, -0.5, 3, 6.5, 10, 13.5, 17, 20.5\}$:
- State the type of sequence.
- Write the explicit formula for the $n$th term.
- State the domain and range of the sequence.
- State the domain and range of the corresponding function.
Differences: $-0.5-(-4)=3.5$, $3-(-0.5)=3.5$, … constant
→ Arithmetic with $d=3.5$
$a_1=-4$, $d=3.5$
$a_n = -4+(n-1)(3.5) = 3.5n - 7.5$
Domain: $\{1, 2, 3, 4, 5, 6, 7, 8\}$ (8 terms, finite 有限数列)
Range: $\{-4, -0.5, 3, 6.5, 10, 13.5, 17, 20.5\}$
Domain: $(-\infty, +\infty)$ — all real numbers
Range: $(-\infty, +\infty)$
Note: the function is continuous; the sequence is just the values at integer inputs 1–8.
Given the sequence $\left\{-15,\; -5,\; -\dfrac{5}{3},\; -\dfrac{5}{9},\; -\dfrac{5}{27}\right\}$:
- State the type of sequence.
- Write the explicit formula.
- State the domain and range of the sequence.
- State the domain and range of the corresponding exponential function.
Ratios: $\dfrac{-5}{-15}=\dfrac{1}{3}$, $\dfrac{-5/3}{-5}=\dfrac{1}{3}$, … constant
→ Geometric with $r=\dfrac{1}{3}$
$a_1=-15$, $r=\dfrac{1}{3}$
$a_n = -15\cdot\left(\dfrac{1}{3}\right)^{n-1}$
Domain: $\{1, 2, 3, 4, 5\}$ (5 terms, finite)
Range: $\left\{-15,\,-5,\,-\dfrac{5}{3},\,-\dfrac{5}{9},\,-\dfrac{5}{27}\right\}$ (all negative)
Domain: $(-\infty,+\infty)$
Range: $(-\infty, 0)$ (since $a=-15<0$, function is always negative)
HA: $y=0$
The arithmetic sequence has $a_1=2$ and $d=4$.
(a) Write the first 5 terms.
(b) State the domain and range of the sequence.
(c) Write the corresponding linear function and state its domain.
(d) What is the key difference between the two?
(b) Sequence domain: $\{1,2,3,4,5\}$; Range: $\{2,6,10,14,18\}$
(c) $f(x)=4x-2$; Domain: $(-\infty,+\infty)$
(d) The sequence only has values at integer inputs; the function is continuous for all real $x$. The graph of the sequence is 5 discrete dots; the function is a continuous line.
6.6 Linear vs. Exponential — Complete Comparison 线性 vs. 指数全面对比
等比数列 ↔ 指数函数（都是"重复乘法"）
| Feature 特性 | Linear / Arithmetic 线性函数 / 等差数列 |
Exponential / Geometric 指数函数 / 等比数列 |
|---|---|---|
| Function form 函数形式 | $f(x)=b+mx$ | $f(x)=a\cdot b^x$ |
| Sequence form 数列形式 | $a_n = a_0 + nd$ (index at 0) $a_n = a_1+(n-1)d$ (index at 1) |
$a_n = a_0\cdot r^n$ (index at 0) $a_n = a_1\cdot r^{n-1}$ (index at 1) |
| Initial value 初始值 | $b$ (y-intercept) or $a_0$ | $a$ (coefficient) or $a_0$ |
| Change factor 变化因子 | $m$ (slope 斜率) / $d$ (common diff. 公差) | $b$ (base 底数) / $r$ (common ratio 公比) |
| Growth type 增长类型 | Constant change 常数变化（加法） | Constant proportional change 等比变化（乘法） |
| AROC 平均变化率 | Constant = $m$ = $d$ | Not constant — increases/decreases |
| Graph 图像 | Straight line 直线 | Curve 曲线（HA at $y=0$） |
| Sequence graph | Collinear discrete points 共线离散点 | Exponentially-spaced discrete points 离散点（指数间隔） |
| Identify from table | First differences constant 一阶差分恒定 | Consecutive ratios constant 连续项之比恒定 |
| Point-$k$ form | $a_n = a_k + (n-k)d$ | $a_n = a_k\cdot r^{n-k}$ |
| Function through a point $(x_1,y_1)$ | $f(x)=y_1+m(x-x_1)$ (point-slope form) | $f(x)=y_1\cdot r^{\,x-x_1}$ (point-ratio form) |
Functions from Two Values & Intervals Longer Than 1 由两个值写函数；间隔不为 1
- Linear through $(x_1,y_1)$ and $(x_2,y_2)$: $m=\dfrac{y_2-y_1}{x_2-x_1}$, then $f(x)=y_1+m(x-x_1)$ — like $a_n=a_k+(n-k)d$.
