Draft v1 — 11 questions across all four IB papers. Question-only practice set; teacher mark-scheme companion to follow if requested.草稿 v1 —— 11 道题，覆盖 IB 四张试卷。仅含题目；如需教师评分细则配套文档，可另行索取。
PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 31 marks不可使用计算器 · 简答题 · 31 分
Section A — Short ResponseA 节 —— 简答题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. No calculator permitted.在每题下方的空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。不可使用计算器。
Q1EASYPaper 1A1.2 Arithmetic Sequences[5 marks]
An arithmetic sequence has $u_5 = 23$ and $u_{15} = 53$.某等差数列（arithmetic sequence）满足 $u_5 = 23$ 与 $u_{15} = 53$。
(a)Find the common difference $d$.求公差（common difference）$d$。[2]
(b)Find $u_1$.求 $u_1$。[2]
(c)Find $S_{10}$.求 $S_{10}$。[1]
Q2EASYPaper 1A1.3 Geometric Sequences[4 marks]
A geometric sequence has $u_2 = 12$ and $u_5 = -96$.某等比数列（geometric sequence）满足 $u_2 = 12$ 与 $u_5 = -96$。
(a)Find the common ratio $r$.求公比（common ratio）$r$。[2]
(b)Find $u_1$ and $u_8$.求 $u_1$ 与 $u_8$。[2]
Q3MEDIUMPaper 1A1.4 Sigma Notation[5 marks]
Evaluate $\displaystyle\sum_{k=1}^{20}(4k - 3)$ by recognising the summand as an arithmetic sequence and applying the appropriate sum formula. Show every step.计算 $\displaystyle\sum_{k=1}^{20}(4k - 3)$：将求和项识别为等差数列，并使用对应的求和公式。写出每一步过程。
Q4MEDIUMPaper 1A1.8 Infinite GP[6 marks]
Express $0.\overline{42} = 0.424242\ldots$ as a fraction in its simplest form. You must justify your answer using the formula for the sum of an infinite geometric series, including a check of the convergence condition.将循环小数 $0.\overline{42} = 0.424242\ldots$ 写为最简分数。必须使用无穷等比级数（infinite GP）的求和公式给出说明，并检验收敛条件（convergence condition）。
Q5MEDIUMPaper 1A1.8 Infinite GP[6 marks]
An infinite geometric series has first term $u_1 = 18$ and sum to infinity $S_\infty = 24$.某无穷等比级数的首项为 $u_1 = 18$，无穷和（sum to infinity）$S_\infty = 24$。
(a)Find the common ratio $r$, expressing your answer as an exact fraction.求公比 $r$，答案以精确分数形式给出。[3]
(b)Find $S_5$ (the sum of the first five terms), expressing your answer as an exact fraction.求 $S_5$（前 5 项之和），答案以精确分数形式给出。[3]
Q6MEDIUMPaper 1A1.2 / 1.3 Mixed[5 marks]
The first three terms of a sequence are $4,\; x,\; 9$.某数列的前三项为 $4,\; x,\; 9$。
(a)If the sequence is arithmetic, find $x$.若该数列为等差数列，求 $x$。[1]
(b)If the sequence is geometric, find all possible values of $x$.若该数列为等比数列，求 $x$ 的所有可能值。[2]
(c)Hence determine, with justification, whether there exists a value of $x$ for which the sequence $4, x, 9$ is both arithmetic and geometric.由此判断（并说明理由）：是否存在某个 $x$ 使数列 $4, x, 9$ 同时是等差与等比数列。[2]
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 16 marks不可使用计算器 · 长答题 · 16 分
Section B — Extended ResponseB 节 —— 长答题
Take time to set up cleanly. Method marks dominate. Show all work; carry expressions in exact form unless told otherwise.认真梳理思路再下笔。方法分（method marks）权重最大。完整展示推导；除特别说明外，表达式保留精确形式（exact form）。
