7.1 Arithmetic Sequences and Sums
Key Takeaways
- An arithmetic sequence (a_n) is defined by a_{n+1} - a_n = d for all n >= 1, yielding the general term a_n = a_1 + (n-1)d.
- The term index property states that if m + n = p + q, then a_m + a_n = a_p + a_q, which provides rapid evaluation paths in Gaokao problems.
- The sum of the first n terms S_n = n(a_1 + a_n)/2 = (d/2)n^2 + (a_1 - d/2)n is a quadratic discrete function with zero constant term.
- Extremum determination of S_n can be analyzed by inspecting the continuous discrete quadratic vertex or finding the term index n where terms transition sign.
- Block partial sums S_k, S_{2k}-S_k, S_{3k}-S_{2k}, ... form another arithmetic sequence with common difference k^2 d.
7.1 Arithmetic Sequences and Sums
1. Mathematical Definition and Recursive Structure
An arithmetic sequence (等差数列) is a sequence of real numbers $(a_n) = (a_1, a_2, a_3, \dots)$ in which the difference between any two consecutive terms is constant. Formally, a sequence $(a_n)$ is an arithmetic sequence if and only if there exists a constant real number $d \in \mathbb{R}$, called the common difference (公差), such that:
Equivalently, the sequence satisfies the three-term relation:
General Term Derivation
To derive the general term $a_n$ from the first term $a_1$ and common difference $d$, we employ the difference accumulation method (累加法). Writing out the definition for successive indices:
a_2 - a_1 &= d \\ a_3 - a_2 &= d \\ a_4 - a_3 &= d \\ &\vdots \\ a_n - a_{n-1} &= d \end{aligned}$$ Summing these $n-1$ equations telescopes the left side: $$(a_2 - a_1) + (a_3 - a_2) + \dots + (a_n - a_{n-1}) = \sum_{k=1}^{n-1} d$$ $$a_n - a_1 = (n-1)d \implies a_n = a_1 + (n-1)d$$ ## 2. Discrete Linear Function Interpretation Re-arranging the general term formula yields: $$a_n = d \cdot n + (a_1 - d)$$ When viewed through the lens of continuous functions, $a_n = f(n)$ where $f(x) = dx + (a_1 - d)$ is a linear function defined on the discrete domain $\mathbb{N}^*$. | Function Property | Continuous Linear Function $f(x) = kx + b$ | Arithmetic Sequence $a_n = dn + (a_1 - d)$ | | :--- | :--- | :--- | | **Domain** | $\mathbb{R}$ | $\mathbb{N}^* = \{1, 2, 3, \dots\}$ | | **Slope / Growth** | $k$ (Slope) | $d$ (Common Difference) | | **Monotonicity** | Strictly increasing if $k>0$, decreasing if $k<0$, constant if $k=0$ | Strictly increasing if $d>0$, decreasing if $d<0$, constant if $d=0$ | | **Geometric Representation** | Continuous straight line in $\mathbb{R}^2$ | Set of collinear discrete points $(n, a_n)$ | ## 3. Core Term Index Properties One of the most powerful tools for solving Gaokao arithmetic sequence problems is the **term index summation property**. ### Theorem (Index Sum Equivalence) If $(a_n)$ is an arithmetic sequence and $m, n, p, q \in \mathbb{N}^*$ satisfy $m + n = p + q$, then: $$a_m + a_n = a_p + a_q$$ *Proof*: Express each term in terms of $a_1$ and $d$: $$a_m + a_n = [a_1 + (m-1)d] + [a_1 + (n-1)d] = 2a_1 + (m+n-2)d$$ $$a_p + a_q = [a_1 + (p-1)d] + [a_1 + (q-1)d] = 2a_1 + (p+q-2)d$$ Since $m + n = p + q$, the two expressions are identically equal. $\blacksquare$ ### Corollary: Arithmetic Mean Property For any term $a_k$ ($k \ge 2$), setting $m = k-1$ and $n = k+1$ gives $(k-1) + (k+1) = 2k$, so: $$a_{k-1} + a_{k+1} = 2 a_k \implies a_k = \frac{a_{k-1} + a_{k+1}}{2}$$ Similarly, for an odd number of terms $2k-1$, the sum of terms symmetric about the middle term $a_k$ satisfies: $$a_1 + a_{2k-1} = a_2 + a_{2k-2} = \dots = 2 a_k$$4. Partial Sum Formulas ($S_n$)
The sum of the first $n$ terms of a sequence is denoted by $S_n = \sum_{k=1}^n a_k = a_1 + a_2 + \dots + a_n$.