- Exponential: $r^{\,x_2-x_1}=\dfrac{y_2}{y_1}$, then $f(x)=y_1\cdot r^{\,x-x_1}$ — like $a_n=a_k\cdot r^{\,n-k}$.
- For a table whose inputs step by $h\neq1$: a constant difference $D$ per step means slope $\dfrac{D}{h}$; a constant ratio $R$ per step means base $R^{1/h}$ per unit (CED 2.2.B.1: "equal-length input-value intervals").
(a) A linear function has $f(3)=10$ and $f(7)=22$. Write it in point-slope form and in the form $b+mx$.
(b) An exponential function has $g(2)=18$ and $g(5)=486$. Write it in point-ratio form and in the form $ab^x$.
(c) Which sequences correspond to $f$ and $g$ if we only use whole-number inputs?
| $x$ | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| $p(x)$ | 5 | 20 | 80 | 320 |
| $x$ | 1 | 4 | 7 | 10 |
|---|---|---|---|---|
| $q(x)$ | 7 | 16 | 25 | 34 |
Decide whether each function is linear or exponential, and write its formula. What is the change per one unit of $x$?
$q$: differences $9$ over each 3-unit interval → linear. Slope $\tfrac{9}{3}=3$, so $q(x)=7+3(x-1)=3x+4$.
Determine whether each sequence is arithmetic, geometric, or neither. If arithmetic or geometric, write the explicit formula.
| $n$ | Seq A | Seq B | Seq C |
|---|---|---|---|
| 1 | 3 | 2 | 1 |
| 2 | 6 | 6 | 4 |
| 3 | 12 | 18 | 9 |
| 4 | 24 | 54 | 16 |
Geometric, $r=2$, $a_1=3$
$a_n = 3\cdot2^{n-1}$
Geometric, $r=3$, $a_1=2$
$a_n = 2\cdot3^{n-1}$
Ratios $4/1=4$, $9/4=2.25$ → not constant
Neither — this is $a_n=n^2$ (perfect squares 完全平方数).
Second differences $5-3=2$, $7-5=2$ are constant, so it follows a quadratic pattern (see Chapter 1, 7.2). 二阶差分恒定 → 二次型。
Given the sequence $\{81, 27, 9, 3, 1, \frac{1}{3}, \ldots\}$:
(a) Identify the type and find $r$.
(b) Write the explicit formula (index starts at 1).
(c) Find $a_{10}$.
(d) Write the corresponding exponential function and compare domains.
(b) $a_n = 81\cdot\left(\dfrac{1}{3}\right)^{n-1} = 3^4\cdot3^{-(n-1)}=3^{5-n}$
(c) $a_{10}=81\cdot\left(\dfrac{1}{3}\right)^9 = \dfrac{81}{3^9}=\dfrac{3^4}{3^9}=3^{-5}=\dfrac{1}{243}$
(d) $f(x)=81\cdot\left(\dfrac{1}{3}\right)^{x-1}$; Domain of function: $(-\infty,+\infty)$; Domain of sequence: $\{1,2,3,\ldots\}$ (positive integers only)
6.7 Sequences in Context 情境中的数列
- Repeated addition of the same amount → arithmetic; repeated multiplication by the same factor (a fixed percent) → geometric. 每次加同样的量 → 等差；每次乘同样的倍数（固定百分比）→ 等比
- Decide what $n$ counts and whether to start at $a_0$ (the starting amount, before any change) or $a_1$ (the first item). 想清楚 $n$ 数的是什么，从 $a_0$ 还是 $a_1$ 开始
- Interpret $d$ or $r$ with units: "$d=3$ seats per row", "$r=0.75$: each bounce is 75% of the previous height". 带单位解释 $d$ 或 $r$
- The domain is whole numbers in a limited range (you cannot have row 2.5 or a negative number of bounces). 定义域是有限范围内的整数
(a) A theater has 30 rows. Row 1 has 22 seats, and each row has 3 more seats than the row in front of it. Write a formula for the number of seats $s_n$ in row $n$. How many seats are in row 30? Which row has 64 seats?
