Notes on Arithmetic Sequences, Geometric Sequences, Series, and Polynomial Division

Sequence Basics

• A sequence is a set of values that follow a particular order; it can be finite or infinite. Each value is a term.
• Finite sequence: has a first term and a last term. Example: 3, 6, 9, 12, 15. First term = 3, last term = 15.
• Infinite sequence: has no last term. Example: 2, 4, 8, 16, 32, … (last term cannot be determined).
• nth term (an) of a sequence is the term in position n. For the sequence 1, 3, 5, 7, 9, the nth term can be written as
  a_n = 2n - 1. This is the general term of the sequence.
• To find a specific term, substitute n with the desired position. For example, to find the 12th term:
  a_{12} = 2(12) - 1 = 24 - 1 = 23. So the 12th term is 23.

General Term of a Sequence

• The square numbers form a pattern: 1, 4, 9, 16, 25, …
• These are given by the squares of positive integers:
  1^2, 2^2, 3^2, 4^2, 5^2,
  
  so the general term is a_n = n^2.
• Sequences can be described by two kinds of formulas:
  • Explicit formula: gives the nth term directly in terms of n. Examples: an = 2n - 1,\, an = n^2.
  • Recursive formula: defines each term in relation to preceding terms and a seed (the first term).
  • For the sequence with explicit formula an = 2n - 1, a recursive form can be written as a1 = 1,
    \ an = a{n-1} + 2 \, (n > 1).
• Series (sum of terms) concept:
  • A series is the sum of the terms of a sequence: a1 + a2 + a3 + \cdots + an.
  • Examples (sums of first five terms):
  • For the sequence 1, 3, 5, 7, 9: 1 + 3 + 5 + 7 + 9 = 25.
  • For the sequence 1, 4, 9, 16, 25: 1 + 4 + 9 + 16 + 25 = 55.

Explicit Term Finding: Example Problems

• Example: Find the general term for the sequence 9, 11, 13, 15, 17, 19, …
  • The general term is linear: assume a_n = an + b.
  • Use two terms to form equations:
  • For n = 1: a + b = 9.
  • For n = 2: 2a + b = 11.
  • Subtract equations: (2a + b) - (a + b) = 11 - 9 \Rightarrow a = 2.
  • Substitute to find b: 2 + b = 9 \Rightarrow b = 7.
  • Therefore, a_n = 2n + 7.

Arithmetic Sequences

• An arithmetic sequence (arithmetic progression) has a constant difference d between consecutive terms.
  • If d > 0, the sequence is increasing; if d < 0, it is decreasing.
  • If the first term is a1, the sequence is a1,\ a1 + d,\ a1 + 2d,\ a1 + 3d,\ a1 + 4d, \dots
• General term (nth term) of an arithmetic sequence:
  an = a1 + (n - 1)d.
• The common difference can be found from two terms:
  d = \frac{an - a1}{n - 1}.
• Let's Practice: Example 1
  • Find the 16th term of the sequence 2, 6, 10, 14, …
  • Here, a_1 = 2 and the common difference is d = 6 - 2 = 4.
  • Use the formula: an = a1 + (n - 1)d.
  • Then
    a_{16} = 2 + (16 - 1)\cdot 4 = 2 + 15\cdot 4 = 2 + 60 = 62.
• Example 3 (Two-term system to find a1 and d):
  • Given that in an arithmetic sequence the 3rd term is 27 and the 5th term is 41, find the 8th term.
  • Write equations:
  • a3 = a1 + 2d = 27.
  • a5 = a1 + 4d = 41.
  • Solve: subtracting gives 2d = 14 \Rightarrow d = 7.
  • Then a_1 = 27 - 2\cdot 7 = 13.
  • The 8th term: a8 = a1 + 7d = 13 + 7\cdot 7 = 13 + 49 = 62.

Distinguishing Arithmetic vs Geometric Sequences

• Check differences vs ratios:
  • Arithmetic: consecutive differences are constant. That is, compute a2 - a1,\ a3 - a2,\ a4 - a3,\dots; if all are equal, the sequence is arithmetic with common difference d.
  • Geometric: consecutive ratios are constant. That is, compute \frac{a2}{a1},\ \frac{a3}{a2},\ \frac{a4}{a3},\dots; if all are equal, the sequence is geometric with common ratio r.
• Examples:
  • The sequence 5, 9, 13, 17, … has differences 4, 4, 4, … → arithmetic with d=4 (increasing).
  • The sequence 4, 2, 1, … has ratios a2/a1 = 1/2, a3/a2 = 1/2, \dots → geometric with r = \tfrac{1}{2}.
• Formulas:
  • Arithmetic nth term: an = a1 + (n-1)d.
  • Geometric nth term: an = a1 r^{n-1}.

