Arithmetic Series

Calculate sums of arithmetic series using formulas. Grade 11 Math 20-1 Alberta.

Lesson 3.2 of Sequences and Series in Math 20-1 — Alberta curriculum lessons.

What you'll learn in this lesson

• What is a Series?. Understand the difference between a sequence and a series, learn partial sum notation using S_n, and meet the two arithmetic series formulas — one for when you know the last term and one for when you know the common difference.
• Using the Formulas. Apply both sum formulas to find S_n for arithmetic series — including examples with fractions, large n values, and series where you must first find n from the given endpoints.
• Working Backwards. Start from given sums and work backwards to find unknowns: the first term, common difference, number of terms, and the full series — including systems of equations from two given sums.
• Reading a Series Graph. Interpret graphs of partial sums (n, S_n), extract individual terms using t_n = S_n minus S_n-1, and use a quick reference table to choose the right approach for any arithmetic series problem.

Inside the lesson — a free preview

Definition

Sequence vs. Series. A sequence is a list of numbers in order. A series is what you get when you add those numbers together.

For the sequence 7, 5, 3, 1, , the corresponding series is:

7 + 5 + 3 + 1 +

Like sequences, series can be finite (stop at some point) or infinite (continue forever).

Notation: S_n

Partial Sum Notation. S_n means the sum of the first n terms of a series. It is called a partial sum because it only adds part of an infinite series.

Examples of what the notation means:

S_3 = sum of the first 3 terms

S_8 = sum of the first 8 terms

S_50 = sum of the first 50 terms

Example: Partial Sums of 7 + 5 + 3 + 1 + . S_1 = 7

S_2 = 7 + 5 = 12

S_3 = 7 + 5 + 3 = 15

S_4 = 7 + 5 + 3 + 1 = 16

Each partial sum includes one more term than the previous.

The Two Sum Formulas

Adding 50 or 100 terms by hand would take forever. Instead, we use formulas. Which one you use depends on what information you have.

Formula 1 — When You Know the First and Last Terms. S_n = (n(t_1 + t_n))/(2)

Use this when you know t_1 (first term), t_n (last term being added), and n (number of terms).

In plain words: average the first and last term, then multiply by how many terms there are.

Formula 2 — When You Know the First Term and Common Difference. S_n = (n[2t_1 + (n-1)d])/(2)

Use this when you know t_1 (first term) and d (common difference) but NOT the last term.

How to Choose. If you know the last term, Formula 1 is faster.

If you only have d but not t_n, use Formula 2. Alternatively, find t_n first using the general term formula t_n = t_1 + d(n-1), then use Formula 1.

Worked example: Formula 2 Examples

These examples show how to apply Formula 2 when you know t_1 and d but not the last term.

Example 1a — Find S_34 for 14 + 20 + 26 + . We know t_1 and d but not the last term, so use Formula 2.

Step 1 — Identify what we know.

t_1 = 14, d = 20 - 14 = 6, n = 34

Step 2 — Plug into Formula 2.

S_n = (n[2t_1 + (n-1)d])/(2)

S_34 = (34[2(14) + (34-1)(6)])/(2)

Step 3 — Simplify inside the brackets.

S_34 = (34[28 + 33(6)])/(2)

S_34 = (34[28 + 198])/(2)

S_34 = (34(226))/(2)

Step 4 — Multiply and divide.

S_34 = (7684)/(2)

Answer: S_34 = 3842

Example 1b — Find S_24 for (1)/(12) + (1)/(2) + (11)/(12) + . Fractions can look intimidating but the process is identical.

Step 1 — Identify what we know.

t_1 = (1)/(12)

To find d, rewrite t_2 with the same denominator: (1)/(2) = (6)/(12).

d = (6)/(12) - (1)/(12) = (5)/(12), n = 24

Step 2 — Plug into Formula 2.

S_24 = (24[2((1)/(12)) + (24-1)((5)/(12))])/(2)

Step 3 — Simplify inside the brackets.

S_24 = (24[(2)/(12) + 23((5)/(12))])/(2)

S_24 = (24[(2)/(12) + (115)/(12)])/(2)

S_24 = (24[(117)/(12)])/(2)

Step 4 — Simplify the fraction.

24 × (117)/(12) = 2 × 117 = 234

S_24 = (234)/(2)

Answer: S_24 = 117

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