[{"id":"math-sequence-basics-001","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":1,"stem":"A sequence is given by $a_{n} = 3n − 2 (n ∈ ℕ⁺)$. What are its first three terms?","stemZh":"已知数列 $a_{n} = 3n − 2（n ∈ ℕ⁺），则它的前三项为？$","options":[{"id":"A","text":"1, 3, 5"},{"id":"B","text":"0, 3, 6"},{"id":"C","text":"1, 4, 7"},{"id":"D","text":"2, 5, 8"}],"answer":"C","explanation":"$a_{1} = 3·1 − 2 = 1, a_{2} = 3·2 − 2 = 4, a_{3} = 3·3 − 2 = 7$. The index of a sequence starts at $n = 1$, so the list beginning with 0 is wrong.","explanationZh":"$a_{1} = 1，a_{2} = 4，a_{3} = 7。数列的项数从 n = 1$ 开始，因此以 0 开头的选项是错误的。","tags":["easy","general-term","indexing"],"status":"published"},{"id":"math-sequence-basics-002","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":1,"stem":"A sequence ${a_{n}}$ can be regarded as a function $f(n) = a_{n}$. Its domain is","stemZh":"数列 ${a_{n}}$ 可以看成函数 $f(n) = a_{n}，它的定义域是$","options":[{"id":"A","text":"$ℕ⁺ (or {1, 2,\\text{ …, }n}\\text{ for }a\\text{ finite }sequence)$"},{"id":"B","text":"$ℝ$"},{"id":"C","text":"$ℤ$"},{"id":"D","text":"$[0, +∞)$"}],"answer":"A","explanation":"A sequence is a function whose domain is the set of positive integers $ℕ⁺$, or the finite set ${1, 2,\\text{ …, }n}$ when the sequence has only n terms.","explanationZh":"数列是定义域为正整数集 $ℕ⁺（有限数列时为 {1, 2, …, n}）的函数。$","tags":["easy","definition","sequence-as-function"],"status":"published"},{"id":"math-sequence-basics-003","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":1,"stem":"How many terms does the sequence 1, 3, 5, 7, …, 99 contain?","stemZh":"数列 1, 3, 5, 7, …, 99 共有多少项？","options":[{"id":"A","text":"49"},{"id":"B","text":"51"},{"id":"C","text":"100"},{"id":"D","text":"50"}],"answer":"D","explanation":"The terms are the odd numbers from 1 to 99. Writing $2k − 1 = 99$ gives $k = 50$, so there are 50 terms.","explanationZh":"这些项是从 1 到 99 的所有奇数。由 $2k − 1 = 99 得 k = 50，因此共有 50$ 项。","tags":["easy","counting-terms","finite-sequence"],"status":"published"},{"id":"math-sequence-basics-004","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"If $a_{n} = n^{2} − n$, then $a_{4} =$","stemZh":"$若 a_{n} = n^{2} −$ n，则 $a_{4} =$","options":[{"id":"A","text":"6"},{"id":"B","text":"12"},{"id":"C","text":"16"},{"id":"D","text":"20"}],"answer":"B","explanation":"$a_{4} = 4^{2} − 4 = 16 − 4 = 12. (16\\text{ is only }n^{2},\\text{ forgetting to subtract }n.)$","explanationZh":"$a_{4} = 4^{2} − 4 = 16 − 4 = 12。（只算 n^{2} 得 16，是漏减 n$ 的错误。）","tags":["easy","general-term","evaluation"],"status":"published"},{"id":"math-sequence-basics-005","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"A sequence satisfies $a_{1} = 1$ and $a_{n+1} = a_{n} + 2$. Its first four terms are","stemZh":"数列满足 $a_{1} = 1，a_{n+1} = a_{n} + 2，则它的前四项为$","options":[{"id":"A","text":"1, 2, 3, 4"},{"id":"B","text":"1, 3, 5, 7"},{"id":"C","text":"1, 3, 6, 10"},{"id":"D","text":"2, 4, 6, 8"}],"answer":"B","explanation":"$a_{2} = 1 + 2 = 3, a_{3} = 3 + 2 = 5, a_{4} = 5 + 2 = 7$, so the terms are 1, 3, 5, 7.","explanationZh":"$a_{2} = 3，a_{3} = 5，a_{4} = 7，所以前四项是 1, 3, 5, 7。$","tags":["easy","recurrence","listing-terms"],"status":"published"},{"id":"math-sequence-basics-006","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"Which of the following is an infinite sequence?","stemZh":"下列哪一项是无穷数列？","options":[{"id":"A","text":"1, 2, 3, 4, 5"},{"id":"B","text":"The first 100 positive even numbers written in order"},{"id":"C","text":"$a_{n} = 1/n$ for $n = 1, 2, …, 10$"},{"id":"D","text":"$a_{n} = 1/n$ for $n ∈ ℕ⁺$"}],"answer":"D","explanation":"Only the last option has domain $ℕ⁺$ with no last term. The others have a finite number of terms $(5, 100$ and $10 respectively)$.","explanationZh":"只有最后一项的定义域是全体正整数 $ℕ⁺，没有末项。其余各项分别只有 5 项、100 项、10$ 项。","tags":["easy","finite-vs-infinite","definition"],"status":"published"},{"id":"math-sequence-basics-007","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"A general term for the sequence 2, 5, 10, 17, 26, … is","stemZh":"数列 2, 5, 10, 17, 26, … 的一个通项公式是","options":[{"id":"A","text":"$a_{n} = n^{2} + 1$"},{"id":"B","text":"$a_{n} = n^{2} − 1$"},{"id":"C","text":"$a_{n} = 2n$"},{"id":"D","text":"$a_{n} = n^{2} + n$"}],"answer":"A","explanation":"Each term is one more than a perfect $square: 2 = 1^{2} + 1, 5 = 2^{2} + 1, 10 = 3^{2} + 1, 17 = 4^{2} + 1$, so $a_{n} = n^{2} + 1$.","explanationZh":"每一项都比完全平方数大 $1：2 = 1^{2} + 1，5 = 2^{2} + 1，10 = 3^{2} + 1，故 a_{n} = n^{2} + 1。$","tags":["easy","general-term","pattern-recognition"],"status":"published"},{"id":"math-sequence-basics-008","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":1,"stem":"For $a_{n} = (−1)^{n}$, the terms $a_{1}, a_{2}, a_{3}$ are","stemZh":"$若 a_{n} = (−1)^{n}，则 a_{1}, a_{2}, a_{3}$ 分别为","options":[{"id":"A","text":"1, 1, 1"},{"id":"B","text":"1, −1, 1"},{"id":"C","text":"−1, 1, −1"},{"id":"D","text":"−1, −1, −1"}],"answer":"C","explanation":"$(−1)^{1} = −1, (−1)^{2} = 1, (−1)^{3} = −1$. The option 1, −1, 1 comes from using $(−1)^{n-1}$ instead of $(−1)^{n}$.","explanationZh":"$(−1)^{1} = −1，(−1)^{2} = 1，(−1)^{3} = −1。选项 1, −1, 1$ 是把通项误作 $(−1)^{n-1}$ 的结果。","tags":["easy","oscillating-sequence","sign"],"status":"published"},{"id":"math-sequence-basics-009","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"The sequence 1, 1/2, 1/3, 1/4, … is","stemZh":"数列 1, 1/2, 1/3, 1/4, … 是","options":[{"id":"A","text":"increasing"},{"id":"B","text":"constant"},{"id":"C","text":"decreasing"},{"id":"D","text":"oscillating"}],"answer":"C","explanation":"$a_{n} = 1/n$ and $a_{n+1} = 1/(n+1) < 1/n = a_{n}$ for every n, so the sequence is strictly decreasing.","explanationZh":"$a_{n} = 1/n，a_{n+1} = 1/(n+1) < 1/n = a_{n}，所以该数列单调递减。$","tags":["easy","monotonicity","classification"],"status":"published"},{"id":"math-sequence-basics-010","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":2,"stem":"If $a_{n} = 2^{n} − 1$, then $a_{5} − a_{3} =$","stemZh":"$若 a_{n} = 2^{n} − 1，则 a_{5} − a_{3} =$","options":[{"id":"A","text":"6"},{"id":"B","text":"24"},{"id":"C","text":"28"},{"id":"D","text":"30"}],"answer":"B","explanation":"$a_{5} = 2^{5} − 1 = 31$ and $a_{3} = 2^{3} − 1 = 7$, so $a_{5} − a_{3} = 31 − 7 = 24$.","explanationZh":"$a_{5} = 2^{5} − 1 = 31，a_{3} = 2^{3} − 1 = 7，故 a_{5} − a_{3} = 31 − 7 = 24。$","tags":["easy","general-term","evaluation"],"status":"published"},{"id":"math-sequence-basics-011","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"Let $S_{n}$ be the sum of the first n terms of a sequence and suppose $S_{n} = n^{2} + 1$. For $n ≥ 2, a_{n} =$","stemZh":"$设 S_{n}$ 为数列前 n 项和，且 $S_{n} = n^{2} + 1，则当 n ≥ 2 时 a_{n} =$","options":[{"id":"A","text":"2n + 1"},{"id":"B","text":"$n^{2} − 1$"},{"id":"C","text":"2n + 2"},{"id":"D","text":"2n − 1"}],"answer":"D","explanation":"For $n ≥ 2, a_{n} = S_{n} − S_{n}-_{1} = (n^{2} + 1) − ((n − 1)^{2} + 1) = 2n − 1$. Note that $a_{1} = S_{1} = 2$ does not fit this formula, which is why the question restricts to $n ≥ 2$.","explanationZh":"$当 n ≥ 2 时，a_{n} = S_{n} − S_{n}-_{1} = (n^{2} + 1) − ((n − 1)^{2} + 1) = 2n − 1。注意 a_{1} = S_{1} = 2$ 不满足该式，所以题目限定 $n ≥ 2。$","tags":["medium","sum-to-term","a_{n}-equals-S_{n}-minus-S_{n}-₁"],"status":"published"},{"id":"math-sequence-basics-012","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"$S_{n} = 3n^{2} − n$ is the sum of the first n terms of a sequence. Then $a_{5} =$","stemZh":"数列前 n 项和 $S_{n} = 3n^{2} −$ n，则 $a_{5} =$","options":[{"id":"A","text":"26"},{"id":"B","text":"24"},{"id":"C","text":"28"},{"id":"D","text":"30"}],"answer":"A","explanation":"$a_{5} = S_{5} − S_{4} = (3·25 − 5) − (3·16 − 4) = 70 − 44 = 26. (Equivalently a_{n} = 6n − 4\\text{ for }n ≥ 2.)