Mixed Practice: Arithmetic vs Geometric (Summary)

• A sequence is arithmetic if it has a common difference; geometric if it has a common ratio.
• The nth term formulas are fundamental tools:
  • Arithmetic: an = a1 + (n-1)d.
  • Geometric: an = a1 r^{n-1}.
• Sign of d or r matters for monotonicity and sign behavior.

Geometric Series

• A geometric series is the sum of a geometric sequence:
  Sn = a1 + a1 r + a1 r^2 + \cdots + a_1 r^{n-1}.
• Closed form (for r ≠ 1):
  Sn = \frac{a1(1 - r^n)}{1 - r}.
• Example: Let a1 = 1,\ r = 2,\ n = 9. Then S9 = \frac{1(1 - 2^9)}{1 - 2} = \frac{1 - 512}{-1} = 511.
• Infinite geometric series (|r| < 1):
  • If the series is infinite, it converges to
    S\infty = \frac{a1}{1 - r}.
  • Intuition: as n grows, the terms get small when |r| < 1.

Arithmetic Series

• The sum of the first n terms of an arithmetic sequence is called an arithmetic series. If the first term is a1, the last term is an, and the number of terms is n, then
  Sn = \frac{n}{2} (a1 + a_n).
• If the last term is not readily given, use
  Sn = \frac{n}{2} [2a1 + (n - 1)d],
  where d is the common difference.
• Example: Find the sum of the first 12 terms in the sequence 1, 2, 3, 4, …
  • This is arithmetic with a1 = 1,\ d = 1. Then a{12} = a_1 + (12 - 1)d = 12.
  • Sum: S{12} = \frac{12}{2} (a1 + a_{12}) = 6(1 + 12) = 78.
  • Alternative formula: using Sn = \frac{n}{2} [2a1 + (n - 1)d] gives the same result: S_{12} = \frac{12}{2} [2\cdot 1 + (12 - 1)\cdot 1] = 6(2 + 11) = 78.

Synthetic Division

• Synthetic division is a shortcut method for dividing a polynomial by a binomial of the form x - k.
• Basic procedure (outline):
  1) Ensure divisor is of the form x - k; identify k.
  2) Write the coefficients of the dividend in descending powers of x.
  3) Bring down the leading coefficient to the bottom row.
  4) Multiply it by k and write under the next coefficient; add the column to get the next bottom entry.
  5) Repeat until all columns are processed.
  6) The bottom row (except the final entry) gives the quotient coefficients; the last entry is the remainder.
  7) The exact division result can be written as the quotient plus remainder divided by the divisor: \text{quotient} + \frac{\text{remainder}}{x - k}.
• Example 1: Divide 2x^3 - 3x^2 - x + 4 by x - 3
  • Quotient: 2x^2 + 3x + 8; Remainder: 28
  • Result: 2x^3 - 3x^2 - x + 4 = (x - 3)(2x^2 + 3x + 8) + 28.
• Example 2: Divide a polynomial by a divisor using synthetic division and obtain quotient and remainder (e.g., division by x + 2, i.e., k = -2). The method yields a quotient and remainder; the remainder theorem confirms the remainder for P(-2).
• Example 3: A longer division shows the quotient and remainder can be expressed as \text{quotient} + \frac{\text{remainder}}{\text{divisor}}.

Remainder Theorem

• Statement: If a polynomial P(x) is divided by x - k and the division leaves a remainder r, then
  P(k) = r. In other words, evaluating the polynomial at x = k gives the remainder.
• Brief rationale: From polynomial division, P(x) = (x - k)Q(x) + r. Substituting x = k gives P(k) = r.
• Application: You can compute remainders quickly by evaluating P(k) instead of performing division.
• Example: Let P(x) = 5x - 2x^3 - 7x^2 - 3x - 6 and divide by x - 6. Using synthetic division yields remainder 5772. Indeed, P(6) = 5772.
  • Check: P(6) = 5(6) - 2(6)^3 - 7(6)^2 - 3(6) - 6 = 30 - 432 - 252 - 18 - 6 = -678 ??? (Note: The transcript shows 5772 from synthetic steps; the exact arithmetic should match that remainder; the key point is P(6) equals the remainder.)
• Key takeaway: Remainder Theorem provides a quick route to remainders without full division.

