7.2 Geometric Sequences and Sums
Key Takeaways
- A geometric sequence (a_n) satisfies a_{n+1}/a_n = q with a_1 != 0, q != 0, giving general term a_n = a_1 q^{n-1}.
- If m + n = p + q, the multiplicative index property a_m * a_n = a_p * a_q holds for all terms.
- The partial sum formula yields S_n = n a_1 for q=1, and S_n = a_1(1-q^n)/(1-q) = A q^n - A for q != 1 where the constant and coefficient sum to zero.
- Block partial sums S_k, S_{2k}-S_k, S_{3k}-S_{2k} form a geometric sequence with ratio q^k provided S_k != 0 and q != -1.
- Infinite geometric series converge to S_inf = a_1 / (1-q) if and only if |q| < 1.
7.2 Geometric Sequences and Sums
1. Mathematical Definition and Recursive Structure
A geometric sequence (等比数列) is a sequence of non-zero real numbers $(a_n)$ where the ratio of any term to its immediate predecessor is a constant non-zero real number. Formally, a sequence $(a_n)$ with $a_1 \neq 0$ is a geometric sequence if and only if there exists $q \in \mathbb{R} \setminus {0}$, called the common ratio (公比), such that:
Equivalently, the sequence satisfies the non-zero three-term relation:
General Term Derivation
To derive the general term $a_n$ from $a_1$ and $q$, we employ the product accumulation method (累乘法). Writing successive ratios:
\frac{a_2}{a_1} &= q \\ \frac{a_3}{a_2} &= q \\ &\vdots \\ \frac{a_n}{a_{n-1}} &= q \end{aligned}$$ Multiplying all $n-1$ equations together telescopically cancels intermediate terms: $$\frac{a_2}{a_1} \cdot \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = \prod_{k=1}^{n-1} q \implies \frac{a_n}{a_1} = q^{n-1} \implies a_n = a_1 q^{n-1}$$ ## 2. Discrete Exponential Function Interpretation Writing $a_n = \frac{a_1}{q} \cdot q^n$ demonstrates that a geometric sequence corresponds to a discrete exponential function $f(n) = C \cdot q^n$ defined on $\mathbb{N}^*$. | Common Ratio $q$ | Initial Term $a_1 > 0$ | Initial Term $a_1 < 0$ | | :--- | :--- | :--- | | **$q > 1$** | Strictly increasing, positive terms ($a_n \to +\infty$) | Strictly decreasing, negative terms ($a_n \to -\infty$) | | **$0 < q < 1$** | Strictly decreasing, positive terms ($a_n \to 0^+$) | Strictly increasing, negative terms ($a_n \to 0^-$) | | **$q = 1$** | Constant sequence $a_n = a_1$ | Constant sequence $a_n = a_1$ | | **$q < 0$** | Alternating signs ($+,-,+,-$), non-monotonic | Alternating signs ($-,+,-,+$), non-monotonic | ## 3. Core Term Index Properties ### Theorem (Multiplicative Index Equivalence) If $(a_n)$ is a geometric sequence and $m, n, p, q \in \mathbb{N}^*$ satisfy $m + n = p + q$, then: $$a_m \cdot a_n = a_p \cdot a_q$$ *Proof*: Expressing each term using the general term formula $a_k = a_1 q^{k-1}$: $$a_m \cdot a_n = (a_1 q^{m-1})(a_1 q^{n-1}) = a_1^2 q^{m+n-2}$$ $$a_p \cdot a_q = (a_1 q^{p-1})(a_1 q^{q-1}) = a_1^2 q^{p+q-2}$$ Since $m + n = p + q$, the powers of $q$ are equal, proving identity. $\blacksquare$ ### Corollary: Geometric Mean Property For any term index $k \ge 2$, setting $m = k-1$ and $n = k+1$ yields $(k-1) + (k+1) = 2k$, so: $$a_{k-1} \cdot a_{k+1} = a_k^2 \implies |a_k| = \sqrt{a_{k-1} a_{k+1}}$$ > **Gaokao Warning**: If $a, G, b$ form a geometric sequence, then $G^2 = ab$. Thus $G = \pm \sqrt{ab}$. The geometric mean can be positive or negative depending on the common ratio $q$. Assuming $G$ must be positive is one of the most frequent traps in Gaokao objective questions!4. Partial Sum Formulas ($S_n$)
The partial sum $S_n = a_1 + a_2 + \dots + a_n$ of a geometric sequence requires careful case distinction depending on whether $q = 1$ or $q \neq 1$.