$","explanationZh":"$a_{5} = S_{5} − S_{4} = (75 − 5) − (48 − 4) = 70 − 44 = 26。（也可先得 n ≥ 2 时 a_{n} = 6n − 4。）$","tags":["medium","sum-to-term","computation"],"status":"published"},{"id":"math-sequence-basics-013","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"A sequence satisfies $a_{1} = 1$ and $a_{n+1} = 2a_{n} + 1$. Then $a_{4} =$","stemZh":"数列满足 $a_{1} = 1，a_{n+1} = 2a_{n} + 1，则 a_{4} =$","options":[{"id":"A","text":"7"},{"id":"B","text":"15"},{"id":"C","text":"16"},{"id":"D","text":"31"}],"answer":"B","explanation":"$a_{2} = 2·1 + 1 = 3, a_{3} = 2·3 + 1 = 7, a_{4} = 2·7 + 1 = 15$.","explanationZh":"$a_{2} = 3，a_{3} = 7，a_{4} = 15。$","tags":["medium","recurrence","iteration"],"status":"published"},{"id":"math-sequence-basics-014","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"Compute $1/(1·2) + 1/(2·3) + … + 1/(9·10)$.","stemZh":"计算 $1/(1·2) + 1/(2·3) + … + 1/(9·10)。$","options":[{"id":"A","text":"1"},{"id":"B","text":"8/9"},{"id":"C","text":"9/10"},{"id":"D","text":"1/10"}],"answer":"C","explanation":"Since $1/(n(n+1)) = 1/n − 1/(n+1)$, the sum $telescopes: (1 − 1/2) + (1/2 − 1/3) + … + (1/9 − 1/10) = 1 − 1/10 = 9/10$.","explanationZh":"因为 $1/(n(n+1)) = 1/n − 1/(n+1)，原式 = (1 − 1/2) + (1/2 − 1/3) + … + (1/9 − 1/10) = 1 − 1/10 = 9/10。$","tags":["medium","telescoping","summation"],"status":"published"},{"id":"math-sequence-basics-015","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"The sequence with general term $a_{n} = n/(n+1)$ is","stemZh":"通项为 $a_{n} = n/(n+1)$ 的数列是","options":[{"id":"A","text":"increasing"},{"id":"B","text":"decreasing"},{"id":"C","text":"constant"},{"id":"D","text":"oscillating"}],"answer":"A","explanation":"$a_{n} = 1 − 1/(n+1)$. As $n grows, 1/(n+1) decreases$, so $a_{n}$ increases $(1/2, 2/3, 3/4, …)$.","explanationZh":"$a_{n} = 1 − 1/(n+1)，随着 n 增大，1/(n+1)$ 减小，所以 $a_{n} 递增（1/2, 2/3, 3/4$, …）。","tags":["medium","monotonicity","fraction"],"status":"published"},{"id":"math-sequence-basics-016","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"A sequence satisfies $a_{1} = 2$ and $a_{n+1} = a_{n}/(a_{n} + 1)$. Then $a_{3} =$","stemZh":"数列满足 $a_{1} = 2，a_{n+1} = a_{n}/(a_{n} + 1)，则 a_{3} =$","options":[{"id":"A","text":"2/3"},{"id":"B","text":"3/5"},{"id":"C","text":"1/2"},{"id":"D","text":"2/5"}],"answer":"D","explanation":"$a_{2} = 2/(2 + 1) = 2/3$, then $a_{3} = (2/3)/(2/3 + 1) = (2/3)/(5/3) = 2/5$.","explanationZh":"$a_{2} = 2/3，a_{3} = (2/3)/(5/3) = 2/5。$","tags":["medium","recurrence","fractions"],"status":"published"},{"id":"math-sequence-basics-017","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"$S_{n} = 2^{n} − 1$ is the sum of the first n terms of a sequence. Then $a_{3} =$","stemZh":"数列前 n 项和 $S_{n} = 2^{n} − 1，则 a_{3} =$","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"7"},{"id":"D","text":"8"}],"answer":"B","explanation":"$a_{3} = S_{3} − S_{2} = (2^{3} − 1) − (2^{2} − 1) = 7 − 3 = 4. (7\\text{ is }S_{3},\\text{ not }a_{3}.)$","explanationZh":"$a_{3} = S_{3} − S_{2} = 7 − 3 = 4。（7 是 S_{3}$ 而不是 $a_{3}。）$","tags":["medium","sum-to-term","computation"],"status":"published"},{"id":"math-sequence-basics-018","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"Which recurrence relation produces the sequence 1, 2, 4, 8, …?","stemZh":"下列哪个递推关系可以得到数列 1, 2, 4, 8, …？","options":[{"id":"A","text":"$a_{1} = 1, a_{n+1} = a_{n} + 2$"},{"id":"B","text":"$a_{1} = 2, a_{n+1} = 2a_{n}$"},{"id":"C","text":"$a_{1} = 1, a_{n+1} = 2a_{n}$"},{"id":"D","text":"$a_{1} = 1, a_{n+1} = a_{n}^{2}$"}],"answer":"C","explanation":"Starting from $a_{1} = 1$ and doubling each time gives 1, 2, 4, 8, …. With $a_{1} = 2$ we would get 2, 4, 8, 16, … instead.","explanationZh":"$从 a_{1} = 1$ 开始每次乘以 2 得 1, 2, 4, 8, …；若 $a_{1} = 2$ 则得到 2, 4, 8, 16, …。","tags":["medium","recurrence","initial-term"],"status":"published"},{"id":"math-sequence-basics-019","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"Compute $1/(1·3) + 1/(3·5) + … + 1/(9·11)$.","stemZh":"计算 $1/(1·3) + 1/(3·5) + … + 1/(9·11)。$","options":[{"id":"A","text":"1/11"},{"id":"B","text":"10/11"},{"id":"C","text":"1/2"},{"id":"D","text":"5/11"}],"answer":"D","explanation":"Because $1/((2k−1)(2k+1)) = (1/2)(1/(2k−1) − 1/(2k+1))$, the sum is $(1/2)(1 − 1/11) = 5/11$. Forgetting the factor 1/2 gives 10/11.","explanationZh":"因为 $1/((2k−1)(2k+1)) = (1/2)(1/(2k−1) − 1/(2k+1))，原式 = (1/2)(1 − 1/11) = 5/11。漏掉系数 1/2$ 就会得到 10/11。","tags":["medium","telescoping","summation"],"status":"published"},{"id":"math-sequence-basics-020","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":3,"stem":"The sequence with general term $a_{n} = n^{2} − 5n + 6$ has a term equal to 0 when $n =$","stemZh":"通项为 $a_{n} = n^{2} − 5n + 6$ 的数列中，等于 0 的项对应的 n 为","options":[{"id":"A","text":"2 or 3"},{"id":"B","text":"1 or 6"},{"id":"C","text":"2"},{"id":"D","text":"3"}],"answer":"A","explanation":"$n^{2} − 5n + 6 = (n − 2)(n − 3) = 0$ gives $n = 2$ or $n = 3$, so both $a_{2}$ and $a_{3}$ are $0. (1$ and 6 are the factors of the constant term, not the roots.)","explanationZh":"$n^{2} − 5n + 6 = (n − 2)(n − 3) = 0，得 n = 2 或 n = 3，即 a_{2} = a_{3} = 0。（1 和 6$ 是常数项的因数，不是方程的根。）","tags":["medium","general-term","quadratic"],"status":"published"},{"id":"math-sequence-basics-021","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"Let $S_{n} = n^{2} + 2n$ be the sum of the first n terms of a sequence. Which statement is correct?","stemZh":"设数列前 n 项和 $S_{n} = n^{2} + 2n，下列说法正确的是？$","options":[{"id":"A","text":"$a_{n} = 2n + 1$ only for $n ≥ 2; a_{1}$ does not fit the formula"},{"id":"B","text":"$a_{n} = 2n + 1$ for every $n ∈ ℕ⁺$"},{"id":"C","text":"$a_{n} = 2n + 2$ for every $n ∈ ℕ⁺$"},{"id":"D","text":"$a_{n} = n^{2} + 2n$ for every $n ∈ ℕ⁺$"}],"answer":"B","explanation":"For $n ≥ 2, a_{n} = S_{n} − S_{n}-_{1} = (n^{2} + 2n) − ((n − 1)^{2} + 2(n − 1)) = 2n + 1$, and $a_{1} = S_{1} = 3 = 2·1 + 1$ also fits. Because $S_{n}$ has no constant term, the formula holds for all n here.","explanationZh":"$n ≥ 2 时 a_{n} = S_{n} − S_{n}-_{1} = 2n + 1，且 a_{1} = S_{1} = 3 = 2·1 + 1$ 也符合。由于 $S_{n}$ 没有常数项，该式对所有 n 成立。","tags":["hard","sum-to-term","n-equals-1-case"],"status":"published"},{"id":"math-sequence-basics-022","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"A sequence satisfies $a_{1} = 1$ and $a_{n+1} = a_{n} + 2n$. Then $a_{5} =$","stemZh":"数列满足 $a_{1} = 1，a_{n+1} = a_{n} + 2n，则 a_{5} =$","options":[{"id":"A","text":"11"},{"id":"B","text":"19"},{"id":"C","text":"21"},{"id":"D","text":"25"}],"answer":"C","explanation":"$a_{2} = 1 + 2·1 = 3, a_{3} = 3 + 2·2 = 7, a_{4} = 7 + 2·3 = 13, a_{5} = 13 + 2·4 = 21$. The added amount depends on n, so the sequence is not arithmetic.","explanationZh":"$a_{2} = 3，a_{3} = 7，a_{4} = 13，a_{5} = 21。每次加上的量与 n$ 有关，所以该数列不是等差数列。","tags":["hard","recurrence","non-arithmetic"],"status":"published"},{"id":"math-sequence-basics-023","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"Compute 1/2 + 1/6 + 1/12 + … + 1/110.","stemZh":"计算 1/2 + 1/6 + 1/12 + … + 1/110。","options":[{"id":"A","text":"10/11"},{"id":"B","text":"9/10"},{"id":"C","text":"11/12"},{"id":"D","text":"1/11"}],"answer":"A","explanation":"$1\\cdot 2$, $2\\cdot 3$, …, $10\\cdot 11$, so the sum telescopes to $1 − 1/11 = 10/11$. ($9/10$ would stop at $1/90$.)","explanationZh":"分母依次为 $1·2, 2·3, …, 10·11，故原式 = 1 − 1/11 = 10/11。（若只加到 1/90，结果为 9/10。）