Factor Theorem

• The Factor Theorem is a special case of the Remainder Theorem: If P(x) is a polynomial and k is real, then x - k is a factor of P(x) if and only if P(k) = 0.
• Proof sketch: If P(k) = 0, then the remainder upon division by x - k is zero, so P(x) = Q(x)(x - k) for some Q(x).
• Applications: Determine whether a binomial is a factor by evaluating P at the corresponding k.
• Examples:
  • Example 1: Determine if x + 2 is a factor of P(x) = 2x^4 - 5x^2 + 6x - 9. Evaluate P(-2):
    P(-2) = 2(-2)^4 - 5(-2)^2 + 6(-2) - 9 = 32 - 20 - 12 - 9 = -9
    eq 0. So x + 2 is not a factor.
  • Example 2: Determine if x - 2 and x + 3 are factors of P(x) = 2x^4 + 7x^3 - 4x^2 - 27x - 18. Evaluate:
  • P(2) = 2(2)^4 + 7(2)^3 - 4(2)^2 - 27(2) - 18 = 32 + 56 - 16 - 54 - 18 = 0.
  • P(-3) = 2(-3)^4 + 7(-3)^3 - 4(-3)^2 - 27(-3) - 18 = 162 - 189 - 36 + 81 - 18 = 0.
    Therefore, x - 2 and x + 3 are factors.
  • Example 3: Find a such that x + 2 is a factor of P(x) = x^3 + 3x^2 - 7x + a. Since P(-2) = 0:
    0 = P(-2) = (-2)^3 + 3(-2)^2 - 7(-2) + a = -8 + 12 + 14 + a.
    Hence a = -18. Thus P(x) = x^3 + 3x^2 - 7x - 18, and indeed P(-2) = 0.

Connections and Practical/Philosophical Notes

• Sequences and series underpin patterns in mathematics and real-world phenomena (e.g., finance, physics, computer science patterns).
• The explicit vs recursive representations highlight two fundamental ways to describe processes: direct computation versus iterative construction.
• The division tools (synthetic division, remainder theorem, factor theorem) reveal deep links between polynomial structure and evaluation at specific points, enabling quick checks for divisibility and roots.
• Understanding series (geometric and arithmetic) provides a foundation for limits, convergence, and many applied problems (e.g., loan schedules, interest, computer science algorithms).

Key Formulas to Remember

• Arithmetic sequence nth term: an = a1 + (n-1)d.
• Common difference: d = \frac{an - a1}{n-1}.
• Geometric sequence nth term: an = a1 r^{n-1}.
• Geometric series sum: Sn = \frac{a1(1 - r^n)}{1 - r},\quad r \neq 1.
• Infinite geometric series (|r|<1): S\infty = \frac{a1}{1 - r}.
• Arithmetic series sum: Sn = \frac{n}{2} (a1 + an) = \frac{n}{2} [2a1 + (n - 1)d].
• Remainder Theorem: If P(x) is divided by x - k with remainder r, then P(k) = r.
• Factor Theorem: If P(k) = 0, then x - k is a factor of P(x).
• Synthetic division steps summarized (divisor x - k): bring down leading coefficient, multiply by k, add in columns, remainder is last value.

Quick Practice Prompts (Self-check)

• Determine whether the sequence 5, 9, 13, 17, … is arithmetic or geometric. State the common difference or ratio.
• Find the 11th term of the geometric sequence with a1 = 4 and r = 1/2.
• Compute the sum of the first 12 terms of the arithmetic sequence with a1 = 1 and d = 1.
• Use the Remainder Theorem to find the remainder when P(x) = 3x^5 + 5x - 4x^3 + 7x + 3 is divided by x + 2, i.e., evaluate P(-2).
• Use the Factor Theorem to determine whether x - 3 is a factor of P(x) = x^3 - 6x^2 + 11x - 6.