Arithmetic Series#
Arithmetic Sequences Revision#
Theory#
Before we dive into arithmetic series, let’s quickly refresh our understanding of arithmetic sequences. Remember, an arithmetic sequence is a pattern where each term increases (or decreases) by the same amount - what we call the common difference, \(d\).
For any arithmetic sequence with first term \(a\) and common difference \(d\):
General term: \(T_n = a + (n-1)d\)
The sequence looks like: \(a, a+d, a+2d, a+3d, ...\)
Application#
Examples#
Example: Reviewing Sequence Patterns#
Let’s identify the pattern in the sequence: 3, 7, 11, 15, 19, …
\(T_1 = 3, \quad T_2 = 7, \quad T_3 = 11\)
\(d = 7 - 3 = 4 \quad \text{(common difference)}\)
\(T_n = 3 + (n-1) \times 4 = 3 + 4n - 4 = 4n - 1\)
Interactive Visualization: Arithmetic Sequence Visualizer#
Multiple Choice Questions#
Arithmetic Series#
Theory#
Now here’s where things get really interesting! An arithmetic series is what we get when we add up the terms of an arithmetic sequence. It’s like taking all those individual stepping stones and asking, “What’s the total distance we’ve covered?”
The Big Question: If we have an arithmetic sequence, how can we find the sum of the first \(n\) terms without adding them all individually?
Let’s explore this step by step. Imagine you’re adding the first \(n\) terms of an arithmetic sequence:
Here’s a brilliant insight discovered by young Gauss: What happens if we write this sum backwards and add it to itself?
Adding these two expressions term by term: $\(2S_n = [a + (a+(n-1)d)] + [(a+d) + (a+(n-2)d)] + ... + [(a+(n-1)d) + a]\)$
Notice how each bracket simplifies to the same value! Each bracket equals \(2a + (n-1)d\), and we have \(n\) such brackets:
Therefore: $\(S_n = \frac{n}{2}[2a + (n-1)d]\)$
Alternative Form: Since the last term \(l = a + (n-1)d\), we can also write: $\(S_n = \frac{n}{2}(a + l)\)$
This beautifully shows that the sum equals the average of first and last terms, multiplied by the number of terms!
Sigma Notation: We can express arithmetic series using sigma notation: $\(S_n = \sum_{r=1}^{n} [a + (r-1)d]\)$
Key Properties:
The sum grows quadratically with \(n\) (it’s a quadratic function in \(n\))
When graphed against \(n\), \(S_n\) forms a parabola
The formula works for both increasing and decreasing sequences
Special case: Sum of first \(n\) natural numbers = \(\frac{n(n+1)}{2}\)
Interactive Visualization: Arithmetic Series Explorer#
Application#
Examples#
Example 1: Basic Series Calculation#
Let’s find the sum of the first 20 terms of the arithmetic sequence: 3, 7, 11, 15, …
Method 1: Using the standard formula
First, let’s identify what we know:
First term: \(a = 3\)
Common difference: \(d = 7 - 3 = 4\)
Number of terms: \(n = 20\)
\(S_{20} = \frac{n}{2}[2a + (n-1)d] \quad \text{(applying our formula)}\)
\(S_{20} = \frac{20}{2}[2(3) + (20-1)(4)] \quad \text{(substituting values)}\)
\(S_{20} = 10[6 + 19 \times 4] \quad \text{(simplifying)}\)
\(S_{20} = 10[6 + 76] = 10 \times 82 = 820\)
Method 2: Using first and last terms
Let’s find the last term first: \(T_{20} = a + (n-1)d = 3 + 19 \times 4 = 3 + 76 = 79\)
\(S_{20} = \frac{n}{2}(a + l) \quad \text{(using alternative formula)}\)
\(S_{20} = \frac{20}{2}(3 + 79) = 10 \times 82 = 820 \quad \text{(same answer!)}\)
Example 2: Finding Missing Information#
The sum of the first 15 terms of an arithmetic series is 555. If the first term is 7, find the common difference.
Let’s work through this systematically:
Given information:
\(S_{15} = 555\)
\(a = 7\)
\(n = 15\)
\(d = ?\)
\(S_n = \frac{n}{2}[2a + (n-1)d] \quad \text{(starting with our formula)}\)
\(555 = \frac{15}{2}[2(7) + (15-1)d] \quad \text{(substituting known values)}\)
\(555 = \frac{15}{2}[14 + 14d] \quad \text{(simplifying)}\)
\(555 = \frac{15}{2} \times 14(1 + d) \quad \text{(factoring out 14)}\)
\(555 = 105(1 + d) \quad \text{(calculating)}\)
\(\frac{555}{105} = 1 + d \quad \text{(dividing both sides by 105)}\)
\(5.286 = 1 + d\)
\(d = 4.286 \approx 4.29 \quad \text{(to 2 decimal places)}\)
Example 3: Real-World Application - Stadium Seating#
A stadium has 25 rows of seats. The first row has 20 seats, and each subsequent row has 3 more seats than the previous row. What’s the total seating capacity?
This is a perfect arithmetic series problem! Let’s identify the components:
First term (first row): \(a = 20\)
Common difference: \(d = 3\)
Number of terms (rows): \(n = 25\)
Method 1: Direct calculation
\(S_{25} = \frac{n}{2}[2a + (n-1)d]\)
\(S_{25} = \frac{25}{2}[2(20) + (25-1)(3)]\)
\(S_{25} = \frac{25}{2}[40 + 24 \times 3]\)
\(S_{25} = \frac{25}{2}[40 + 72] = \frac{25}{2} \times 112 = 25 \times 56 = 1400\)
Method 2: Understanding the pattern
Let’s think about what’s happening:
Row 1: 20 seats
Row 2: 23 seats
Row 3: 26 seats
…
Row 25: \(20 + 24 \times 3 = 92\) seats
Using first and last terms: \(S_{25} = \frac{25}{2}(20 + 92) = \frac{25}{2} \times 112 = 1400\) seats
The stadium has a total capacity of 1,400 seats!
Multiple Choice Questions#
Sector Specific Questions: Arithmetic Series Applications#
Key Takeaways#
Important
Arithmetic series is the sum of terms in an arithmetic sequence
Two key formulas: \(S_n = \frac{n}{2}[2a + (n-1)d]\) or \(S_n = \frac{n}{2}(a + l)\)
Gauss’s insight: Pairing terms from opposite ends gives constant sums
Growth pattern: The sum grows quadratically with the number of terms
Real-world applications: From physics (motion) to finance (savings) to engineering (structures)
Connection to other topics: Sigma notation, quadratic functions, and later, calculus integration
Problem-solving tip: Choose the formula based on what information you have
Special case to remember: Sum of first n natural numbers = \(\frac{n(n+1)}{2}\)