$","tags":["hard","telescoping","summation"],"status":"published"},{"id":"math-sequence-basics-024","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"Let $a_{n} = 1/(n(n+1))$ and let $S_{n}$ be the sum of the first n terms. The least n with $S_{n} > 10/11$ is","stemZh":"$设 a_{n} = 1/(n(n+1))，S_{n}$ 为前 n 项和，则使 $S_{n} > 10/11$ 成立的最小 n 为","options":[{"id":"A","text":"9"},{"id":"B","text":"10"},{"id":"C","text":"12"},{"id":"D","text":"11"}],"answer":"D","explanation":"$S_{n} = 1 − 1/(n+1)$. Then $S_{n} > 10/11$ means $1/(n+1) < 1/11$, i.e. $n + 1 > 11$, so $n ≥ 11$. Note $S_{10} = 10/11 exactly$, so $n = 10$ is not enough for a strict inequality.","explanationZh":"$S_{n} = 1 − 1/(n+1)。由 1 − 1/(n+1) > 10/11 得 1/(n+1) < 1/11，即 n + 1 > 11，所以 n ≥ 11。注意 S_{10}$ 恰好等于 $10/11，故 n = 10$ 不满足严格大于。","tags":["hard","telescoping","inequality"],"status":"published"},{"id":"math-sequence-basics-025","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":4,"stem":"A sequence satisfies $a_{1} = 3$ and $a_{n+1} − a_{n} = 2^{n}$. Then $a_{4} =$","stemZh":"数列满足 $a_{1} = 3，a_{n+1} − a_{n} = 2^{n}，则 a_{4} =$","options":[{"id":"A","text":"15"},{"id":"B","text":"17"},{"id":"C","text":"19"},{"id":"D","text":"11"}],"answer":"B","explanation":"$a_{2} = 3 + 2^{1} = 5, a_{3} = 5 + 2^{2} = 9, a_{4} = 9 + 2^{3} = 17$.","explanationZh":"$a_{2} = 3 + 2 = 5，a_{3} = 5 + 4 = 9，a_{4} = 9 + 8 = 17。$","tags":["hard","recurrence","indexing"],"status":"published"},{"id":"math-sequence-basics-026","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"For the sequence $a_{n} = (n − 3)/(n + 1)$, which statement is correct?","stemZh":"对于数列 $a_{n} = (n − 3)/(n + 1)，下列说法正确的是？$","options":[{"id":"A","text":"It is strictly decreasing for all $n ∈ ℕ⁺$"},{"id":"B","text":"It decreases while $n ≤ 3$ and increases afterwards"},{"id":"C","text":"It is strictly increasing for all $n ∈ ℕ⁺$"},{"id":"D","text":"It increases first and then decreases"}],"answer":"C","explanation":"$a_{n} = (n + 1 − 4)/(n + 1) = 1 − 4/(n + 1)$. As $n increases, 4/(n + 1) decreases$, so $a_{n}$ increases for every $n: a_{1} = −1, a_{2} = −1/3, a_{3} = 0, a_{4} = 1/5$, …. The sign change at $n = 3$ does not change the monotonicity.","explanationZh":"$a_{n} = 1 − 4/(n + 1)。n$ 增大时 $4/(n + 1)$ 减小，故 $a_{n}$ 对一切 $n 递增：a_{1} = −1，a_{2} = −1/3，a_{3} = 0，a_{4} = 1/5$, …。n $= 3$ 处的变号并不改变单调性。","tags":["hard","monotonicity","algebraic-rewrite"],"status":"published"},{"id":"math-sequence-basics-027","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"$S_{n} = n^{2} + n + 1$ is the sum of the first n terms of a sequence. Then $a_{2} =$","stemZh":"数列前 n 项和 $S_{n} = n^{2} + n + 1，则 a_{2} =$","options":[{"id":"A","text":"4"},{"id":"B","text":"3"},{"id":"C","text":"5"},{"id":"D","text":"2"}],"answer":"A","explanation":"$a_{2} = S_{2} − S_{1} = (4 + 2 + 1) − (1 + 1 + 1) = 7 − 3 = 4$. Because $S_{n}$ has a non-zero constant term, the sequence is not arithmetic and $a_{n} ≠ S_{n} − S_{n}-_{1}$ at $n = 1$; here $a_{1} = 3$.","explanationZh":"$a_{2} = S_{2} − S_{1} = 7 − 3 = 4。由于 S_{n}$ 含有非零常数项，该数列不是等差数列，且 $n = 1$ 时不能用 $a_{n} = S_{n} − S_{n}-_{1}，此时 a_{1} = S_{1} = 3。$","tags":["hard","sum-to-term","constant-term-trap"],"status":"published"},{"id":"math-sequence-basics-028","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"Compute $(2 + 4 + 6 + … + 100) − (1 + 3 + 5 + … + 99)$.","stemZh":"计算 $(2 + 4 + 6 + … + 100) − (1 + 3 + 5 + … + 99)。$","options":[{"id":"A","text":"100"},{"id":"B","text":"0"},{"id":"C","text":"25"},{"id":"D","text":"50"}],"answer":"D","explanation":"Pair the $terms: (2 − 1) + (4 − 3) + … + (100 − 99)$. There are 50 pairs and each pair equals 1, so the difference is 50.","explanationZh":"$配对相减：(2 − 1) + (4 − 3) + … + (100 − 99)，共 50$ 对，每对等于 1，故差为 50。","tags":["hard","pairing","summation"],"status":"published"},{"id":"math-sequence-basics-029","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"A sequence satisfies $a_{1} = 2$ and $a_{n+1} = a_{n} + 3$. What is the least n for which $S_{n} > 100$?","stemZh":"数列满足 $a_{1} = 2，a_{n+1} = a_{n} + 3，则使 S_{n} > 100$ 成立的最小 n 为","options":[{"id":"A","text":"8"},{"id":"B","text":"9"},{"id":"C","text":"10"},{"id":"D","text":"7"}],"answer":"B","explanation":"$a_{n} = 3n − 1$ and $S_{n} = n(2 + 3n − 1)/2 = n(3n + 1)/2. S_{8} = 8·25/2 = 100$ and $S_{9} = 9·28/2 = 126$, so the least n with $S_{n}$ strictly greater than 100 is 9.","explanationZh":"$a_{n} = 3n − 1，S_{n} = n(3n + 1)/2。S_{8} = 100，S_{9} = 126，因此满足 S_{n} > 100$ 的最小 n 是 9。","tags":["hard","strict-inequality","summation"],"status":"published"},{"id":"math-sequence-basics-030","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"sequence-basics","type":"single-choice","difficulty":5,"stem":"$S_{n} = 3^{n} − 2$ is the sum of the first n terms of a sequence. For $n ≥ 2, a_{n} =$","stemZh":"数列前 n 项和 $S_{n} = 3^{n} − 2，则当 n ≥ 2 时 a_{n} =$","options":[{"id":"A","text":"$2·3^{n}$"},{"id":"B","text":"$3^{n-1}$"},{"id":"C","text":"$2·3^{n-1}$"},{"id":"D","text":"$3^{n} − 2$"}],"answer":"C","explanation":"For $n ≥ 2, a_{n} = S_{n} − S_{n}-_{1} = (3^{n} − 2) − (3^{n-1} − 2) = 3^{n} − 3^{n-1} = 2·3^{n-1}$. Note $a_{1} = S_{1} = 1$, which does not fit this formula — that is why $n = 1$ is excluded.","explanationZh":"$n ≥ 2 时，a_{n} = S_{n} − S_{n}-_{1} = (3^{n} − 2) − (3^{n-1} − 2) = 3^{n} − 3^{n-1} = 2·3^{n-1}。注意 a_{1} = S_{1} = 1$ 不满足该式，所以需排除 $n = 1。$","tags":["hard","sum-to-term","exponential"],"status":"published"},{"id":"math-arithmetic-sequence-001","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"In the arithmetic sequence 3, 7, 11, …, the common difference d is","stemZh":"等差数列 3, 7, 11, … 的公差 d 为","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"7"},{"id":"D","text":"11"}],"answer":"B","explanation":"$d = 7 − 3 = 11 − 7 = 4$, the constant difference between consecutive terms.","explanationZh":"$d = 7 − 3 = 11 − 7 = 4，即相邻两项的差。$","tags":["easy","common-difference","definition"],"status":"published"},{"id":"math-arithmetic-sequence-002","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"The 10th term of the arithmetic sequence with $a_{1} = 2$ and $d = 3$ is","stemZh":"首项 $a_{1} = 2、公差 d = 3$ 的等差数列的第 10 项为","options":[{"id":"A","text":"30"},{"id":"B","text":"32"},{"id":"C","text":"27"},{"id":"D","text":"29"}],"answer":"D","explanation":"$a_{10} = a_{1} + 9d = 2 + 9·3 = 29. (Using a_{1} + 10d\\text{ would give }32.)$","explanationZh":"$a_{10} = a_{1} + 9d = 2 + 27 = 29。（若误用 a_{1} + 10d$ 会得到 32。）","tags":["easy","general-term","indexing"],"status":"published"},{"id":"math-arithmetic-sequence-003","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"Which of the following is an arithmetic sequence?","stemZh":"下列哪一项是等差数列？","options":[{"id":"A","text":"1, 4, 7, 10"},{"id":"B","text":"1, 2, 4, 8"},{"id":"C","text":"1, 4, 9, 16"},{"id":"D","text":"1, 1/2, 1/3, 1/4"}],"answer":"A","explanation":"Only 1, 4, 7, 10 has a constant difference $(d = 3)$. The others are geometric, squares and reciprocals.","explanationZh":"只有 1, 4, 7, 10 相邻两项的差恒定（d $= 3），其余分别是等比数列、平方数列和倒数数列。$","tags":["easy","recognition","definition"],"status":"published"},{"id":"math-arithmetic-sequence-004","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"An arithmetic sequence has $a_{1} = 1, a_{5} = 9$. Its sum $S_{5}$ is","stemZh":"等差数列中 $a_{1} = 1，a_{5} = 9，则 S_{5} 为$","options":[{"id":"A","text":"20"},{"id":"B","text":"45"},{"id":"C","text":"25"},{"id":"D","text":"50"}],"answer":"C","explanation":"$S_{5} = 5(a_{1} + a_{5})/2 = 5(1 + 9)/2 = 25$.","explanationZh":"$S_{5} = 5(a_{1} + a_{5})/2 = 5 \\times 10/2 = 25。