Arithmetic progressions are the backbone of many mathematical sequences and series. They're simple yet powerful, with a constant difference between terms that unlocks a world of patterns and calculations. From finding specific terms to summing series, APs are essential tools in your mathematical toolkit.

Understanding APs isn't just about memorizing formulas. It's about recognizing patterns, solving problems, and connecting concepts. Whether you're working with basic sequences or tackling complex combinatorics, mastering APs will give you a solid foundation for deeper mathematical exploration.

Arithmetic progressions

Definition and characteristics

• An arithmetic progression (AP) is a sequence of numbers where the difference between consecutive terms remains constant
  • This constant difference is called the common difference, denoted by $d$
  • Example: 2, 5, 8, 11, 14 is an AP with a common difference of 3
• The general term of an AP is given by the formula $a_n = a_1 + (n-1)d$
  • $a_n$ represents the $n$th term
  • $a_1$ is the first term
  • $n$ is the term number
  • Example: In the AP 3, 7, 11, 15, 19, the 4th term ($a_4$) is $3 + (4-1)2 = 15$
• The sum of the first $n$ terms of an AP is calculated using the formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$
  • $S_n$ is the sum of the first $n$ terms
  • Example: The sum of the first 5 terms of the AP 1, 4, 7, 10, 13 is $\frac{5}{2}(2(1) + (5-1)3) = 35$
• An AP can be increasing ($d > 0$), decreasing ($d < 0$), or constant ($d = 0$)
  • Example: 1, 3, 5, 7, 9 is an increasing AP; 10, 8, 6, 4, 2 is a decreasing AP; 4, 4, 4, 4, 4 is a constant AP
• The arithmetic mean (AM) between two numbers $a$ and $b$ is defined as $\frac{a + b}{2}$, which is the middle term of the AP $a$, AM, $b$
  • Example: The arithmetic mean of 3 and 9 is $\frac{3 + 9}{2} = 6$, forming the AP 3, 6, 9

Properties and proofs

• The sum of any two terms equidistant from the ends of a finite AP is constant and equal to the sum of the first and last terms
  • Proof: Let $a_1$, $a_2$, ..., $a_n$ be an AP. Consider $a_k$ and $a_{n-k+1}$. Using the general term formula, $a_k + a_{n-k+1} = (a_1 + (k-1)d) + (a_1 + (n-k)d) = 2a_1 + (n-1)d = a_1 + a_n$
  • Example: In the AP 2, 5, 8, 11, 14, the sum of the 2nd and 4th terms (5 + 11 = 16) equals the sum of the 1st and 5th terms (2 + 14 = 16)
• The arithmetic mean of two terms in an AP is equal to the arithmetic mean of their positions in the sequence
  • Proof: Let $a_i$ and $a_j$ be two terms in an AP. Their arithmetic mean is $\frac{a_i + a_j}{2} = \frac{(a_1 + (i-1)d) + (a_1 + (j-1)d)}{2} = a_1 + (\frac{i+j}{2} - 1)d = a_{\frac{i+j}{2}}$
  • Example: In the AP 1, 4, 7, 10, 13, the arithmetic mean of the 2nd and 4th terms (4 and 10) is 7, which is the 3rd term (the arithmetic mean of positions 2 and 4)
• If $a$, $b$, and $c$ are in AP, then $b - a = c - b = d$ (the common difference)
  • Proof: By definition, if $a$, $b$, and $c$ are in AP, then $b - a = c - b = d$
  • Example: In the AP 3, 6, 9, we have 6 - 3 = 9 - 6 = 3 (the common difference)
• If $a$, $b$, $c$, and $d$ are in AP, then $a + d = b + c$
  • Proof: Using the general term formula, $a + d = (a_1 + (1-1)d) + (a_1 + (4-1)d) = 2a_1 + 3d$ and $b + c = (a_1 + (2-1)d) + (a_1 + (3-1)d) = 2a_1 + 3d$
  • Example: In the AP 2, 5, 8, 11, we have 2 + 11 = 5 + 8 = 13
• If $a$, $b$, and $c$ are in AP, then $2b = a + c$
  • Proof: Using the general term formula, $2b = 2(a_1 + (2-1)d) = 2a_1 + 2d$ and $a + c = (a_1 + (1-1)d) + (a_1 + (3-1)d) = 2a_1 + 2d$
  • Example: In the AP 1, 4, 7, we have 2(4) = 1 + 7 = 8