(b) A ball is dropped from 200 cm. After each bounce it rises to 75% of its previous height. Let $h_n$ be the height after the $n$th bounce, with $h_0=200$. Write a formula, find $h_4$, and find the first bounce that stays below 20 cm.
Same amount added each row: $d=3$ seats per row, $s_1=22$. $\;s_n=22+3(n-1)$, domain $n=1,2,\dots,30$.
$s_{30}=22+3(29)=109$ seats. $22+3(n-1)=64\Rightarrow n=15$: row 15.
Same factor each bounce: $r=0.75$, and $h_0=200$ is the height before any bounce, so start at $n=0$: $\;h_n=200(0.75)^n$.
$h_4=200(0.75)^4\approx63.281$ cm.
$h_8=200(0.75)^8\approx20.023$ cm (still above 20), $h_9\approx15.017$ cm. The 9th bounce is the first below 20 cm.
Offer A pays \$52,000 in year 1 with a 4% raise every year. Offer B pays \$52,000 in year 1 with a \$2,500 raise every year.
(a) Write formulas $A_n$ and $B_n$ for the salary in year $n$, and identify each type of sequence.
(b) Compare the raises in year 2 and in year 11.
(c) Compare the salaries in year 10 and year 15. In which year does Offer A first pay more?
(b) Offer A's raise grows: year 2 raise $=\$2{,}080$ (less than B's \$2,500), year 11 raise $\approx\$2{,}960.49$ (more than \$2,500). B's raise is always \$2,500 (CED 2.1.B.3).
(c) Year 10: $A_{10}\approx\$74{,}012.21$, $B_{10}=\$74{,}500$. Year 15: $A_{15}\approx\$90{,}047.18$, $B_{15}=\$87{,}000$. Offer A first pays more in year 12 ($\approx\$80{,}051.61$ vs \$79,500).
Chapter 6 Summary 本章总结
Formulas at a Glance 公式速查
| Type | Formula (index at 1) | Formula (index at 0) | Point-$k$ form |
|---|---|---|---|
| Arithmetic 等差 | $a_n=a_1+(n-1)d$ | $a_n=a_0+nd$ | $a_n=a_k+(n-k)d$ |
| Geometric 等比 | $a_n=a_1\cdot r^{n-1}$ | $a_n=a_0\cdot r^n$ | $a_n=a_k\cdot r^{n-k}$ |
| Linear fn | $f(x)=b+mx$ | $f(x)=y_1+m(x-x_1)$ | |
| Exponential fn | $f(x)=ab^x$ | $f(x)=y_1\cdot r^{\,x-x_1}$ | |
- ➕ 等差数列 = 重复加法 = 公差 $d$ = 类比线性斜率 $m$
- ✖️ 等比数列 = 重复乘法 = 公比 $r$ = 类比指数底数 $b$
- 🔍 判断方法：一阶差分恒定→等差；连续项之比恒定→等比
- 📍 点到点公式：等差用加法 $+(n-k)d$；等比用乘法 $\cdot r^{n-k}$
- 🔵 数列图像：离散点（只有整数输入），函数图像是连续曲线
- ⚠️ $r=1$ 与 $r=-1$ 是特殊等比数列：前者保持常数，后者正负交替
- 📏 表格间隔为 $h$：差 $D$ → 斜率 $D/h$；比 $R$ → 每单位因子 $R^{1/h}$
- 🎭 情境题：加同样的量 → 等差；乘同样的百分比 → 等比；先确定从 $a_0$ 还是 $a_1$ 开始
Arithmetic: $d = \dfrac{a_m - a_k}{m-k}$ (total change ÷ number of steps)
Geometric: $r^{\,m-k} = \dfrac{a_m}{a_k}$, so $r = \left(\dfrac{a_m}{a_k}\right)^{1/(m-k)}$ (take the $(m-k)$th root of the ratio)
⚠️ If $m-k$ is even, $r$ can be positive or negative: take $\pm$ (see T-4). 若 $m-k$ 为偶数，$r$ 有正负两个值。
Chapter 6 Review 章末习题
Terms in a sequence are given by $s_n=3-5n$ for $n=0,1,2,3,\ldots$. Which of the following describes the sequence?