$","tags":["easy","sum-formula","computation"],"status":"published"},{"id":"math-arithmetic-sequence-005","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"If 2, x, 8 form an arithmetic sequence, then $x =$","stemZh":"若 2, x, 8 成等差数列，则 $x =$","options":[{"id":"A","text":"4"},{"id":"B","text":"6"},{"id":"C","text":"5"},{"id":"D","text":"10"}],"answer":"C","explanation":"For three consecutive $terms, 2x = 2 + 8 = 10$, so $x = 5. (4\\text{ would come from }a\\text{ geometric sequence, where }x^{2} = 16.)$","explanationZh":"三数成等差，则 $2x = 2 + 8 = 10，x = 5。（4$ 是等比数列 $x^{2} = 16$ 的结果。）","tags":["easy","arithmetic-mean","computation"],"status":"published"},{"id":"math-arithmetic-sequence-006","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"For the arithmetic sequence with general term $a_{n} = 3n − 1$, the common difference is","stemZh":"通项为 $a_{n} = 3n − 1$ 的等差数列，其公差为","options":[{"id":"A","text":"3"},{"id":"B","text":"−1"},{"id":"C","text":"2"},{"id":"D","text":"3n"}],"answer":"A","explanation":"$d = a_{n+1} − a_{n} = (3n + 2) − (3n − 1) = 3$, which is the coefficient of n.","explanationZh":"$d = a_{n+1} − a_{n} = 3，也就是 n$ 的系数。","tags":["easy","common-difference","general-term"],"status":"published"},{"id":"math-arithmetic-sequence-007","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"Compute 1 + 2 + 3 + … + 100.","stemZh":"计算 1 + 2 + 3 + … + 100。","options":[{"id":"A","text":"5000"},{"id":"B","text":"4950"},{"id":"C","text":"5500"},{"id":"D","text":"5050"}],"answer":"D","explanation":"$S_{100} = 100(1 + 100)/2 = 5050$.","explanationZh":"$S_{100} = 100(1 + 100)/2 = 5050。$","tags":["easy","sum-formula","computation"],"status":"published"},{"id":"math-arithmetic-sequence-008","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":1,"stem":"In an arithmetic sequence $a_{3} = 7$ and $a_{5} = 11$. Then $d =$","stemZh":"等差数列中 $a_{3} = 7，a_{5} = 11，则 d =$","options":[{"id":"A","text":"4"},{"id":"B","text":"2"},{"id":"C","text":"1"},{"id":"D","text":"3"}],"answer":"B","explanation":"$d = (a_{5} − a_{3})/(5 − 3) = (11 − 7)/2 = 2$. There are two steps from $a_{3}$ to $a_{5}$, so divide by 2, not by 1.","explanationZh":"$d = (a_{5} − a_{3})/(5 − 3) = 4/2 = 2。从 a_{3} 到 a_{5}$ 相隔 2 个公差，要除以 2。","tags":["easy","common-difference","indexing"],"status":"published"},{"id":"math-arithmetic-sequence-009","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"An arithmetic sequence has $d = 0$ and $a_{1} = 5$. Then $a_{10} =$","stemZh":"等差数列的公差 $d = 0，首项 a_{1} = 5，则 a_{10} =$","options":[{"id":"A","text":"0"},{"id":"B","text":"50"},{"id":"C","text":"15"},{"id":"D","text":"5"}],"answer":"D","explanation":"With $d = 0$ every term equals $a_{1} = 5$, so $a_{10} = 5. A$ constant sequence is a special arithmetic sequence.","explanationZh":"$d = 0$ 时每一项都等于 $a_{1} = 5，故 a_{10} = 5。常数列是公差为 0$ 的等差数列。","tags":["easy","constant-sequence","edge-case"],"status":"published"},{"id":"math-arithmetic-sequence-010","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":2,"stem":"An arithmetic sequence has $a_{1} = 10$ and $d = −2$. Its first four terms are","stemZh":"等差数列 $a_{1} = 10，d = −2，则它的前四项为$","options":[{"id":"A","text":"10, 8, 6, 4"},{"id":"B","text":"10, 12, 14, 16"},{"id":"C","text":"10, 9, 8, 7"},{"id":"D","text":"8, 6, 4, 2"}],"answer":"A","explanation":"Each term is 2 less than the previous one: 10, 8, 6, 4. A negative d means the sequence decreases.","explanationZh":"每一项比前一项少 $2：10, 8, 6, 4。d < 0$ 时数列递减。","tags":["easy","listing-terms","negative-d"],"status":"published"},{"id":"math-arithmetic-sequence-011","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"An arithmetic sequence has $a_{1} = 2$ and $d = 4$. If $S_{n} = 200$, then $n =$","stemZh":"等差数列 $a_{1} = 2，d = 4，若 S_{n} = 200，则 n =$","options":[{"id":"A","text":"8"},{"id":"B","text":"9"},{"id":"C","text":"10"},{"id":"D","text":"12"}],"answer":"C","explanation":"$S_{n} = na_{1} + n(n−1)d/2 = 2n + 2n(n − 1) = 2n^{2}$. Setting $2n^{2} = 200$ gives $n^{2} = 100$, so $n = 10$.","explanationZh":"$S_{n} = na_{1} + n(n−1)d/2 = 2n + 2n(n − 1) = 2n^{2}。由 2n^{2} = 200 得 n^{2} = 100，故 n = 10。$","tags":["medium","sum-formula","solve-for-n"],"status":"published"},{"id":"math-arithmetic-sequence-012","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"The sum of the first n terms of an arithmetic sequence is $S_{n} = 2n^{2} −$ n. Its first term and common difference are","stemZh":"等差数列前 n 项和 $S_{n} = 2n^{2} −$ n，则它的首项和公差为","options":[{"id":"A","text":"$a_{1} = 1, d = 4$"},{"id":"B","text":"$a_{1} = 2, d = 4$"},{"id":"C","text":"$a_{1} = 1, d = 2$"},{"id":"D","text":"$a_{1} = 2, d = 2$"}],"answer":"A","explanation":"$a_{1} = S_{1} = 2 − 1 = 1; a_{2} = S_{2} − S_{1} = 6 − 1 = 5$, so $d = a_{2} − a_{1} = 4. Check: S_{n} = n·1 + n(n−1)·4/2 = 2n^{2} −$ n.","explanationZh":"$a_{1} = S_{1} = 1，a_{2} = S_{2} − S_{1} = 6 − 1 = 5，故 d = 4。验证：S_{n} = n + 2n(n − 1) = 2n^{2} −$ n。","tags":["medium","sum-as-quadratic","find-a_{1}-and-d"],"status":"published"},{"id":"math-arithmetic-sequence-013","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"In an arithmetic sequence $a_{2} + a_{8} = 20$. Then $a_{5} =$","stemZh":"等差数列中 $a_{2} + a_{8} = 20，则 a_{5} =$","options":[{"id":"A","text":"20"},{"id":"B","text":"5"},{"id":"C","text":"40"},{"id":"D","text":"10"}],"answer":"D","explanation":"Since $2 + 8 = 5 + 5$, the equidistant-term property gives $a_{2} + a_{8} = 2a_{5}$, so $a_{5} = 10$.","explanationZh":"因为 $2 + 8 = 5 + 5，由等距离项性质 a_{2} + a_{8} = 2a_{5}，得 a_{5} = 10。$","tags":["medium","equidistant-terms","property"],"status":"published"},{"id":"math-arithmetic-sequence-014","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"An arithmetic sequence has $a_{1} = 20$ and $d = −3$. The largest n with $a_{n} > 0$ is","stemZh":"等差数列 $a_{1} = 20，d = −3，则使 a_{n} > 0$ 成立的最大 n 为","options":[{"id":"A","text":"6"},{"id":"B","text":"7"},{"id":"C","text":"8"},{"id":"D","text":"9"}],"answer":"B","explanation":"$a_{n} = 20 − 3(n − 1) = 23 − 3n$. Then $23 − 3n > 0$ gives $n < 23/3 ≈ 7.67$, so the largest integer n is $7 (a_{7} = 2, a_{8} = −1)$.","explanationZh":"$a_{n} = 23 − 3n，由 23 − 3n > 0 得 n < 23/3 ≈ 7.67，故最大整数 n = 7（a_{7} = 2，a_{8} = −1）。$","tags":["medium","general-term","inequality"],"status":"published"},{"id":"math-arithmetic-sequence-015","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"The sum of the first n terms of an arithmetic sequence is $S_{n} = n^{2} + 3n$. Then $a_{4} =$","stemZh":"等差数列前 n 项和 $S_{n} = n^{2} + 3n，则 a_{4} =$","options":[{"id":"A","text":"8"},{"id":"B","text":"10"},{"id":"C","text":"12"},{"id":"D","text":"14"}],"answer":"B","explanation":"$a_{4} = S_{4} − S_{3} = (16 + 12) − (9 + 9) = 28 − 18 = 10$. Equivalently $a_{n} = 2n + 2$ for all n here.","explanationZh":"$a_{4} = S_{4} − S_{3} = 28 − 18 = 10。（也可先得 a_{n} = 2n + 2。）$","tags":["medium","sum-to-term","computation"],"status":"published"},{"id":"math-arithmetic-sequence-016","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"In an arithmetic sequence $a_{5} = 12$ and $a_{10} = 27$. Then $a_{15} =$","stemZh":"等差数列中 $a_{5} = 12，a_{10} = 27，则 a_{15} =$","options":[{"id":"A","text":"37"},{"id":"B","text":"40"},{"id":"C","text":"42"},{"id":"D","text":"45"}],"answer":"C","explanation":"$d = (27 − 12)/(10 − 5) = 3$, so $a_{15} = a_{10} + 5d = 27 + 15 = 42. (Equivalently a_{5}, a_{10}, a_{15}\\text{ are themselves in arithmetic }progression.)$","explanationZh":"$d = (27 − 12)/5 = 3，故 a_{15} = a_{10} + 5d = 27 + 15 = 42。