Sum of Arithmetic Series
Description
An Arithmetic Series is the sum of the terms in an arithmetic sequence. The formula $S_n = \frac{n}{2}(a_1 + a_n)$ offers a shortcut to summing long lists of numbers without adding them one by one.
The intuition behind the formula is simple: you take the average of the first and last terms $\frac{a_1 + a_n}{2}$ and multiply it by the number of terms ($n$). Alternatively, you can think of it as pairing the first and last numbers, the second and second-to-last, and so on. Each pair sums to the same value ($a_1 + a_n$), and there are $n/2$ such pairs.
If you don't know the last term ($a_n$), you can substitute the sequence formula to get $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.
History & Origins
The formula is famously associated with Carl Friedrich Gauss. The Legend of Gauss: As the story goes, when Gauss was a young schoolboy (around age 7 or 10), his teacher asked the class to sum the numbers from 1 to 100 to keep them busy. While other students struggled with manual addition, Gauss instantly wrote "5050" on his slate. He realized that $1+100=101$, $2+99=101$, $3+98=101$, and so on. Since there are 50 such pairs, the sum is $50 \times 101 = 5050$. Aryabhata (476–550 AD): The Indian mathematician Aryabhata also gave this rule in his treatise Aryabhatiya, referring to the sum as the "heap" or "pile" calculation.
Proof by "Reverse and Add"
We write the sum forwards and backwards, then add the two equations together.
Let $S_n = a_1 + (a_1+d) + ... + (a_n-d) + a_n$.
Write it in reverse: $S_n = a_n + (a_n-d) + ... + (a_1+d) + a_1$.
Add the two equations vertically: $2S_n = (a_1+a_n) + (a_1+a_n) + ... + (a_1+a_n)$.
Notice that every vertical pair sums to $(a_1 + a_n)$ because the $+d$ and $-d$ cancel out.
Since there are $n$ terms, we have $n$ copies of $(a_1 + a_n)$.
So, $2S_n = n(a_1 + a_n)$.
Divide by 2: $S_n = \frac{n}{2}(a_1 + a_n)$.
Variables
| Symbol | Meaning |
|---|---|
Sₙ | Sum of the first n terms |
n | Number of terms to sum |
a₁ | First term |
aₙ | nth (Last) term |
d | Common difference (optional form) |
Examples
Basic Calculation
Problem: Sum the integers from 1 to 100.
Solution:
Stack of Pipes
Problem: A stack of pipes has 20 pipes in the bottom row, 19 in the next, and so on, until the top row has 5 pipes. How many pipes are there in total?
Solution: 200 pipes
- Identify sequence: 20, 19, ..., 5.
- Identify parameters: $a_1 = 5$ (top), $a_n = 20$ (bottom).
- Find count ($n$): The sequence goes from 5 to 20. Count = $(20 - 5) + 1 = 16$ rows.
- Use formula: $S_{16} = \frac{16}{2}(5 + 20)$.
- Calculate: $8(25) = 200$.
Salary Savings
Problem: You save $1000 in the first year, and increase your savings by $500 each year. How much have you saved after 10 years?
Solution: $32,500
- Identify: $a_1 = 1000$, $d = 500$, $n = 10$.
- Find last term ($a_{10}$): $a_{10} = 1000 + (10-1)500 = 1000 + 4500 = 5500$.
- Sum formula: $S_{10} = \frac{10}{2}(1000 + 5500)$.
- Calculate: $5(6500) = 32,500$.
Common Mistakes
Confusing n with an
$n$ is the *count* of numbers (e.g., 50 terms). $a_n$ is the *value* of the last number. In the sum 10, 20, 30... $n=3$, but $a_n=30$.
Using it for Geometric Series
This formula only works if the difference between terms is constant (addition). If terms are multiplied (e.g., 2, 4, 8...), you must use the Geometric Series formula.
Real-World Applications
Stadium Seating
Architects often design amphitheaters or stadiums where each row back has more seats than the one in front to accommodate the widening curve. Calculating the total capacity involves summing an arithmetic series where the common difference is the number of extra seats per row.
Computer Science
When analyzing the time complexity of algorithms (Big O), loops often involve sums like $1 + 2 + ... + n$. This sum is $n(n+1)/2$, which is $O(n^2)$. Knowing this series helps engineers predict how slow a program will get as input size grows.
Frequently Asked Questions
What if I don't know the last term?
You can combine the sum formula with the sequence formula ($a_n = a_1 + (n-1)d$). This gives $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.
Does this work for negative numbers?
Yes! The formula works perfectly for decreasing sequences (negative difference) or sequences with negative terms (e.g., -10, -8, -6...).