Properties of arithmetic progressions

Applications of general term and sum formulas

• Use the general term formula $a_n = a_1 + (n-1)d$ to find specific terms or the common difference in an AP
  • Example: Find the 10th term of the AP 3, 7, 11, 15, ...
    • Solution: $a_{10} = 3 + (10-1)4 = 3 + 36 = 39$
  • Example: Find the common difference of the AP 2, 5, 8, 11, 14, ...
    • Solution: Using the general term formula, $5 = 2 + (2-1)d$, so $d = 3$
• Apply the sum formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$ to calculate the sum of a specific number of terms in an AP
  • Example: Find the sum of the first 20 terms of the AP 1, 4, 7, 10, ...
    • Solution: $S_{20} = \frac{20}{2}(2(1) + (20-1)3) = 10(2 + 57) = 590$
• Utilize the properties of APs to solve problems involving equidistant terms, arithmetic means, or relationships between terms
  • Example: In the AP 3, x, 15, 21, the sum of the 2nd and 4th terms is 36. Find the value of x.
    • Solution: Using the property that the sum of two terms equidistant from the ends is constant, $x + 21 = 3 + 15$, so $x = 18 - 21 = -3$
  • Example: Find the arithmetic mean of the 5th and 9th terms of the AP 2, 5, 8, 11, ...
    • Solution: Using the property that the arithmetic mean of two terms is equal to the term in the position of their arithmetic mean, the required term is the 7th term. $a_7 = 2 + (7-1)3 = 20$

Combining AP concepts with other additive combinatorics techniques

• Combine AP concepts with the pigeonhole principle or Cauchy-Davenport theorem to solve more complex problems
  • Example: Prove that any set of 10 distinct positive integers contains an AP of length 3.
    • Solution: Consider the residues of the 10 integers modulo 9. By the pigeonhole principle, there must be two integers with the same residue. Let these integers be $a$ and $b$, with $b > a$. Then $b - a$ is divisible by 9. Now consider the integer $b + (b - a)$. This integer is also in the set, as $b + (b - a) \leq b + 9 \leq b + (b - 1) \leq 2b - 1 \leq 20 - 1 = 19$. Therefore, $a$, $b$, and $b + (b - a)$ form an AP of length 3.
• Recognize and apply APs in various contexts, such as sequences, series, or number theory problems
  • Example: Prove that the sum of the first $n$ odd positive integers is equal to $n^2$.
    • Solution: The first $n$ odd positive integers form an AP with $a_1 = 1$ and $d = 2$. Using the sum formula, $S_n = \frac{n}{2}(2(1) + (n-1)2) = n(1 + (n-1)) = n^2$.

Arithmetic progressions in combinatorics

Solving problems using AP properties

• Identify APs within a given set of numbers or a sequence
  • Example: In the set {1, 3, 4, 6, 8, 9, 11}, identify all APs of length 3 or more.
    • Solution: The APs of length 3 or more are {1, 3, 6, 9}, {1, 4, 8}, {3, 6, 9}, and {3, 6, 8, 11}.
• Apply AP properties to solve problems involving counting or combinatorial arguments
  • Example: How many APs of length 5 with integer terms exist such that all terms are between 1 and 100 (inclusive)?
    • Solution: For an AP of length 5, the common difference $d$ must be an integer. The smallest possible first term is 1, and the largest possible first term is 96 (to ensure the last term does not exceed 100). For each first term $a_1$, the common difference must satisfy $a_1 + 4d \leq 100$, or $d \leq \frac{99 - a_1}{4}$. The total number of APs is the sum of the number of possible common differences for each first term: $\sum_{a_1=1}^{96} \lfloor \frac{99 - a_1}{4} \rfloor = 1,015$.

Connecting APs with other mathematical concepts

• Explore the connections between APs and other mathematical concepts, such as geometric progressions, Fibonacci numbers, or quadratic equations
  • Example: Prove that the sum of the first $n$ terms of the Fibonacci sequence is one less than the $(n+2)$th Fibonacci number.
    • Solution: Let $F_n$ denote the $n$th Fibonacci number. Consider the sequence $S_n = F_1 + F_2 + ... + F_n$. We can write $S_n = F_1 + F_2 + ... + F_{n-1} + F_n$ and $S_{n-1} = F_1 + F_2 + ... + F_{n-1}$. Subtracting, we get $S_n - S_{n-1} = F_n$. This recursion is similar to the Fibonacci recursion, with the initial values $S_1 = F_1 = 1$ and $S_2 = F_1 + F_2 = 1 + 1 = 2 = F_3 - 1$. By induction, $S_n = F_{n+2} - 1$ for all $n \geq 1$.
• Use APs to derive or prove identities involving sums or series
  • Example: Prove that $\sum_{k=1}^{n} k^3 = (\sum_{k=1}^{n} k)^2$ for all positive integers $n$.
    • Solution: The sum of the first $n$ positive integers is an AP with $a_1 = 1$ and $d = 1$, so $\sum_{k=1}^{n} k = \frac{n}{2}(2(1) + (n-1)1) = \frac{n(n+1)}{2}$. The sum of the first $n$ cubes can be found using the AP $1^3, 2^3, 3^3, ..., n^3$ with $a_1 = 1$ and $d = 7$ (the common difference between consecutive cubes). Thus, $\sum_{k=1}^{n} k^3 = \frac{n}{2}(2(1) + (n-1)7) = \frac{n(n+1)}{2} \cdot \frac{2 + 7(n-1)}{2} = (\frac{n(n+1)}{2})^2 = (\sum_{k=1}^{n} k)^2$.