答案：(A) $s_n=3+(-5)n$ 是"首项 + 公差 × $n$"的形式：$s_0=3$，$s_1=-2$，$s_2=-7$，每次加 $-5$。所以是等差数列，首项（$n=0$ 时）3，公差 $-5$。
(B) 把 $s_1=-2$ 当成了首项——这里 $n$ 从 0 开始；(C)、(D) 相邻两项的比 $\dfrac{-2}{3}$、$\dfrac{-7}{-2}$ 不相等，不是等比。
| $n$ | $4$ | $7$ | $10$ |
|---|---|---|---|
| $a_n$ | $29$ | $41$ | $53$ |
The table gives values of an arithmetic sequence $a_n$ for selected values of $n$. Which of the following linear functions is constructed from the initial value $a_0$ and the common difference of the sequence?
答案：(B) $n$ 每增加 3，$a_n$ 增加 12，所以公差 $d=\dfrac{12}{3}=4$（不是 12）。从 $a_4=29$ 往回退 4 步：$a_0=29-4\times4=13$。所以 $f(x)=13+4x$。
(A) 把每 3 项的变化量 12 当成了公差；(C) 用的是 $a_1=17$；(D) 用的是 $a_4$。
The first term of a geometric sequence is $g_1=4$. For $n\ge2$, the terms are given by $g_n=1.5\cdot g_{n-1}$. For which of the following functions is $g_n=f(n)$ for all positive integers $n$?
答案：(D) 从 $g_1$ 出发乘了 $n-1$ 次 1.5：$g_n=4(1.5)^{n-1}$。再用指数法则改写：$4(1.5)^{n-1}=4\cdot\dfrac{(1.5)^n}{1.5}=\dfrac83(1.5)^n$。检查：$f(1)=\dfrac83\times1.5=4$ ✓。
(A) $f(1)=6$，多乘了一次；(B) 底数和系数弄反了；(C) $f(1)=6$。
An exponential function $f$ satisfies $f(0)=5$ and $f(x+2)=1.21\cdot f(x)$ for all real numbers $x$. Which of the following could define $f$?
答案：(C) 输入每增加 2，输出乘 1.21，所以 $b^2=1.21$，$b=1.1$（$b>0$）。$f(x)=5(1.1)^x$。检查：$f(x+2)=5(1.1)^x(1.1)^2=1.21f(x)$ ✓。
(A) 1.21 是每 2 个单位的倍数，不是每 1 个单位的；(B) 把倍数"平均分"成 $1.21\div2$——倍数要开方，不能除；(D) $5(1.21)^{2x}$ 每增加 2 乘 $1.21^4$。
The fourth term of a geometric sequence is $20$, the fifth term is $30$, and the $n$th term is $151.875$. The solution to which of the following equations gives the value of $n$?
答案：(A) 公比 $r=\dfrac{30}{20}=1.5$。从第 4 项走到第 $n$ 项要乘 $n-4$ 次，所以方程是 $20(1.5)^{n-4}=151.875$，解得 $n=9$。
(B) 把 20 当成了第 1 项；(C) 30 是第 5 项，指数应是 $n-5$；(D) 是等差数列的写法。
In a simulation, the population of a bacteria colony is modeled by a geometric sequence, where the first day of the simulation is day $1$. The population on day $2$ was $1500$ bacteria, and the population on day $7$ was $18{,}000$ bacteria. To the nearest whole number, what was the population on day $5$ according to the model?
答案：(B) 从第 2 天到第 7 天乘了 5 次公比：$r^5=\dfrac{18000}{1500}=12$，$r=12^{1/5}\approx1.644$。第 5 天：$1500r^3\approx6662$。
(A) 是第 2 天和第 7 天的几何平均（对应第 4.5 天）；(C) 按等差数列算的；(D) 把公比当成了 $12\div5=2.4$。
A ball is dropped from a height of $12$ feet. After each bounce, the maximum height of the ball is $75\%$ of the previous maximum height. On which bounce is the maximum height of the ball first less than $1$ foot?
