Arithmetic Series Study Notes

Key Concepts of Arithmetic Series

• An arithmetic series is the sum of the first n terms of an arithmetic sequence.

• An arithmetic sequence has a first term a_1 and a constant difference d between consecutive terms.

• The nth term of an arithmetic sequence is given by:
  an = a1 + (n-1)\,d

• The sum of the first n terms (arithmetic series) has two common formulas:

  • Using first and last terms:
    $S<em>n = \frac{n}{2}\,(a</em>1 + a_n)$

  • Using first term and common difference:
    $S<em>n = \frac{n}{2}\,[2a</em>1 + (n-1)\,d]$

• Quick pairing insight: summing symmetric pairs (a1 + an), (a2 + a{n-1}), etc., yields a constant pair sum if the sequence is arithmetic; number of pairs is n/2, so $S<em>n = \frac{n}{2}\,(a</em>1 + a_n)$ when n is even.

• The pairing idea also explains the classic 1+2+…+n result: for 1 through 100, there are 50 pairs each summing to 101, giving 5050.

Essential Formulas (Cheat Sheet)

• Arithmetic sequence term:
  $a<em>n = a</em>1 + (n-1)\,d$

• Sum of first n terms (using first and last):
  $S<em>n = \frac{n}{2}\,(a</em>1 + a_n)$

• Sum of first n terms (using first term and common difference):
  $S<em>n = \frac{n}{2}\,[2a</em>1 + (n-1)\,d]$

• Last term in terms of a1 and d: $a</em>n = a_1 + (n-1)\,d$

Quick Method: Pairing (From Page 1 Activity)

• Sum of the first 100 counting numbers by pairing: (1+100), (2+99), …, (50+51)

• Each pair sums to 101; there are 50 pairs.

• Total sum:
  $S_{100} = 50 \times 101 = 5050$

• Implication: The pairing method leads to the same result as $S<em>n = \frac{n}{2}\,(a</em>1 + a_n)$ when n is even.

Worked Examples

Example 1: Sum of the first 50 positive odd numbers

• Given: a_1 = 1, d = 2, n = 50

• Last term: $a<em>n = a</em>1 + (n-1)d = 1 + 49\cdot 2 = 99$

• Using last term formula:
  $S<em>{50} = \frac{50}{2}\,(a</em>1 + a_{50}) = 25\,(1 + 99) = 25\times 100 = 2500$

• Using first-term/diff form:
  $S<em>{50} = \frac{50}{2}\,[2a</em>1 + (n-1)d] = 25\,[2\cdot 1 + 49\cdot 2] = 25\,[2 + 98] = 2500$

Example 2: Sum of the first 100 multiples of 3

• Given: a_1 = 3, d = 3, n = 100

• Last term: $a<em>n = a</em>1 + (n-1)d = 3 + 99\cdot 3 = 300$

• Sum:
  $S<em>{100} = \frac{100}{2}\,(a</em>1 + a_{100}) = 50\,(3 + 300) = 50\times 303 = 15\,150$

• Alternate check:
  $S<em>{100} = \frac{n}{2}\,[2a</em>1 + (n-1)d] = 50\,[2\cdot 3 + 99\cdot 3] = 50\,[6 + 297] = 15\,150$

Example 3: Find a_{50} using the formula and sum

• Given: a_1 = 1, n = 50, d = 2

• a{50} = a1 + (50-1)d = 1 + 49\cdot 2 = 99

• Sum:
  $S<em>{50} = \frac{50}{2}\,(a</em>1 + a_{50}) = 25\,(1 + 99) = 2500$

Example 4: Sum of the first 100 multiples of 3 (alternate method)

• a1 = 3, an = 99, n = 33 (since 3,6,…,99)

• Sum:
  $S<em>{33} = \frac{33}{2}\,(a</em>1 + a_n) = \frac{33}{2}\,(3 + 99) = \frac{33}{2}\cdot 102 = 1683$

• Also verify a{33} = a1 + (33-1)d = 3 + 32\cdot 3 = 99

Practice Problems and Solutions (Selected)