（a_{5}、a_{10}、a_{15}$ 也成等差数列。）","tags":["medium","general-term","property"],"status":"published"},{"id":"math-arithmetic-sequence-017","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"Which expression gives $S_{n}$ for an arithmetic sequence with first term $a_{1}$ and common difference d?","stemZh":"首项为 $a_{1}、公差为 d$ 的等差数列，其前 n 项和 $S_{n}$ 的表达式是","options":[{"id":"A","text":"$na_{1} + n(n−1)d/2$"},{"id":"B","text":"$na_{1} + n(n+1)d/2$"},{"id":"C","text":"$n(a_{1} + a_{n})d/2$"},{"id":"D","text":"$n(a_{1} + d)/2$"}],"answer":"A","explanation":"$S_{n} = n(a_{1} + a_{n})/2 = n(a_{1} + a_{1} + (n − 1)d)/2 = na_{1} + n(n−1)d/2$.","explanationZh":"$S_{n} = n(a_{1} + a_{n})/2 = n(2a_{1} + (n − 1)d)/2 = na_{1} + n(n−1)d/2。$","tags":["medium","sum-formula","formula-recall"],"status":"published"},{"id":"math-arithmetic-sequence-018","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"Three consecutive terms of an arithmetic sequence satisfy $a_{1} + a_{2} + a_{3} = 12$ and $a_{1}a_{2}a_{3} = 48$. Then $a_{2} =$","stemZh":"等差数列连续三项满足 $a_{1} + a_{2} + a_{3} = 12，a_{1}a_{2}a_{3} = 48，则 a_{2} =$","options":[{"id":"A","text":"3"},{"id":"B","text":"6"},{"id":"C","text":"12"},{"id":"D","text":"4"}],"answer":"D","explanation":"By the arithmetic mean property $a_{1} + a_{3} = 2a_{2}$, so $3a_{2} = 12$ and $a_{2} = 4$. The product condition is consistent $(a_{1} = 2, a_{3} = 6)$ but is not needed.","explanationZh":"由等差中项性质 $a_{1} + a_{3} = 2a_{2}，故 3a_{2} = 12，a_{2} = 4。积的条件与之相符（a_{1} = 2，a_{3} = 6），但并非必需。$","tags":["medium","arithmetic-mean","property"],"status":"published"},{"id":"math-arithmetic-sequence-019","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"An arithmetic sequence has $a_{1} = −5$ and $d = 2$. Then $S_{10} =$","stemZh":"等差数列 $a_{1} = −5，d = 2，则 S_{10} =$","options":[{"id":"A","text":"30"},{"id":"B","text":"35"},{"id":"C","text":"40"},{"id":"D","text":"50"}],"answer":"C","explanation":"$S_{10} = 10a_{1} + 10·9·d/2 = −50 + 90 = 40$.","explanationZh":"$S_{10} = 10 \\times (−5) + 45 \\times 2 = −50 + 90 = 40。$","tags":["medium","sum-formula","negative-a_{1}"],"status":"published"},{"id":"math-arithmetic-sequence-020","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":3,"stem":"An arithmetic sequence satisfies $S_{5} = 25$ and $S_{10} = 100$. Then $S_{15} =$","stemZh":"等差数列满足 $S_{5} = 25，S_{10} = 100，则 S_{15} =$","options":[{"id":"A","text":"200"},{"id":"B","text":"225"},{"id":"C","text":"250"},{"id":"D","text":"175"}],"answer":"B","explanation":"$S_{5}, S_{10} − S_{5}, S_{15} − S_{10}$ are also in arithmetic progression: 25, 75, 125. Hence $S_{15} = 100 + 125 = 225$.","explanationZh":"$S_{5}、S_{10} − S_{5}、S_{15} − S_{10} 也成等差数列：25, 75, 125，故 S_{15} = 100 + 125 = 225。$","tags":["medium","block-sums","property"],"status":"published"},{"id":"math-arithmetic-sequence-021","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"An arithmetic sequence has $S_{n} = −n^{2} + 20n$. The maximum value of $S_{n}$ is","stemZh":"等差数列 $S_{n} = −n^{2} + 20n，则 S_{n}$ 的最大值为","options":[{"id":"A","text":"20"},{"id":"B","text":"90"},{"id":"C","text":"110"},{"id":"D","text":"100"}],"answer":"D","explanation":"$S_{n} = −(n − 10)^{2} + 100$, so the maximum is 100 at $n = 10$. Equivalently $a_{n} = −2n + 21$ stays positive up to $n = 10$ and turns negative at $n = 11$.","explanationZh":"$S_{n} = −(n − 10)^{2} + 100，故最大值为 100（n = 10）。也可由 a_{n} = −2n + 21$ 知前 10 项为正、第 11 项起为负。","tags":["hard","maximum-of-S_{n}","quadratic"],"status":"published"},{"id":"math-arithmetic-sequence-022","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"An arithmetic sequence has $a_{1} = 21$ and $d = −3$. The value(s) of n at which $S_{n}$ is maximal:","stemZh":"等差数列 $a_{1} = 21，d = −3，则 S_{n}$ 取最大值时的 n 为","options":[{"id":"A","text":"$n = 7$ only"},{"id":"B","text":"$n = 7$ or $n = 8$"},{"id":"C","text":"$n = 8$ only"},{"id":"D","text":"$n = 9$"}],"answer":"B","explanation":"$a_{n} = 24 − 3n$, so $a_{7} = 3 > 0, a_{8} = 0$ and $a_{9} = −3 < 0$. Since $a_{8} = 0$ adds $nothing, S_{7} = S_{8} = 84$ is the maximum, attained at both $n = 7$ and $n = 8$.","explanationZh":"$a_{n} = 24 − 3n，a_{7} = 3 > 0，a_{8} = 0，a_{9} = −3 < 0。由于第 8$ 项为 $0，S_{7} = S_{8} = 84$ 同为最大值，故 $n = 7 或 8。$","tags":["hard","maximum-of-S_{n}","zero-term-trap"],"status":"published"},{"id":"math-arithmetic-sequence-023","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"In a non-constant arithmetic sequence $S_{7} = S_{12}$. Which statement must be true?","stemZh":"在非恒等的等差数列中 $S_{7} = S_{12}，则下列一定正确的是$","options":[{"id":"A","text":"$S_{19} = 0$"},{"id":"B","text":"$S_{18} = 0$"},{"id":"C","text":"$a_{9} = 0$"},{"id":"D","text":"$a_{11} = 0$"}],"answer":"A","explanation":"Write $S_{n} = An^{2} +$ Bn with $A ≠ 0. S_{7} = S_{12}$ gives $19A + B = 0$, hence $S_{19} = 19(19A + B) = 0$. The other options give −18A, −2A and 2A, which are not zero.","explanationZh":"$设 S_{n} = An^{2} +$ Bn（A $≠ 0）。由 S_{7} = S_{12} 得 19A + B = 0，所以 S_{19} = 19(19A + B) = 0。其余各项分别为 −18A、−2A、2A，均不为 0。$","tags":["hard","sum-as-quadratic","symmetry"],"status":"published"},{"id":"math-arithmetic-sequence-024","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"In an arithmetic sequence $a_{1} + a_{4} + a_{7} = 39$ and $a_{2} + a_{5} + a_{8} = 33$. Then $a_{3} + a_{6} + a_{9} =$","stemZh":"等差数列中 $a_{1} + a_{4} + a_{7} = 39，a_{2} + a_{5} + a_{8} = 33，则 a_{3} + a_{6} + a_{9} =$","options":[{"id":"A","text":"21"},{"id":"B","text":"24"},{"id":"C","text":"27"},{"id":"D","text":"30"}],"answer":"C","explanation":"Let $T_{1} = a_{1} + a_{4} + a_{7}, T_{2} = a_{2} + a_{5} + a_{8}, T_{3} = a_{3} + a_{6} + a_{9}$. Each is obtained from the previous by adding 3d, so $T_{1}, T_{2}, T_{3}$ are arithmetic: 39, 33, 27.","explanationZh":"$记 T_{1} = a_{1} + a_{4} + a_{7}，T_{2} = a_{2} + a_{5} + a_{8}，T_{3} = a_{3} + a_{6} + a_{9}，后一个比前一个多 3d，故 T_{1}、T_{2}、T_{3} 成等差数列：39, 33, 27。$","tags":["hard","block-sums","property"],"status":"published"},{"id":"math-arithmetic-sequence-025","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"An arithmetic sequence has $a_{1} > 0, d < 0$ and $S_{5} = S_{11}$. For which n is $S_{n}$ maximal?","stemZh":"等差数列 $a_{1} > 0，d < 0 且 S_{5} = S_{11}，则 S_{n}$ 取最大值时 $n =$","options":[{"id":"A","text":"7"},{"id":"B","text":"9"},{"id":"C","text":"10"},{"id":"D","text":"8"}],"answer":"D","explanation":"With $S_{n} = An^{2} + Bn, S_{5} = S_{11}$ gives $16A + B = 0$ and $A = d/2 < 0$. Then $a_{n} = 2An + B − A = A(2n − 17)$, which is positive for $n ≤ 8$ and negative from $n = 9$. Hence the maximum is at $n = 8$.","explanationZh":"$设 S_{n} = An^{2} +$ Bn，由 $S_{5} = S_{11} 得 16A + B = 0，且 A = d/2 < 0。a_{n} = A(2n − 17)，当 n ≤ 8$ 时为正，n $≥ 9$ 时为负，故最大值在 $n = 8$ 处取得。","tags":["hard","maximum-of-S_{n}","quadratic"],"status":"published"},{"id":"math-arithmetic-sequence-026","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"Two arithmetic sequences ${a_{n}}$ and ${b_{n}}$ have sums $S_{n}$ and $T_{n}$ with $S_{n}/T_{n} = (2n + 1)/(3n + 2)$. Then $a_{5}/b_{5} =$","stemZh":"两个等差数列 ${a_{n}}、{b_{n}}$ 的前 n 项和分别为 $S_{n}、T_{n}，且 S_{n}/T_{n} = (2n + 1)/(3n + 2)，则 a_{5}/b_{5} =$","options":[{"id":"A","text":"19/29"},{"id":"B","text":"11/17"},{"id":"C","text":"9/14"},{"id":"D","text":"3/5"}],"answer":"A","explanation":"$a_{5}/b_{5} = 2a_{5}/2b_{5} = (a_{1} + a_{9})/(b_{1} + b_{9}) = (2S_{9}/9)/(2T_{9}/9) = S_{9}/T_{9} = (2·9 + 1)/(3·9 + 2) = 19/29$. Using $n = 5$ instead of $n = 9$ gives the distractor 11/17.","explanationZh":"$a_{5}/b_{5} = (a_{1} + a_{9})/(b_{1} + b_{9}) = S_{9}/T_{9} = (2\\times9 + 1)/(3\\times9 + 2) = 19/29。