Q7HARDPaper 1B1.2 / 1.4 Multi-skill[16 marks]
The sum of the first $n$ terms of a sequence $\{u_n\}$ is given by数列 $\{u_n\}$ 的前 $n$ 项和（partial sum）为
$$ S_n = n^2 + 3n. $$
(a)Find $u_1$, $u_2$, and $u_3$ directly from the definition $u_n = S_n - S_{n-1}$ for $n \ge 2$, and $u_1 = S_1$.根据定义 $u_n = S_n - S_{n-1}$（$n \ge 2$）以及 $u_1 = S_1$，直接求出 $u_1$、$u_2$、$u_3$。[3]
(b)Show that for $n \ge 2$, $u_n = 2n + 2$.证明：当 $n \ge 2$ 时，$u_n = 2n + 2$。[3]
(c)Verify that the formula in part (b) also gives the correct value of $u_1$.验证 (b) 中的公式同样给出正确的 $u_1$ 值。[1]
(d)Hence show that $\{u_n\}$ is an arithmetic sequence, stating its common difference.由此证明 $\{u_n\}$ 是等差数列，并写出其公差。[2]
(e)Find a closed-form expression in $n$ for $\displaystyle\sum_{k=1}^{n} u_{2k}$ (the sum of the first $n$ even-indexed terms).求 $\displaystyle\sum_{k=1}^{n} u_{2k}$（前 $n$ 个偶数下标项之和）关于 $n$ 的闭合表达式（closed-form）。[4]
PART III · PAPER 2第三部分 · 第二卷Calculator · mixed response · 24 marks可使用计算器 · 混合题型 · 24 分
Paper 2 — Calculator Permitted第二卷 —— 允许使用计算器
A graphing calculator is required. Give numerical answers correct to 3 significant figures unless otherwise specified.需要图形计算器（GDC）。除特别说明外，数值答案保留 3 位有效数字。
Q8MEDIUMPaper 21.4 Compound Interest[8 marks]
On 1 January 2025, Lily invests €$5000$ in an account paying $4.2\%$ per annum compounded quarterly.2025 年 1 月 1 日，Lily 将 €$5000$ 存入一个年利率 $4.2\%$、按季度复利（compounded quarterly）的账户。
(a)Find, to the nearest cent, the value of Lily's investment on 1 January 2031.求 2031 年 1 月 1 日 Lily 的投资金额（精确到分）。[2]
(b)Lily wants the value to first exceed €$7500$. Find the year in which this first occurs (i.e. the first 1 January date on which the balance exceeds €$7500$).Lily 希望账户余额首次超过 €$7500$。求该情况首次出现的年份（即首个使余额超过 €$7500$ 的 1 月 1 日所在年份）。[3]
(c)On the same date, Jenny invests €$5000$ at $4.2\%$ compounded annually instead. Find the difference, to the nearest cent, between Lily's and Jenny's account values on 1 January 2031, and state in one sentence why Lily's value is the larger of the two.同日，Jenny 也存入 €$5000$，但年利率 $4.2\%$ 改为按年复利（compounded annually）。求 2031 年 1 月 1 日 Lily 与 Jenny 账户金额的差（精确到分），并用一句话说明为何 Lily 的金额更大。[3]
Q9MEDIUMPaper 21.4 / 1.8 Real-World GP[7 marks]
A ball is dropped from a height of $10~\mathrm{m}$. After each bounce, it rises to $75\%$ of the previous height. Air resistance is negligible.一个小球从 $10~\mathrm{m}$ 高处自由下落。每次反弹后上升到上一次高度的 $75\%$。忽略空气阻力。
(a)Find the height of the third bounce, in metres.求第三次反弹的高度（米）。[2]
(b)Find the total vertical distance the ball travels before coming to rest, giving your answer in metres.求小球最终静止前经过的总垂直距离（米）。[4]
(c)State, in one sentence, why this total distance is finite even though the ball bounces infinitely many times.用一句话说明：尽管小球反弹无穷多次，为何这一总距离仍为有限值。[1]