答案：(C) 每次乘 0.75（固定比例），第 $n$ 次弹起的最大高度是 $h_n=12(0.75)^n$——等比数列。用计算器的表格：$h_8\approx1.201$，$h_9\approx0.901$，所以第 9 次首次低于 1 英尺。
(A) 当成每次减少 $12\times25\%=3$ 英尺（等差）；(B) $h_8$ 还大于 1；(D) 把落下的 12 英尺也算成了"第 1 次"。
The function $f$ is exponential. If $f(2)=7.2$ and $f(7)=21.6$, what is the value of $f(10)$?
答案：(B) 输入每增加 5，输出乘 $\dfrac{21.6}{7.2}=3$，所以每增加 1 乘 $3^{1/5}$。从 $x=7$ 到 $x=10$ 增加 3：$f(10)=21.6\cdot3^{3/5}\approx41.757$。
(A) 按线性函数外推；(C) 把倍数按比例"分"成 $1+2\times\tfrac35=2.2$；(D) 以为再增加 3 也乘 3。
Directions: Solve each part without a calculator. Show the work that leads to your answers.
The sequence $a_n$ is arithmetic with $a_3=17$ and $a_8=42$. The sequence $g_n$ is geometric with $g_3=24$ and $g_6=81$.
(a) Find the common difference of $a_n$, and write an expression for $a_n$.
(b) Is $2024$ a term of the sequence $a_n$? Explain. Then find the first term of $a_n$ that is greater than $2024$.
(c) Find the common ratio of $g_n$, and write an expression for $g_n$. What is $g_1$?
(d) The exponential function $f$ is given by $f(x)=24(1.5)^{x-3}$, so $g_n=f(n)$. Describe one way the graph of the sequence $g_n$ differs from the graph of $f$.
(a) 从第 3 项到第 8 项加了 5 次公差：$5d=42-17=25$，$d=5$。$a_n=17+5(n-3)=2+5n$。
(b) $2+5n=2024\Rightarrow n=\dfrac{2022}{5}=404.4$，不是整数，所以 2024 不是这个数列的项（数列只在整数 $n$ 上有定义）。$a_{404}=2022$，$a_{405}=2027$，所以第一个大于 2024 的项是 $a_{405}=2027$。
(c) 从第 3 项到第 6 项乘了 3 次公比：$r^3=\dfrac{81}{24}=\dfrac{27}{8}$，$r=\dfrac32$。$g_n=24\left(\dfrac32\right)^{n-3}$，$g_1=24\div\left(\dfrac32\right)^2=\dfrac{32}{3}$。
(d) 数列的定义域是正整数，图像是一个个孤立的点 $(1,\tfrac{32}{3}),(2,16),(3,24),\ldots$；$f$ 的定义域是全体实数，图像是一条连续的曲线，数列的点都落在这条曲线上。
The graph shows the first six terms of the sequence $b_n$, for $n=1,2,\ldots,6$. The coordinates of the first four points are labeled. A calculator is allowed.
(a) Is $b_n$ arithmetic, geometric, or neither? Justify your answer using the labeled values.
(b) Write an expression for $b_n$, and find $b_8$.
(c) Find the average rate of change of $b_n$ from $n=1$ to $n=3$ and from $n=3$ to $n=5$. Explain how the two values are consistent with your answer to (a).
(d) Is there a value of $n$ for which $b_n=10$? Explain. What is the first term of $b_n$ that is less than $10$?