若误取 n = 5$ 则得 11/17。","tags":["hard","sum-ratio","middle-term"],"status":"published"},{"id":"math-arithmetic-sequence-027","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"In an arithmetic sequence $a_{1} + … + a_{10} = 100$ and $a_{11} + … + a_{20} = 300$. Then $d =$","stemZh":"等差数列中 $a_{1} + … + a_{10} = 100，a_{11} + … + a_{20} = 300，则 d =$","options":[{"id":"A","text":"1"},{"id":"B","text":"2"},{"id":"C","text":"3"},{"id":"D","text":"4"}],"answer":"B","explanation":"$S_{10} = 10a_{1} + 45d = 100$ and $S_{20} − S_{10} = 10a_{1} + 145d = 300$. Subtracting gives $100d = 200$, so $d = 2$.","explanationZh":"$S_{10} = 10a_{1} + 45d = 100，S_{20} − S_{10} = 10a_{1} + 145d = 300，两式相减得 100d = 200，d = 2。$","tags":["hard","block-sums","solve-for-d"],"status":"published"},{"id":"math-arithmetic-sequence-028","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"An arithmetic sequence has $a_{1} = 13$ and $S_{3} = S_{11}$. The value of n at which $S_{n}$ is maximal is","stemZh":"等差数列 $a_{1} = 13 且 S_{3} = S_{11}，则 S_{n}$ 取最大值时 $n =$","options":[{"id":"A","text":"6"},{"id":"B","text":"8"},{"id":"C","text":"7"},{"id":"D","text":"9"}],"answer":"C","explanation":"$3a_{1} + 3d = 11a_{1} + 55d$ gives $8a_{1} + 52d = 0$, so with $a_{1} = 13$ we get $d = −2$. Then $a_{n} = 13 − 2(n − 1) = 15 − 2n: a_{7} = 1 > 0$ and $a_{8} = −1 < 0$, so the maximum is at $n = 7$.","explanationZh":"$由 3a_{1} + 3d = 11a_{1} + 55d 得 8a_{1} + 52d = 0，代入 a_{1} = 13 得 d = −2。a_{n} = 15 − 2n，a_{7} = 1 > 0，a_{8} = −1 < 0，故最大值在 n = 7。$","tags":["hard","maximum-of-S_{n}","two-conditions"],"status":"published"},{"id":"math-arithmetic-sequence-029","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":4,"stem":"An arithmetic sequence satisfies $S_{10} = 100$ and $S_{20} = 400$. Then $S_{30} =$","stemZh":"等差数列满足 $S_{10} = 100，S_{20} = 400，则 S_{30} =$","options":[{"id":"A","text":"900"},{"id":"B","text":"800"},{"id":"C","text":"700"},{"id":"D","text":"1000"}],"answer":"A","explanation":"$S_{10}, S_{20} − S_{10}, S_{30} − S_{20}$ are in arithmetic progression: 100, 300, 500, so $S_{30} = 400 + 500 = 900$.","explanationZh":"$S_{10}、S_{20} − S_{10}、S_{30} − S_{20} 成等差数列：100, 300, 500，故 S_{30} = 400 + 500 = 900。$","tags":["hard","block-sums","property"],"status":"published"},{"id":"math-arithmetic-sequence-030","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"arithmetic-sequence","type":"single-choice","difficulty":5,"stem":"In an arithmetic sequence $a_{m} = n$ and $a_{n} = m$ with $m ≠$ n. Then $a_{m+n} =$","stemZh":"等差数列中 $a_{m} = n，a_{n} =$ m（m $≠$ n），则 $a_{m+n} =$","options":[{"id":"A","text":"m + n"},{"id":"B","text":"m − n"},{"id":"C","text":"1"},{"id":"D","text":"0"}],"answer":"D","explanation":"$d = (a_{n} − a_{m})/(n − m) = (m − n)/(n − m) = −1$, and $a_{1} = a_{m} − (m − 1)d = n + m − 1$. Thus $a_{m+n} = a_{1} + (m + n − 1)d = (m + n − 1) − (m + n − 1) = 0$.","explanationZh":"$d = (m − n)/(n − m) = −1，a_{1} = n + m − 1，故 a_{m+n} = a_{1} + (m + n − 1)d = 0。$","tags":["hard","general-term","symbolic"],"status":"published"},{"id":"math-geometric-sequence-001","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":1,"stem":"In the geometric sequence 2, 6, 18, …, the common ratio q is","stemZh":"等比数列 2, 6, 18, … 的公比 q 为","options":[{"id":"A","text":"3"},{"id":"B","text":"2"},{"id":"C","text":"6"},{"id":"D","text":"1/3"}],"answer":"A","explanation":"$q = 6/2 = 18/6 = 3$, the constant ratio of consecutive terms.","explanationZh":"$q = 6/2 = 18/6 = 3，即相邻两项的比。$","tags":["easy","common-ratio","definition"],"status":"published"},{"id":"math-geometric-sequence-002","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":1,"stem":"The 5th term of the geometric sequence with $a_{1} = 2$ and $q = 3$ is","stemZh":"首项 $a_{1} = 2、公比 q = 3$ 的等比数列第 5 项为","options":[{"id":"A","text":"54"},{"id":"B","text":"81"},{"id":"C","text":"162"},{"id":"D","text":"243"}],"answer":"C","explanation":"$a_{5} = a_{1}q^{4} = 2·3^{4} = 2·81 = 162$.","explanationZh":"$a_{5} = a_{1}q^{4} = 2 \\times 81 = 162。$","tags":["easy","general-term","indexing"],"status":"published"},{"id":"math-geometric-sequence-003","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":1,"stem":"Which of the following is a geometric sequence?","stemZh":"下列哪一项是等比数列？","options":[{"id":"A","text":"1, 3, 5, 7"},{"id":"B","text":"2, 4, 8, 16"},{"id":"C","text":"1, 4, 9, 16"},{"id":"D","text":"1, 2, 3, 5"}],"answer":"B","explanation":"Only 2, 4, 8, 16 has a constant ratio $(q = 2)$. The first is arithmetic, the third is squares, the fourth has no fixed ratio.","explanationZh":"只有 2, 4, 8, 16 相邻两项的比恒定（q $= 2）。第一个是等差数列，第三个是平方数列，第四个比值不固定。$","tags":["easy","recognition","definition"],"status":"published"},{"id":"math-geometric-sequence-004","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"For a geometric sequence the common ratio q must satisfy","stemZh":"等比数列的公比 q 必须满足","options":[{"id":"A","text":"$q > 0$"},{"id":"B","text":"$q ≥ 1$"},{"id":"C","text":"$q ∈ ℤ$"},{"id":"D","text":"$q ≠ 0$"}],"answer":"D","explanation":"The only requirement is $q ≠ 0 (and\\text{ every term must be }non-zero); q$ may be negative or a fraction.","explanationZh":"唯一要求是 $q ≠ 0（且各项均不为 0）；q$ 可以是负数或分数。","tags":["easy","common-ratio","edge-case"],"status":"published"},{"id":"math-geometric-sequence-005","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"A geometric sequence has $a_{1} = 1$ and $q = 1$. Then $S_{10} =$","stemZh":"等比数列 $a_{1} = 1，q = 1，则 S_{10} =$","options":[{"id":"A","text":"1"},{"id":"B","text":"0"},{"id":"C","text":"$2^{10}$"},{"id":"D","text":"10"}],"answer":"D","explanation":"When $q = 1$ the sum formula $a_{1}(1 − q^{n})/(1 − q)$ is not valid; instead every term is 1 and $S_{10} = 10$.","explanationZh":"$q = 1$ 时不能使用 $a_{1}(1 − q^{n})/(1 − q)，此时每项都是 1，故 S_{10} = 10。$","tags":["easy","q-equals-1","edge-case"],"status":"published"},{"id":"math-geometric-sequence-006","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"If 2, x, 8 form a geometric sequence with $x > 0$, then $x =$","stemZh":"若 2, x, 8 成等比数列且 $x > 0，则 x =$","options":[{"id":"A","text":"$\\pm4$"},{"id":"B","text":"4"},{"id":"C","text":"−4"},{"id":"D","text":"5"}],"answer":"B","explanation":"$x^{2} = 2·8 = 16$ gives $x = \\pm4$, and the condition $x > 0$ selects $x = 4. (Without$ the condition both signs are possible.)","explanationZh":"$x^{2} = 2 \\times 8 = 16，得 x = \\pm4，再由 x > 0 取 x = 4。（没有该条件时两解都成立。）$","tags":["easy","geometric-mean","sign-care"],"status":"published"},{"id":"math-geometric-sequence-007","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":1,"stem":"In a geometric sequence $a_{2} = 6$ and $a_{3} = 18$. Then $a_{1} =$","stemZh":"等比数列中 $a_{2} = 6，a_{3} = 18，则 a_{1} =$","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"6"},{"id":"D","text":"1"}],"answer":"A","explanation":"$q = a_{3}/a_{2} = 3$, so $a_{1} = a_{2}/q = 6/3 = 2$.","explanationZh":"$q = 18/6 = 3，故 a_{1} = a_{2}/q = 2。$","tags":["easy","common-ratio","backwards"],"status":"published"},{"id":"math-geometric-sequence-008","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"A geometric sequence has general term $a_{n} = 3·2^{n-1}$. Then $a_{1} =$","stemZh":"等比数列通项 $a_{n} = 3·2^{n-1}，则 a_{1} =$","options":[{"id":"A","text":"6"},{"id":"B","text":"2"},{"id":"C","text":"3"},{"id":"D","text":"1"}],"answer":"C","explanation":"$a_{1} = 3·2^{0} = 3$. The first term is the coefficient in front of the power, not the base 2.","explanationZh":"$a_{1} = 3·2^{0} = 3。首项是幂前面的系数，而不是底数 2。