Q10HARDPaper 21.2 / 1.3 / 1.4 Hybrid[9 marks]
A worker's annual salary in year $n$ forms an arithmetic sequence $\{u_n\}$. The year-1 salary is €$42\,000$ and the year-30 salary is €$71\,000$.某劳动者在第 $n$ 年的年薪构成等差数列 $\{u_n\}$。第 1 年年薪为 €$42\,000$，第 30 年年薪为 €$71\,000$。
(a)Find the annual increment $d$.求每年的年薪增量（annual increment）$d$。[2]
(b)Find the worker's total earnings over the 30-year career.求该劳动者 30 年职业生涯的总收入。[3]
(c)Suppose instead the worker's annual salary $\{v_n\}$ formed a geometric sequence with the same year-1 and year-30 salaries.现假设该劳动者的年薪 $\{v_n\}$ 改为等比数列，且第 1 年与第 30 年的年薪保持不变。
(i)Find the common ratio $r$, correct to four decimal places.求公比 $r$，保留 4 位小数。[2]
(ii)Without computing the GP total in full, state with justification whether the geometric scheme produces higher or lower total earnings than the arithmetic scheme.无需算出等比方案的总和具体数值，判断（并说明理由）等比方案的总收入比等差方案高还是低。[2]
PART IV · PAPER 3第四部分 · 第三卷Calculator · HL extended exploration · 16 marks可使用计算器 · HL 长题探究 · 16 分
Paper 3 — HL Extended Problem第三卷 —— HL 长题探究
A graphing calculator is required. Method marks are heavily weighted. Show full reasoning at every step; partial credit is generous.需要图形计算器（GDC）。方法分权重很高。每一步都要写出完整推理；过程分（partial credit）给得较为宽松。
Q11HARDPaper 31.8 Infinite GP & Power Series[16 marks]
This question explores the link between the infinite geometric series and certain related power-series identities.本题探究无穷等比级数与若干相关幂级数（power series）恒等式之间的联系。
(a)State the formula for the sum to infinity $S_\infty$ of a convergent infinite geometric series with first term $u_1$ and common ratio $r$, and give the exact convergence condition on $r$.写出首项为 $u_1$、公比为 $r$ 的收敛无穷等比级数的无穷和 $S_\infty$ 公式，并精确给出关于 $r$ 的收敛条件。[2]
(b)Hence write down, for $|x| < 1$, the closed-form value of由此写出当 $|x| < 1$ 时下式的闭合形式（closed-form）：
$$ f(x) = 1 + x + x^2 + x^3 + x^4 + \cdots $$
[1]
(c)By differentiating both sides of the identity in (b) term-by-term (treating $f$ as a function of $x$), derive a closed-form expression for将 (b) 中恒等式两边对 $x$ 逐项求导（视 $f$ 为 $x$ 的函数），推导下式的闭合形式表达式：
$$ g(x) = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \cdots, \qquad |x| < 1. $$
Show every step of the differentiation.写出每一步求导过程。[4]
(d)Use your closed-form expression for $g(x)$ to compute the exact value of利用 $g(x)$ 的闭合表达式，求下式的精确值：
$$ \sum_{k=1}^{\infty} k\!\left(\tfrac{1}{2}\right)^{\!k-1}. $$
[3]
(e)Verify your answer to (d) numerically by computing the partial sum $\displaystyle\sum_{k=1}^{10} k\!\left(\tfrac{1}{2}\right)^{\!k-1}$ on your GDC. State the value to 4 decimal places and comment briefly on its proximity to the predicted limit.在图形计算器上计算部分和（partial sum）$\displaystyle\sum_{k=1}^{10} k\!\left(\tfrac{1}{2}\right)^{\!k-1}$ 来对 (d) 的答案作数值验证。给出 4 位小数的结果，并简要评论其与所预测极限值的接近程度。[2]