Gauss Reverse Pairing Derivation
To derive $S_n$, write $S_n$ in forward and reverse order:
S_n &= a_1 + a_2 + a_3 + \dots + a_{n-1} + a_n \\ S_n &= a_n + a_{n-1} + a_{n-2} + \dots + a_2 + a_1 \end{aligned}$$ Adding these two equations column by column: $$2 S_n = (a_1 + a_n) + (a_2 + a_{n-1}) + (a_3 + a_{n-2}) + \dots + (a_n + a_1)$$ By the term index summation property, $a_1 + a_n = a_2 + a_{n-1} = a_3 + a_{n-2} = \dots$. Thus, all $n$ paired sums are equal to $a_1 + a_n$: $$2 S_n = n(a_1 + a_n) \implies S_n = \frac{n(a_1 + a_n)}{2}$$ Substituting $a_n = a_1 + (n-1)d$ yields the secondary quadratic sum formula: $$S_n = \frac{n[2a_1 + (n-1)d]}{2} = n a_1 + \frac{n(n-1)}{2} d = \frac{d}{2} n^2 + \left(a_1 - \frac{d}{2}\right) n$$ ### Quadratic Functional Structure of $S_n$ Let $A = \frac{d}{2}$ and $B = a_1 - \frac{d}{2}$. Then $S_n = A n^2 + B n$. > **Key Gaokao Insight**: A sequence $(a_n)$ is an arithmetic sequence if and only if its partial sum $S_n$ can be expressed as a quadratic function of $n$ with **zero constant term** ($S_n = A n^2 + B n$). If a constant term $C \neq 0$ is present ($S_n = A n^2 + B n + C$), the sequence is arithmetic starting only from $n=2$, while $a_1 = S_1 = A + B + C \neq A + B$. ## 5. Partial Block Sum Property For any positive integer $k$, if we partition the terms of an arithmetic sequence $(a_n)$ into consecutive blocks of size $k$ and sum each block: $$T_1 = S_k = a_1 + a_2 + \dots + a_k$$ $$T_2 = S_{2k} - S_k = a_{k+1} + a_{k+2} + \dots + a_{2k}$$ $$T_3 = S_{3k} - S_{2k} = a_{2k+1} + a_{2k+2} + \dots + a_{3k}$$ Then the sequence of block sums $(T_1, T_2, T_3, \dots)$ forms another **arithmetic sequence** with common difference $D = k^2 d$. *Proof*: Note that $a_{k+i} - a_i = k d$ for each $i \in \{1, 2, \dots, k\}$. Summing over $i=1 \dots k$: $$T_2 - T_1 = \sum_{i=1}^k (a_{k+i} - a_i) = \sum_{i=1}^k k d = k \cdot (k d) = k^2 d. \quad \blacksquare$$ ## 6. Extremum of Partial Sum $S_n$ When $d \neq 0$, $S_n = \frac{d}{2} n^2 + \left(a_1 - \frac{d}{2}\right) n$ is a quadratic function of discrete integer variable $n$. ### Determination Methods for Maximum / Minimum of $S_n$: 1. **Term Sign Analysis Method (项正负分析法)**: - If $a_1 > 0$ and $d < 0$, terms decrease. $S_n$ reaches its **maximum** value when $n$ is the index of the last non-negative term, i.e., $a_n \ge 0$ and $a_{n+1} \le 0$. - If $a_1 < 0$ and $d > 0$, terms increase. $S_n$ reaches its **minimum** value when $n$ is the index of the last non-positive term, i.e., $a_n \le 0$ and $a_{n+1} \ge 0$. 2. **Symmetry of Quadratic Vertex Method (二次函数对称轴法)**: - Consider continuous function $f(x) = \frac{d}{2} x^2 + \left(a_1 - \frac{d}{2}\right) x$. The axis of symmetry is $x_0 = -\frac{B}{2A} = \frac{1}{2} - \frac{a_1}{d}$. - The integer $n \in \mathbb{N}^*$ closest to $x_0$ maximizes or minimizes $S_n$. If $x_0$ is a half-integer ($k + 0.5$), then $S_k = S_{k+1}$ are dual extrema.Worked Gaokao Exam Examples
Example 1 (Term Index and General Term Determination)
Problem: In an arithmetic sequence $(a_n)$, it is known that $a_3 + a_8 = 22$ and $S_5 = 35$.
- Find the general term formula for $a_n$.
- Find the partial sum formula for $S_n$.
Solution:
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Express the given conditions in terms of $a_1$ and $d$. Using $a_n = a_1 + (n-1)d$: Using $S_5 = \frac{5(a_1 + a_5)}{2} = 5a_1 + 10d = 35$, which simplifies to: Now solve the system of linear equations: Multiplying the second equation by 2 gives $2a_1 + 4d = 14$. Subtracting this from the first equation: Substituting $d = 2$ back into $a_1 + 2d = 7$ yields $a_1 = 3$. Therefore, the general term is:
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The partial sum formula $S_n$ is:
Example 2 (Extremum of Partial Sum $S_n$)
Problem: Let $(a_n)$ be an arithmetic sequence with initial term $a_1 = 25$ and common difference $d = -4$.
- Find the term index $n$ for which the partial sum $S_n$ attains its maximum value.
- Calculate the maximum value of $S_n$.
Solution:
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Method 1: Term Sign Analysis The general term is $a_n = 25 + (n-1)(-4) = 29 - 4n$. We solve $a_n \ge 0$: Since $n \in \mathbb{N}^*$, the terms are positive for $n \le 7$ and negative for $n \ge 8$: Because adding positive terms increases $S_n$ and adding negative terms decreases $S_n$, $S_n$ reaches its maximum at $n = 7$.
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Calculation of Maximum Value: Alternatively, using $S_n = \frac{d}{2}n^2 + (a_1 - \frac{d}{2})n = -2n^2 + 27n$: The maximum value of $S_n$ is $91$ at $n = 7$.
In an arithmetic sequence (a_n), if a_2 + a_5 + a_8 = 18, what is the value of a_1 + a_9?
For an arithmetic sequence (a_n) with a_1 = -11 and common difference d = 2, at which term index n does the partial sum S_n achieve its minimum value?
Let (a_n) be an arithmetic sequence. If the sum of the first 4 terms is S_4 = 12 and the sum of the first 8 terms is S_8 = 40, what is the sum of terms from a_9 to a_12 (i.e., S_12 - S_8)?