$","tags":["easy","general-term","indexing"],"status":"published"},{"id":"math-geometric-sequence-009","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"For a geometric sequence with $a_{1} = 1$ and $q = 2, S_{5} =$","stemZh":"等比数列 $a_{1} = 1，q = 2，则 S_{5} =$","options":[{"id":"A","text":"16"},{"id":"B","text":"31"},{"id":"C","text":"32"},{"id":"D","text":"63"}],"answer":"B","explanation":"$S_{5} = a_{1}(1 − q^{5})/(1 − q) = (1 − 32)/(−1) = 31$, which also equals 1 + 2 + 4 + 8 + 16.","explanationZh":"$S_{5} = (1 − 2^{5})/(1 − 2) = 31，即 1 + 2 + 4 + 8 + 16 = 31。$","tags":["easy","sum-formula","computation"],"status":"published"},{"id":"math-geometric-sequence-010","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":2,"stem":"A geometric sequence has $a_{1} = 3$ and $q = −1$. Its first four terms are","stemZh":"等比数列 $a_{1} = 3，q = −1，则它的前四项为$","options":[{"id":"A","text":"3, 3, 3, 3"},{"id":"B","text":"−3, 3, −3, 3"},{"id":"C","text":"3, −3, 3, −3"},{"id":"D","text":"3, −1, 1/3, −1/9"}],"answer":"C","explanation":"Each term is the previous one multiplied by −1, so the signs alternate while the absolute value stays 3: 3, −3, 3, −3.","explanationZh":"每次乘以 −1，符号交替而绝对值不变：3, −3, 3, −3。","tags":["easy","negative-ratio","listing-terms"],"status":"published"},{"id":"math-geometric-sequence-011","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence has $a_{1} = 1$ and $q = 3$. If $a_{n} = 81$, then $n =$","stemZh":"等比数列 $a_{1} = 1，q = 3，若 a_{n} = 81，则 n =$","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"6"},{"id":"D","text":"5"}],"answer":"D","explanation":"$a_{n} = 3^{n-1} = 81 = 3^{4}$, so $n − 1 = 4$ and $n = 5$.","explanationZh":"$3^{n-1} = 81 = 3^{4}，故 n − 1 = 4，n = 5。$","tags":["medium","solve-for-n","general-term"],"status":"published"},{"id":"math-geometric-sequence-012","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence with all terms positive has $a_{3} = 4$ and $a_{5} = 16$. Then $a_{4} =$","stemZh":"各项均为正的等比数列中 $a_{3} = 4，a_{5} = 16，则 a_{4} =$","options":[{"id":"A","text":"$\\pm8$"},{"id":"B","text":"8"},{"id":"C","text":"−8"},{"id":"D","text":"10"}],"answer":"B","explanation":"$a_{4}^{2} = a_{3}a_{5} = 64$, so $a_{4} = \\pm8$; since all terms are $positive, a_{4} = 8$.","explanationZh":"$a_{4}^{2} = a_{3}a_{5} = 64，得 a_{4} = \\pm8；由各项为正知 a_{4} = 8。$","tags":["medium","geometric-mean","sign-care"],"status":"published"},{"id":"math-geometric-sequence-013","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"The sum of the first n terms of a geometric sequence is $S_{n} = 3(2^{n} − 1)$. Then $a_{1} =$","stemZh":"等比数列前 n 项和 $S_{n} = 3(2^{n} − 1)，则 a_{1} =$","options":[{"id":"A","text":"3"},{"id":"B","text":"6"},{"id":"C","text":"2"},{"id":"D","text":"1"}],"answer":"A","explanation":"$a_{1} = S_{1} = 3(2 − 1) = 3. (Indeed a_{1} = 3\\text{ and }q = 2\\text{ give }S_{n} = 3(2^{n} − 1)/(2 − 1).)$","explanationZh":"$a_{1} = S_{1} = 3(2 − 1) = 3。（a_{1} = 3、q = 2$ 时确有 $S_{n} = 3(2^{n} − 1)。）$","tags":["medium","sum-formula","find-a_{1}"],"status":"published"},{"id":"math-geometric-sequence-014","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence has $a_{1} = 2$ and $q = 3$. Then $S_{4} =$","stemZh":"等比数列 $a_{1} = 2，q = 3，则 S_{4} =$","options":[{"id":"A","text":"40"},{"id":"B","text":"54"},{"id":"C","text":"80"},{"id":"D","text":"162"}],"answer":"C","explanation":"$S_{4} = 2(3^{4} − 1)/(3 − 1) = 2·80/2 = 80$, which agrees with 2 + 6 + 18 + 54.","explanationZh":"$S_{4} = 2(3^{4} − 1)/(3 − 1) = 80，与 2 + 6 + 18 + 54$ 一致。","tags":["medium","sum-formula","computation"],"status":"published"},{"id":"math-geometric-sequence-015","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"Compute 1 + 1/2 + 1/4 + 1/8 + 1/16.","stemZh":"计算 1 + 1/2 + 1/4 + 1/8 + 1/16。","options":[{"id":"A","text":"31/16"},{"id":"B","text":"31/32"},{"id":"C","text":"15/8"},{"id":"D","text":"63/32"}],"answer":"A","explanation":"With $a_{1} = 1$ and $q = 1/2: S_{5} = (1 − (1/2)^{5})/(1 − 1/2) = 2(1 − 1/32) = 31/16$.","explanationZh":"$a_{1} = 1，q = 1/2，S_{5} = (1 − (1/2)^{5})/(1 − 1/2) = 2 \\times 31/32 = 31/16。$","tags":["medium","sum-formula","fractions"],"status":"published"},{"id":"math-geometric-sequence-016","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"In a geometric sequence $a_{1}a_{9} = 16$. Then $a_{5} =$","stemZh":"等比数列中 $a_{1}a_{9} = 16，则 a_{5} =$","options":[{"id":"A","text":"4"},{"id":"B","text":"−4"},{"id":"C","text":"8"},{"id":"D","text":"$\\pm4$"}],"answer":"D","explanation":"Since $1 + 9 = 5 + 5, a_{1}a_{9} = a_{5}^{2} = 16$, so $a_{5} = \\pm4$. Without extra sign information both signs are possible (for example $q = 1$ gives $a_{5} = 4$, while alternating-sign sequences can give $−4)$.","explanationZh":"$由 1 + 9 = 5 + 5 得 a_{1}a_{9} = a_{5}^{2} = 16，故 a_{5} = \\pm4。没有附加条件时两种符号都可能。$","tags":["medium","equidistant-terms","sign-care"],"status":"published"},{"id":"math-geometric-sequence-017","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence has $a_{1} = 1$ and $q = 2$. If $S_{n} = 127$, then $n =$","stemZh":"等比数列 $a_{1} = 1，q = 2，若 S_{n} = 127，则 n =$","options":[{"id":"A","text":"6"},{"id":"B","text":"8"},{"id":"C","text":"7"},{"id":"D","text":"9"}],"answer":"C","explanation":"$S_{n} = (2^{n} − 1)/(2 − 1) = 2^{n} − 1 = 127$, so $2^{n} = 128$ and $n = 7$.","explanationZh":"$2^{n} − 1 = 127，得 2^{n} = 128，故 n = 7。$","tags":["medium","solve-for-n","sum-formula"],"status":"published"},{"id":"math-geometric-sequence-018","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"The geometric sequence with $a_{1} = 1$ and $q = −2$ is","stemZh":"$a_{1} = 1、q = −2$ 的等比数列是","options":[{"id":"A","text":"increasing"},{"id":"B","text":"oscillating"},{"id":"C","text":"decreasing"},{"id":"D","text":"constant"}],"answer":"B","explanation":"The terms 1, −2, 4, −8, … alternate in sign while growing in absolute value, so the sequence is oscillating — it is neither increasing nor decreasing.","explanationZh":"各项为 1, −2, 4, −8, …，符号交替而绝对值递增，因此是摆动数列，既非递增也非递减。","tags":["medium","negative-ratio","monotonicity"],"status":"published"},{"id":"math-geometric-sequence-019","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"For a geometric sequence with $q ≠ 1$, the sum of the first n terms is","stemZh":"公比 $q ≠ 1$ 的等比数列，其前 n 项和为","options":[{"id":"A","text":"$a_{1}(1 − q^{n})/(1 + q)$"},{"id":"B","text":"$a_{1}(1 − q^{n})/(1 − q)$"},{"id":"C","text":"$na_{1}$"},{"id":"D","text":"$a_{1}q^{n}/(1 − q)$"}],"answer":"B","explanation":"$S_{n} = a_{1}(1 − q^{n})/(1 − q)$ for $q ≠ 1 (equivalently a_{1}(q^{n} − 1)/(q − 1)); na_{1}$ only applies when $q = 1$.","explanationZh":"$q ≠ 1 时 S_{n} = a_{1}(1 − q^{n})/(1 − q)（也可写成 a_{1}(q^{n} − 1)/(q − 1)）；na_{1}$ 只适用于 $q = 1。$","tags":["medium","sum-formula","formula-recall"],"status":"published"},{"id":"math-geometric-sequence-020","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":3,"stem":"A geometric sequence has $a_{1} = 3$ and $q = 1/2$. Then $S_{3} =$","stemZh":"等比数列 $a_{1} = 3，q = 1/2，则 S_{3} =$","options":[{"id":"A","text":"21/8"},{"id":"B","text":"7/2"},{"id":"C","text":"15/4"},{"id":"D","text":"21/4"}],"answer":"D","explanation":"$S_{3} = 3 + 3/2 + 3/4 = 21/4$. Using the $formula: 3(1 − 1/8)/(1 − 1/2) = 3·(7/8)·2 = 21/4$.","explanationZh":"$S_{3} = 3 + 3/2 + 3/4 = 21/4；用公式得 3(1 − 1/8)/(1 − 1/2) = 21/4。