A sequence is an ordered collection of numbers arranged in a definite order according to a specific rule. Formally, a sequence can be regarded as a function whose domain is the set of natural numbers $\mathbb{N}$ or a subset of the form $\{1, 2, 3, \dots, k\}$.
Some sequences do not have an obvious explicit algebraic formula for their $n$-th term but are instead generated by a recurrence relation, where subsequent terms are defined using preceding terms.
The Fibonacci sequence is a classic example defined by the recurrence relation:
$$a_1 = a_2 = 1$$ $$a_n = a_{n-2} + a_{n-1} \text{ for } n > 2$$Let $a_1, a_2, a_3, \dots, a_n$ be a given sequence. The algebraic expression representing the sum of these terms:
$$a_1 + a_2 + a_3 + \dots + a_n + \dots$$is called the series associated with the given sequence. A series is finite or infinite depending on whether the underlying sequence is finite or infinite.
Series are compactly represented using the Greek letter $\sum$ (sigma notation):
$$\sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \dots + a_n$$Remark: The term "series" refers to the indicated sum expression, whereas the "sum of a series" refers to the single numerical value resulting from the addition of those terms.
A sequence $a_1, a_2, a_3, \dots, a_n, \dots$ is called a geometric progression (or G.P.) if each term is non-zero and the ratio of any term to its immediately preceding term is a constant throughout the sequence.
Mathematically, for all $k \ge 1$:
$$\frac{a_{k+1}}{a_k} = r \text{ (constant)}$$Where:
The general form of a G.P. is written as:
$$a, ar, ar^2, ar^3, \dots, ar^{n-1}, \dots$$The $n$-th term ($a_n$) of a geometric progression is given by the formula:
$$a_n = a r^{n-1}$$This is also referred to as the general term of the G.P.
Let $S_n$ denote the sum of the first $n$ terms of a G.P. with first term $a$ and common ratio $r$. The sum is evaluated under two distinct cases:
| Condition on Common Ratio ($r$) | Sum Formula ($S_n$) | Derivation Methodology |
|---|---|---|
| $r = 1$ | $S_n = na$ | Direct summation of $n$ identical terms: $a + a + \dots + a$. |
| $r \neq 1$ | $S_n = \frac{a(r^n - 1)}{r - 1} = \frac{a(1 - r^n)}{1 - r}$ | Obtained by subtracting $r S_n$ from $S_n$ to eliminate intermediate terms. |
The geometric mean (G.M.) of two positive real numbers $a$ and $b$ is defined as the positive square root of their product:
$$G = \sqrt{ab}$$If $G$ is the geometric mean of $a$ and $b$, then the sequence $a, G, b$ forms a geometric progression because:
$$\frac{G}{a} = \frac{b}{G} \implies G^2 = ab$$Given any two positive numbers $a$ and $b$, we can insert $n$ numbers $G_1, G_2, \dots, G_n$ between them such that the resulting sequence $a, G_1, G_2, \dots, G_n, b$ is a geometric progression.
Since $b$ is the $(n+2)$-th term of this G.P., we have:
$$b = a r^{n+1} \implies r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$$The inserted geometric means are calculated as:
Let $A$ and $G$ be the Arithmetic Mean and Geometric Mean of two positive real numbers $a$ and $b$, respectively:
$$A = \frac{a + b}{2}, \quad G = \sqrt{ab}$$Theorem: For any two positive real numbers $a$ and $b$, the Arithmetic Mean is always greater than or equal to the Geometric Mean:
$$A \ge G$$Proof:
$$A - G = \frac{a + b}{2} - \sqrt{ab} = \frac{a + b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2}$$Since the square of any real number is non-negative, we have:
$$\frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0 \implies A - G \ge 0 \implies A \ge G$$Equality holds if and only if $a = b$.
Mathematical Problem Statement: The sum of the first three terms of a G.P. is $\frac{13}{12}$ and their product is $-1$. Find the common ratio and the terms.