1) Find the sum of the 1st 100 counting numbers

• a1 = 1, an = 100, n = 100

• $S_{100} = \frac{100}{2}\,(1 + 100) = 50\cdot 101 = 5050$

2) Find the sum of the 1st 100 even numbers

• a1 = 2, an = 200, n = 100

• $S_{100} = \frac{100}{2}\,(2 + 200) = 50\cdot 202 = 10100$

3) Find the sum of the 1st 90 odd numbers

• a1 = 1, a{90} = 179, n = 90

• $S_{90} = \frac{90}{2}\,(1 + 179) = 45\cdot 180 = 8100$

4) Find the sum of the multiples of 3 between 1 and 100

• a1 = 3, an = 99, n = 33

• $S_{33} = \frac{33}{2}\,(3 + 99) = \frac{33}{2}\cdot 102 = 1683$

5) Find the sum of the multiples of 8 between 1 and 100

• a1 = 8, an = 96, n = 12

• $S_{12} = \frac{12}{2}\,(8 + 96) = 6\cdot 104 = 624$

Problem Solving (Page 18) – Worked Solutions

Problem 1

• Scenario: A starting gift of P3000 on the 10th birthday, increasing by 200 each year, through the 20th birthday.

• Terms: a_1 = 3000, d = 200, n = 11 (years 10 through 20 inclusive)

• Last term: a{11} = a1 + (11-1)d = 3000 + 10\cdot 200 = 5000

• Sum:
  $S<em>{11} = \frac{11}{2}\,(a</em>1 + a_{11}) = \frac{11}{2}\,(3000 + 5000) = \frac{11}{2}\cdot 8000 = 44{,}000$

• Answer: P44,000 total by the 20th birthday

Problem 2

• Scenario: Starting monthly salary of P30{,}000, with an annual increase of P500. What is the total salary over 10 years?

• Interpretation A (annual salary model):

  • Annual salaries form an arithmetic sequence with a_1 = 360{,}000 (12×30,000) and d = 500 per year; n = 10

  • a_{10} = 360{,}000 + 9\cdot 500 = 364{,}500

  • Sum:
    $S_{10} = \frac{10}{2}\,(360{,}000 + 364{,}500) = 5 \cdot 724{,}500 = 3{,}622{,}500$

• Interpretation B (monthly payments with annual increase):

  • Monthly salary in year i: m_i = 30{,}000 + (i-1)\cdot 500

  • Total over 10 years (12 months per year):
    $S = 12 \, \sum<em>{i=1}^{10} \left[ 30{,}000 + (i-1)\cdot 500 \right]$ = 12 \left[ 10\cdot 30{,}000 + 500 \sum{i=0}^{9} i \right] = 12 \left[ 300{,}000 + 500 \cdot 45
    ight] = 12 \cdot 322{,}500 = 3{,}870{,}000

• Summary:

  • If the problem intends annual salaries only, total = P3{,}622{,}500
    n - If the problem intends monthly payments across 10 years, total = P3{,}870{,}000

• Note: Concrete interpretation depends on whether the increase applies to yearly salary or monthly pay within each year.

Connections to Principles and Real-World Relevance

• Arithmetic series models are common in budgeting, payroll planning, loan amortization, and cumulative totals with steady growth.

• Pairing method illustrates a quick mental check and helps verify formulas like Sn = (n/2)(a1 + a_n).

• Understanding an helps compute sums when the last term is known or needed for the Sn formula.

• Real-world caveat: Salary steps may be irregular or non-arithmetic; this material uses idealized arithmetic progressions for practice and modelling.

Summary of Practice Results (Quick Reference)

• Sum of first 100 counting numbers: $S_{100} = \frac{100}{2}\,(1+100) = 5050$

• Sum of first 100 even numbers: $S_{100} = \frac{100}{2}\,(2+200) = 10100$

• Sum of first 90 odd numbers: $S_{90} = \frac{90}{2}\,(1+179) = 8100$

• Sum of multiples of 3 between 1 and 100: $S_{33} = \frac{33}{2}\,(3+99) = 1683$

• Sum of multiples of 8 between 1 and 100: $S_{12} = \frac{12}{2}\,(8+96) = 624$

Final Check: Key Takeaways

• Always identify a1, d, and n; compute an when needed and use the appropriate S_n formula.

• For sums with known last term an, use $S</em>n = \frac{n}{2}\,(a<em>1 + a</em>n)$.

• For sums where an is not readily known, compute an from $a<em>n = a</em>1 + (n-1)d$ then apply the standard sum formulas.

• In problems involving money or counts over time, clarify whether the values refer to monthly, yearly, or per-period sums to choose the correct interpretation of n, a_1, and d.