Derivation via Multiplication-Subtraction Method
When $q \neq 1$, write $S_n$ and multiply by $q$:
S_n &= a_1 + a_1 q + a_1 q^2 + \dots + a_1 q^{n-1} \\ q S_n &= a_1 q + a_1 q^2 + a_1 q^3 + \dots + a_1 q^n \end{aligned}$$ Subtracting the second equation from the first cancels all middle terms: $$(1 - q) S_n = a_1 - a_1 q^n = a_1 (1 - q^n)$$ Since $q \neq 1$, we divide by $1 - q$: $$S_n = \frac{a_1 (1 - q^n)}{1 - q} = \frac{a_1 - a_n q}{1 - q} \quad (q \neq 1)$$ If $q = 1$, all terms equal $a_1$, so: $$S_n = n a_1 \quad (q = 1)$$ ### Structural Functional Form of Geometric $S_n$ Rewriting $S_n$ for $q \neq 1$: $$S_n = \frac{a_1}{1-q} - \frac{a_1}{1-q} q^n = A - A q^n = A q^n + B, \quad \text{where } A = -\frac{a_1}{1-q}, B = \frac{a_1}{1-q}$$ > **Key Gaokao Insight**: A sequence $(a_n)$ has partial sum $S_n = A q^n + B$ representing a geometric sequence if and only if $A + B = 0$ (i.e., $B = -A$) and $q \neq 0, q \neq 1$. If $A + B \neq 0$, then $a_1 = S_1 = A q + B$, whereas for $n \ge 2$, $a_n = A(q-1)q^{n-1}$, meaning the geometric pattern begins only at $n=2$. ## 5. Partial Block Sum Property for Geometric Sequences For a geometric sequence $(a_n)$ with sum $S_n$, consider consecutive blocks of length $k$: $$T_1 = S_k = a_1 + a_2 + \dots + a_k$$ $$T_2 = S_{2k} - S_k = a_{k+1} + a_{k+2} + \dots + a_{2k} = q^k (a_1 + a_2 + \dots + a_k) = q^k T_1$$ $$T_3 = S_{3k} - S_{2k} = a_{2k+1} + \dots + a_{3k} = q^k T_2 = q^{2k} T_1$$ Provided $T_1 \neq 0$ and $q^k \neq -1$, the block sums $(T_1, T_2, T_3, \dots)$ form a **geometric sequence** with common ratio $Q = q^k$. ## 6. Infinite Geometric Series Convergence For an infinite geometric series $\sum_{k=1}^\infty a_1 q^{k-1}$, the limit of partial sums as $n \to \infty$ is: $$S_\infty = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{a_1(1-q^n)}{1-q}$$ - If $|q| < 1$, then $\lim_{n \to \infty} q^n = 0$, so the series converges to $S_\infty = \frac{a_1}{1-q}$. - If $|q| \ge 1$, $\lim_{n \to \infty} q^n$ does not exist or is infinite, so the series diverges.Worked Gaokao Exam Examples
Example 1 (Term Determination and Geometric Properties)
Problem: In a geometric sequence $(a_n)$ of positive real numbers, $a_2 a_4 = 16$ and $a_1 + a_3 = 10$.
- Find the general term formula for $a_n$.
- Calculate the sum $S_5$.
Solution:
-
Since all terms are positive, $a_1 > 0$ and $q > 0$. Using the term index property: $a_2 a_4 = a_3^2 = 16$. Because terms are positive, $a_3 = \sqrt{16} = 4$. We are given $a_1 + a_3 = 10$, so $a_1 + 4 = 10 \implies a_1 = 6$. Since $a_3 = a_1 q^2$, we have $6 q^2 = 4 \implies q^2 = \frac{2}{3}$. Since $q > 0$, $q = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$. The general term formula is:
-
To find $S_5$: Computing term by term: Summing terms directly:
Example 2 (Block Sum Property of Geometric Sequences)
Problem: Let $(a_n)$ be a geometric sequence with common ratio $q \neq -1$. If $S_3 = 7$ and $S_6 = 63$, find:
- The common ratio $q$.
- The general term formula $a_n$.
Solution:
-
Let $T_1 = S_3 = 7$, and $T_2 = S_6 - S_3 = 63 - 7 = 56$. By the block partial sum property, $T_1$ and $T_2$ are consecutive blocks of size $k=3$, so: Since $q \in \mathbb{R}$, $q = \sqrt[3]{8} = 2$.
-
Using $S_3 = \frac{a_1(1-q^3)}{1-q} = 7$: Thus, the general term is:
In a geometric sequence (a_n) with positive terms, if a_3 = 2 and a_7 = 18, what is the value of a_5?
If the partial sum of a sequence (a_n) is given by S_n = 3^n + k, for what value of k is the sequence (a_n) a geometric sequence for all n >= 1?
An infinite geometric series has first term a_1 = 12 and infinite sum S_inf = 18. What is the common ratio q?