Step-by-Step Algorithmic Solution Trace:
- Let the first three terms of the G.P. be $\frac{a}{r}$, $a$, and $ar$. This choice simplifies the product calculation.
- Set up the product equation:$$\left(\frac{a}{r}\right)(a)(ar) = -1 \implies a^3 = -1$$Taking the real root, we find $a = -1$.
- Set up the sum equation:$$\frac{a}{r} + a + ar = \frac{13}{12}$$Substitute $a = -1$:$$-\frac{1}{r} - 1 - r = \frac{13}{12} \implies -\left(\frac{1 + r + r^2}{r}\right) = \frac{13}{12}$$
- Clear fractions to form a quadratic equation:$$-12(1 + r + r^2) = 13r \implies 12r^2 + 25r + 12 = 0$$
- Solve the quadratic equation using factorization:$$12r^2 + 16r + 9r + 12 = 0 \implies 4r(3r + 4) + 3(3r + 4) = 0 \implies (4r + 3)(3r + 4) = 0$$Thus, $r = -\frac{3}{4}$ or $r = -\frac{4}{3}$.
- Calculate the terms for each value of $r$:
- For $r = -\frac{3}{4}$ and $a = -1$, the terms are: $\frac{4}{3}, -1, \frac{3}{4}$.
- For $r = -\frac{4}{3}$ and $a = -1$, the terms are: $\frac{3}{4}, -1, \frac{4}{3}$.
Final Evaluation Output: The common ratio is $r = -\frac{3}{4}$ or $r = -\frac{4}{3}$, and the three terms are $\frac{4}{3}, -1, \frac{3}{4}$ (or vice versa).
Mathematical Problem Statement: Find the sum of the sequence $7, 77, 777, 7777, \dots$ to $n$ terms.
Step-by-Step Algorithmic Solution Trace:
- Write the sum expression:$$S_n = 7 + 77 + 777 + 7777 + \dots \text{ to } n \text{ terms}$$
- Factor out $7$ to obtain a series of $1$s:$$S_n = 7(1 + 11 + 111 + 1111 + \dots \text{ to } n \text{ terms})$$
- Multiply and divide by $9$ to introduce powers of $10$:$$S_n = \frac{7}{9}(9 + 99 + 999 + 9999 + \dots \text{ to } n \text{ terms})$$
- Express each term as a power of $10$ minus $1$:$$S_n = \frac{7}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$$
- Regroup the terms into a geometric progression and a constant sum:$$S_n = \frac{7}{9}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + 1 \text{ up to } n \text{ terms})]$$
- Apply the G.P. sum formula to the first part ($a = 10, r = 10$):$$S_n = \frac{7}{9}\left[\frac{10(10^n - 1)}{10 - 1} - n\right] = \frac{7}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$$
Final Evaluation Output: $S_n = \frac{70}{81}(10^n - 1) - \frac{7}{9}n$
Mathematical Problem Statement: If $a, b, c, d$ and $p$ are different real numbers such that $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$, show that $a, b, c, d$ are in G.P.
Step-by-Step Algorithmic Solution Trace:
- Expand and regroup the given inequality:$$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \le 0$$
- Express each grouped trinomial as a perfect square:$$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$$
- Since the sum of squares of real numbers is always non-negative, the only way the sum can be less than or equal to $0$ is if each individual square is exactly $0$:$$(ap - b)^2 = 0, \quad (bp - c)^2 = 0, \quad (cp - d)^2 = 0$$
- Solve for $p$ in each equation:$$ap - b = 0 \implies p = \frac{b}{a}$$$$bp - c = 0 \implies p = \frac{c}{b}$$$$cp - d = 0 \implies p = \frac{d}{c}$$
- Equate the expressions for $p$:$$\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$$
Final Evaluation Output: Since the consecutive ratios are equal ($\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$), the numbers $a, b, c, d$ form a geometric progression (G.P.).
Historical Chronicle: Evolution of Sequences and Series