(a) 相邻差：$48-64=-16$，$36-48=-12$，$27-36=-9$，不是常数，不是等差。相邻比：$\dfrac{48}{64}=\dfrac{36}{48}=\dfrac{27}{36}=0.75$，是常数，所以是等比数列，公比 0.75。
(b) $b_n=64(0.75)^{n-1}$，$b_8=64(0.75)^7\approx8.543$。
(c) $\dfrac{b_3-b_1}{3-1}=\dfrac{36-64}{2}=-14$；$\dfrac{b_5-b_3}{5-3}=\dfrac{20.25-36}{2}=-7.875$。两个都是负的（数列递减），而且第二个的绝对值更小：等比数列每次减少的是当前值的 25%，值越来越小，减少得也越来越慢。（两段的平均变化率之比 $\dfrac{-7.875}{-14}=0.5625=0.75^2$，正好是相隔 2 项的倍数。）
(d) 解 $64(0.75)^{n-1}=10$ 得 $n\approx7.45$，不是整数，所以没有 $b_n=10$ 的项。$b_7\approx11.391>10$，$b_8\approx8.543<10$，第一个小于 10 的项是 $b_8$。
A graduate receives two job offers. Offer A pays $\$50{,}000$ in year 1 and increases by $\$2500$ each year. Offer B pays $\$46{,}000$ in year 1 and increases by $6\%$ each year. Let $A_n$ and $B_n$ be the salaries, in dollars, in year $n$. A calculator is allowed.
(a) Identify each sequence as arithmetic or geometric, and write expressions for $A_n$ and $B_n$.
(b) Find the salary from each offer in year $5$.
(c) In which year does Offer B first pay more than Offer A?
(d) Explain why Offer B pays more than Offer A in every year after the year found in (c).
(a) A 每年加固定的 2500 元：等差，$A_n=50000+2500(n-1)$。B 每年乘固定的 1.06：等比，$B_n=46000(1.06)^{n-1}$。
(b) $A_5=50000+2500\times4=60000$；$B_5=46000(1.06)^4\approx58073.94$。
(c) 用计算器列表：$A_6=62500$，$B_6\approx61558.38$；$A_7=65000$，$B_7\approx65251.88$。所以第 7 年 B 第一次更高。
(d) A 每年的涨幅永远是 2500 元；B 每年的涨幅是当年工资的 6%，从第 7 年起 B 的工资已超过 65000，6% 就超过 3900 元，比 2500 大，并且涨幅本身还在逐年变大。所以 B 一旦领先，差距只会越来越大。（一般地，增长因子大于 1 的指数增长最终总会超过线性增长。）
| $x$ | $1$ | $3$ | $4$ | $7$ |
|---|---|---|---|---|
| $f(x)$ | $6.5$ | $10.5$ | $12.5$ | $18.5$ |
| $g(x)$ | $1.6$ | $6.4$ | $12.8$ | $102.4$ |
The table gives values of the functions $f$ and $g$ at selected values of $x$. One of the functions is linear and the other is exponential.
(a) Show that $f$ is linear and $g$ is exponential. Note that the input values are not equally spaced.
(b) Write expressions for $f(x)$ and $g(x)$.
(c) Find $f(9)$ and $g(9)$ without a calculator.
(d) For $x\ge0$, use a calculator to find the value of $x$ at which $g(x)=f(x)$, and describe how $f(x)$ and $g(x)$ compare for larger values of $x$.
(a) 输入间隔分别是 2、1、3，必须换算成"每增加 1"。
$f$：平均变化率 $\dfrac{10.5-6.5}{2}=2$，$\dfrac{12.5-10.5}{1}=2$，$\dfrac{18.5-12.5}{3}=2$，处处相同，所以 $f$ 是线性的。
$g$：各段的比值 $\dfrac{6.4}{1.6}=4$（间隔 2），$\dfrac{12.8}{6.4}=2$（间隔 1），$\dfrac{102.4}{12.8}=8$（间隔 3）。
换算到每增加 1：$4^{1/2}=2$，$2^1=2$，$8^{1/3}=2$，都是 2，所以 $g$ 是指数的。
(b) $f(x)=6.5+2(x-1)=2x+4.5$；$g(x)=1.6\cdot2^{x-1}=0.8\cdot2^x$。
(c) 从 $x=7$ 到 $x=9$ 增加 2：$f(9)=18.5+2\times2=22.5$；$g(9)=102.4\times2^2=409.6$。
(d) 解 $0.8\cdot2^x=2x+4.5$：$x\approx3.955$。在这之前（$0\le x<3.955$）$f(x)>g(x)$；之后 $g(x)>f(x)$，并且差距越来越大：$f$ 每增加 1 只加 2，$g$ 每增加 1 就翻倍。