$","tags":["medium","sum-formula","fractions"],"status":"published"},{"id":"math-geometric-sequence-021","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"In a geometric sequence with positive terms $a_{2}a_{4} + 2a_{3}a_{5} + a_{4}a_{6} = 25$. Then $a_{3} + a_{5} =$","stemZh":"各项为正的等比数列中 $a_{2}a_{4} + 2a_{3}a_{5} + a_{4}a_{6} = 25，则 a_{3} + a_{5} =$","options":[{"id":"A","text":"25"},{"id":"B","text":"$\\pm5$"},{"id":"C","text":"5"},{"id":"D","text":"10"}],"answer":"C","explanation":"Since $a_{2}a_{4} = a_{3}^{2}$ and $a_{4}a_{6} = a_{5}^{2}$, the left side is $a_{3}^{2} + 2a_{3}a_{5} + a_{5}^{2} = (a_{3} + a_{5})^{2} = 25$. With positive terms $a_{3} + a_{5} = 5$.","explanationZh":"$由 a_{2}a_{4} = a_{3}^{2}，a_{4}a_{6} = a_{5}^{2}，左边 = a_{3}^{2} + 2a_{3}a_{5} + a_{5}^{2} = (a_{3} + a_{5})^{2} = 25；各项为正，故 a_{3} + a_{5} = 5。$","tags":["hard","equidistant-terms","algebraic-identity"],"status":"published"},{"id":"math-geometric-sequence-022","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"A geometric sequence satisfies $S_{3} = 7$ and $S_{6} = 63$. Then $q =$","stemZh":"等比数列满足 $S_{3} = 7，S_{6} = 63，则 q =$","options":[{"id":"A","text":"2"},{"id":"B","text":"3"},{"id":"C","text":"1/2"},{"id":"D","text":"−2"}],"answer":"A","explanation":"$S_{6}/S_{3} = (1 − q^{6})/(1 − q^{3}) = 1 + q^{3} = 9$, so $q^{3} = 8$ and $q = 2. (Then a_{1} = 1,\\text{ giving }S_{3} = 7\\text{ and }S_{6} = 63.)$","explanationZh":"$S_{6}/S_{3} = (1 − q^{6})/(1 − q^{3}) = 1 + q^{3} = 9，故 q^{3} = 8，q = 2。（此时 a_{1} = 1，验算 S_{3} = 7，S_{6} = 63。）$","tags":["hard","sum-ratio","solve-for-q"],"status":"published"},{"id":"math-geometric-sequence-023","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"If $S_{n} = 3^{n} + k$ is the sum of the first n terms of a geometric sequence, then $k =$","stemZh":"$若 S_{n} = 3^{n} + k$ 是某等比数列的前 n 项和，则 $k =$","options":[{"id":"A","text":"0"},{"id":"B","text":"1"},{"id":"C","text":"3"},{"id":"D","text":"−1"}],"answer":"D","explanation":"A geometric sum has the form $A(q^{n} − 1)$, so $S_{0} = A(q^{0} − 1) = 0$; here $3^{0} + k = 1 + k = 0$, hence $k = −1$. Indeed $3^{n} − 1 = 2(3^{n} − 1)/(3 − 1)$ is the sum for $a_{1} = 2, q = 3$.","explanationZh":"等比前 n 项和形如 $A(q^{n} − 1)，故 S_{0} = 0，即 3^{0} + k = 0，k = −1。此时 3^{n} − 1$ 正是 $a_{1} = 2、q = 3$ 的前 n 项和。","tags":["hard","sum-form","constant-term-trap"],"status":"published"},{"id":"math-geometric-sequence-024","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"For a geometric sequence with $a_{1} = 1$ and $q = 2$, the three block sums $S_{n}, S_{2n} − S_{n}, S_{3n} − S_{2n}$","stemZh":"等比数列 $a_{1} = 1，q = 2，则 S_{n}、S_{2n} − S_{n}、S_{3n} − S_{2n}$ 这三个分段和","options":[{"id":"A","text":"form an arithmetic sequence"},{"id":"B","text":"form a geometric sequence with ratio $2^{n}$"},{"id":"C","text":"are all equal"},{"id":"D","text":"form a geometric sequence with ratio 2"}],"answer":"B","explanation":"$S_{n} = 2^{n} − 1, S_{2n} − S_{n} = 2^{n}(2^{n} − 1)$ and $S_{3n} − S_{2n} = 2^{2n}(2^{n} − 1)$; each block is $q^{n} = 2^{n}$ times the previous one, so they form a geometric sequence with ratio $2^{n} (not 2)$.","explanationZh":"$S_{n} = 2^{n} − 1，S_{2n} − S_{n} = 2^{n}(2^{n} − 1)，S_{3n} − S_{2n} = 2^{2n}(2^{n} − 1)，每段是前一段的 q^{n} = 2^{n}$ 倍，故公比为 $2^{n}$ 而非 2。","tags":["hard","block-sums","property"],"status":"published"},{"id":"math-geometric-sequence-025","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"A geometric sequence has $a_{1} = 1$ and $q = 3$. The least n with $S_{n} > 100$ is","stemZh":"等比数列 $a_{1} = 1，q = 3，则使 S_{n} > 100$ 成立的最小 n 为","options":[{"id":"A","text":"3"},{"id":"B","text":"4"},{"id":"C","text":"5"},{"id":"D","text":"6"}],"answer":"C","explanation":"$S_{n} = (3^{n} − 1)/2 > 100$ means $3^{n} > 201$. Since $3^{4} = 81$ and $3^{5} = 243$, the least such n is $5 (S_{5} = 121)$.","explanationZh":"$S_{n} = (3^{n} − 1)/2 > 100，即 3^{n} > 201。3^{4} = 81，3^{5} = 243，故最小 n = 5（S_{5} = 121）。$","tags":["hard","logarithm","inequality"],"status":"published"},{"id":"math-geometric-sequence-026","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"A geometric sequence has $a_{1} = 1$ and $q = −2$. Then $S_{6} =$","stemZh":"等比数列 $a_{1} = 1，q = −2，则 S_{6} =$","options":[{"id":"A","text":"21"},{"id":"B","text":"−63"},{"id":"C","text":"63"},{"id":"D","text":"−21"}],"answer":"D","explanation":"$S_{6} = a_{1}(1 − q^{6})/(1 − q) = (1 − 64)/(1 + 2) = −63/3 = −21$. Direct $check: 1 − 2 + 4 − 8 + 16 − 32 = −21. (Using 1 − q = 1\\text{ instead of }3\\text{ gives }−63.)$","explanationZh":"$S_{6} = (1 − (−2)^{6})/(1 − (−2)) = (1 − 64)/3 = −21。直接相加：1 − 2 + 4 − 8 + 16 − 32 = −21。（若把分母误作 1$ 则得 −63。）","tags":["hard","negative-ratio","sign-error"],"status":"published"},{"id":"math-geometric-sequence-027","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"In a geometric sequence with all terms $positive, a_{1}a_{2}a_{3} = 8$ and $a_{7}a_{8}a_{9} = 512$. Then $a_{4}a_{5}a_{6} =$","stemZh":"各项为正的等比数列中 $a_{1}a_{2}a_{3} = 8，a_{7}a_{8}a_{9} = 512，则 a_{4}a_{5}a_{6} =$","options":[{"id":"A","text":"64"},{"id":"B","text":"32"},{"id":"C","text":"128"},{"id":"D","text":"216"}],"answer":"A","explanation":"$a_{1}a_{2}a_{3} = a_{2}^{3} = 8$ gives $a_{2} = 2$, and $a_{7}a_{8}a_{9} = a_{8}^{3} = 512$ gives $a_{8} = 8$. The terms $a_{2}, a_{5}, a_{8}$ are themselves geometric, so $a_{5}^{2} = a_{2}a_{8} = 16$ and $a_{5} = 4$. Hence $a_{4}a_{5}a_{6} = a_{5}^{3} = 64$.","explanationZh":"$a_{1}a_{2}a_{3} = a_{2}^{3} = 8，得 a_{2} = 2；a_{7}a_{8}a_{9} = a_{8}^{3} = 512，得 a_{8} = 8。a_{2}、a_{5}、a_{8} 成等比，a_{5}^{2} = a_{2}a_{8} = 16，a_{5} = 4，故 a_{4}a_{5}a_{6} = a_{5}^{3} = 64。$","tags":["hard","product-property","equidistant-terms"],"status":"published"},{"id":"math-geometric-sequence-028","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":4,"stem":"For a geometric sequence with $a_{1} > 0$, which condition makes it decreasing?","stemZh":"首项 $a_{1} > 0$ 的等比数列，在什么条件下单调递减？","options":[{"id":"A","text":"$q > 1$"},{"id":"B","text":"$0 < q < 1$"},{"id":"C","text":"$q < 0$"},{"id":"D","text":"$q = 1$"}],"answer":"B","explanation":"With $a_{1} > 0: q > 1$ gives an increasing $sequence, 0 < q < 1$ gives a decreasing $one, q = 1$ gives a constant sequence, and $q < 0$ makes the terms alternate in sign (not monotone).","explanationZh":"$a_{1} > 0$ 时：q $> 1 递增，0 < q < 1$ 递减，q $= 1$ 为常数列，q $< 0$ 时符号交替（不是单调）。","tags":["hard","monotonicity","sign-of-q"],"status":"published"},{"id":"math-geometric-sequence-029","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"A geometric sequence has $q = 2$ and $a_{1} + a_{3} + a_{5} = 21$. Then $a_{2} + a_{4} + a_{6} =$","stemZh":"等比数列 $q = 2，且 a_{1} + a_{3} + a_{5} = 21，则 a_{2} + a_{4} + a_{6} =$","options":[{"id":"A","text":"21"},{"id":"B","text":"63"},{"id":"C","text":"42"},{"id":"D","text":"84"}],"answer":"C","explanation":"$a_{2} + a_{4} + a_{6} = q(a_{1} + a_{3} + a_{5}) = 2·21 = 42$, since each even-indexed term is q times the preceding odd-indexed term.","explanationZh":"$a_{2} + a_{4} + a_{6} = q(a_{1} + a_{3} + a_{5}) = 2 \\times 21 = 42，因为每个偶数项等于前一个奇数项乘以$ q。","tags":["hard","sub-sequence","property"],"status":"published"},{"id":"math-geometric-sequence-030","subject":"math","moduleId":"sequences","unitId":"sequences","topicId":"geometric-sequence","type":"single-choice","difficulty":5,"stem":"How many terms of the geometric sequence 1, 2, 4, … must be taken for the sum first to exceed 1000?","stemZh":"等比数列 1, 2, 4, … 至少取多少项时，其和首次超过 1000？","options":[{"id":"A","text":"10"},{"id":"B","text":"9"},{"id":"C","text":"11"},{"id":"D","text":"12"}],"answer":"A","explanation":"$S_{n} = 2^{n} − 1$. Since $S_{9} = 511 < 1000$ and $S_{10} = 1023 > 1000$, ten terms are needed; equivalently $n = \\log_{_{2}}1024 = 10$.","explanationZh":"$S_{n} = 2^{n} − 1。S_{9} = 511 < 1000，S_{10} = 1023 > 1000，故需取 10$ 项（即 $n = \\log_{_{2}}1024 = 10）。$","tags":["hard","logarithm","inequality"],"status":"published"}]

Sequences