7.1 Arithmetic Sequences and Sums

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:

$$a_{n+1} - a_n = d, \quad \forall n \in \mathbb{N}^*$$

Equivalently, the sequence satisfies the three-term relation:

$$2 a_n = a_{n-1} + a_{n+1}, \quad \forall n \ge 2, n \in \mathbb{N}^*$$

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:

$$\begin{aligned} a_2 - a_1 \&= d \\ a_3 - a_2 \&= d \\ a_4 - a_3 \&= d \\ \&\vdots \\ a_n - a_{n-1} \&= d \end{aligned}$$

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 $\mathbb{N}^*$.

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:

$$\begin{aligned} S_n \&= a_1 + a_2 + a_3 + \dots + a_{n-1} + a_n \\ S_n \&= a_n + a_{n-1} + a_{n-2} + \dots + a_2 + a_1 \end{aligned}$$

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 $S_5 = 35$.

1. Find the general term formula for $a_n$.
2. Find the partial sum formula for $S_n$.

Solution:

1. Express the given conditions in terms of $a_1$ and $d$. Using $a_n = a_1 + (n-1)d$:

   $$a_3 + a_8 = (a_1 + 2d) + (a_1 + 7d) = 2a_1 + 9d = 22$$

   Using $S_5 = \frac{5(a_1 + a_5)}{2} = 5a_1 + 10d = 35$, which simplifies to:

   $$a_1 + 2d = 7$$

   Now solve the system of linear equations:

   $$\begin{cases} 2a_1 + 9d = 22 \\ a_1 + 2d = 7 \end{cases}$$

   Multiplying the second equation by 2 gives $2a_1 + 4d = 14$. Subtracting this from the first equation:

   $$5d = 8 \implies d = 2$$

   Substituting $d = 2$ back into $a_1 + 2d = 7$ yields $a_1 = 3$. Therefore, the general term is:

   $$a_n = 3 + (n-1) \cdot 2 = 2n + 1$$

2. The partial sum formula $S_n$ is:

   $$S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(3 + 2n + 1)}{2} = \frac{n(2n + 4)}{2} = n^2 + 2n$$

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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 $d = -4$.

1. Find the term index $n$ for which the partial sum $S_n$ attains its maximum value.
2. Calculate the maximum value of $S_n$.

Solution:

1. Method 1: Term Sign Analysis The general term is $a_n = 25 + (n-1)(-4) = 29 - 4n$. We solve $a_n \ge 0$:

   $$29 - 4n \ge 0 \implies 4n \le 29 \implies n \le 7.25$$

   Since $n \in \mathbb{N}^*$, the terms are positive for $n \le 7$ and negative for $n \ge 8$:

   $$a_7 = 29 - 4(7) = 1 > 0$$ $$a_8 = 29 - 4(8) = -3 < 0$$

   Because adding positive terms increases $S_n$ and adding negative terms decreases $S_n$, $S_n$ reaches its maximum at $n = 7$.

2. Calculation of Maximum Value:

   $$S_7 = \frac{7(a_1 + a_7)}{2} = \frac{7(25 + 1)}{2} = \frac{7 \times 26}{2} = 91$$

   Alternatively, using $S_n = \frac{d}{2}n^2 + (a_1 - \frac{d}{2})n = -2n^2 + 27n$:

   $$S_7 = -2(7)^2 + 27(7) = -98 + 189 = 91$$

   The maximum value of $S_n$ is $91$ at $n = 7$.