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)\,dThe sum of the first n terms (arithmetic series) has two common formulas:
Using first and last terms:
Sn = \frac{n}{2}\,(a1 + a_n)Using first term and common difference:
Sn = \frac{n}{2}\,[2a1 + (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 Sn = \frac{n}{2}\,(a1 + 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:
an = a1 + (n-1)\,dSum of first n terms (using first and last):
Sn = \frac{n}{2}\,(a1 + a_n)Sum of first n terms (using first term and common difference):
Sn = \frac{n}{2}\,[2a1 + (n-1)\,d]Last term in terms of a1 and d: an = 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 = 5050Implication: The pairing method leads to the same result as Sn = \frac{n}{2}\,(a1 + 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: an = a1 + (n-1)d = 1 + 49\cdot 2 = 99
Using last term formula:
S{50} = \frac{50}{2}\,(a1 + a_{50}) = 25\,(1 + 99) = 25\times 100 = 2500Using first-term/diff form:
S{50} = \frac{50}{2}\,[2a1 + (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: an = a1 + (n-1)d = 3 + 99\cdot 3 = 300
Sum:
S{100} = \frac{100}{2}\,(a1 + a_{100}) = 50\,(3 + 300) = 50\times 303 = 15\,150Alternate check:
S{100} = \frac{n}{2}\,[2a1 + (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{50} = \frac{50}{2}\,(a1 + 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{33} = \frac{33}{2}\,(a1 + a_n) = \frac{33}{2}\,(3 + 99) = \frac{33}{2}\cdot 102 = 1683Also 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{11} = \frac{11}{2}\,(a1 + a_{11}) = \frac{11}{2}\,(3000 + 5000) = \frac{11}{2}\cdot 8000 = 44{,}000Answer: 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{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 Sn = \frac{n}{2}\,(a1 + an).
For sums where an is not readily known, compute an from an = a1 + (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.