Math 10 · Lesson 2

Arithmetic Series.

An arithmetic series is just the running total of an arithmetic sequence, and a clever pairing trick lets you add hundreds of terms with a single line of arithmetic.

By the end, you can

1I can tell the difference between a sequence (the list) and a series (the sum of the list).
2I can derive and use Sₙ = n/2 (t₁ + tₙ) to add the first n terms of an arithmetic sequence.
3I can use Sₙ = n/2 [2t₁ + (n − 1)d] when the last term is not given.
4I can find a missing piece (t₁, d, n, tₙ, or Sₙ) by turning the facts into equations and solving.
5I can model real situations (stacked cans, multiples, repeated schedules) as arithmetic series.

01 / CONCEPTS

The core ideas.

Sequence vs series

An arithmetic sequence is an ordered list of terms t₁, t₂, t₃, …. An arithmetic series is what you get when you add those terms together. The sum of the first n terms is written Sₙ.

S_n = t_1 + t_2 + t_3 + \cdots + t_n

Why it works

The sequence is the list of numbers; the series is the running total. The capital S with a little n means 'the sum, stopping after n terms.' Same numbers, different question: list them, or add them up.

The Gauss trick: reverse and add

Write the sum forwards, then write it again backwards underneath, and add the two lines term by term. Every vertical pair adds to the same value, t₁ + tₙ, and there are n pairs. So twice the sum is n times (t₁ + tₙ).

2S_n = n\,(t_1 + t_n)

Why it works

As a boy, Gauss was told to add 1 + 2 + … + 100. He paired 1 with 100, 2 with 99, 3 with 98, and so on: fifty pairs that each add to 101. Fifty times 101 is 5050, in seconds. Reverse-and-add is that same idea for any arithmetic series.

The sum formula

Divide the reverse-and-add result by two and you get the main formula for an arithmetic series: the number of terms, divided by two, times the sum of the first and last term.

S_n = \frac{n}{2}\,(t_1 + t_n)

Why it works

Read it as the average term times how many terms. (t₁ + tₙ)/2 is the average of the first and last term, and because the terms are evenly spaced that is the average of the whole list. Multiply the average by n and you have the total.

When you do not know the last term

If you are not given tₙ, replace it using the general term tₙ = t₁ + (n − 1)d. That turns the sum formula into a version built only from t₁, d, and n.

S_n = \frac{n}{2}\,[\,2t_1 + (n-1)d\,]

Why it works

It is the very same formula. We just swapped tₙ for t₁ + (n − 1)d so you can use it when the last term is missing but you know the common difference.

A sum needs to know how many

To add a series you must know n, the number of terms. If a problem gives you the value of the last term but not its position, find n first from tₙ = t₁ + (n − 1)d, then sum.

t_n = t_1 + (n-1)d \;\Rightarrow\; n

Why it works

You cannot total a list until you know how long it is. Counting the terms is step zero of almost every series problem, especially the 'add all the multiples between A and B' type.

02 / WORKED EXAMPLES

From easy to tricky.

EX 1easy

Find the sum 5 + 8 + 11 + … + 53.

Identify the parts

t₁ = 5, d = 3, tₙ = 53.

Find how many terms

53 = 5 + (n − 1)(3) ⇒ 48 = 3(n − 1) ⇒ n = 17.

Use the sum formula

S₁₇ = 17/2 (5 + 53) = 17/2 (58) = 17 × 29.

EX 2easy

An arithmetic series has t₁ = 12, d = 8, and n = 9 terms. Find S₉.

Last term is unknown, use the second form

Sₙ = n/2 [2t₁ + (n − 1)d].

Substitute

S₉ = 9/2 [2(12) + (9 − 1)(8)] = 9/2 [24 + 64].

Simplify

S₉ = 9/2 (88) = 9 × 44.

EX 3medium

Young Gauss is told to add 1 + 2 + 3 + … + 100. Do it in one line.

Identify the parts

t₁ = 1, tₙ = 100, n = 100.

Pair first with last

Each pair (1+100), (2+99), … adds to 101, and there are 100/2 = 50 pairs.

Sum formula

S₁₀₀ = 100/2 (1 + 100) = 50 × 101.

EX 4medium

For an arithmetic series, t₁ = 8, the last term is 68, and the sum is 608. How many terms are there?

Use the sum formula

Sₙ = n/2 (t₁ + tₙ) ⇒ 608 = n/2 (8 + 68).

Simplify

608 = n/2 (76) = 38n.

Solve for n

n = 608 / 38 = 16.

EX 5hard

An arithmetic series has d = 6, S₁₄ = 574, and n = 14. Find the first term t₁.

Use the second form

Sₙ = n/2 [2t₁ + (n − 1)d] ⇒ 574 = 14/2 [2t₁ + 13(6)].

Simplify

574 = 7 (2t₁ + 78) = 14t₁ + 546.

Solve for t₁

14t₁ = 28 ⇒ t₁ = 2.

EX 6challenge

A store stacks cans: 1 can in the top row, and each row below has one more can. With 18 rows, how many cans are in the display?

Model it as a series

Rows have 1, 2, 3, …, 18 cans. This is an arithmetic series with t₁ = 1, tₙ = 18, n = 18.

Sum formula

S₁₈ = 18/2 (1 + 18) = 9 × 19.

Total

9 × 19 = 171 cans.

03 / COMMON TRAPS

Where students slip.

Trap

Treating a series like a sequence and just listing terms instead of adding them.

Fix

A series is a sum. Sₙ is one number, the total of the first n terms, not a list. If the question says 'find the sum' or uses Sₙ, you must add, using Sₙ = n/2 (t₁ + tₙ).

Trap

Plugging the value of the last term into the formula where the number of terms n belongs.

Fix

n counts how many terms there are; tₙ is the value of the last one. They are different. For 5 + 8 + … + 53, tₙ = 53 but n = 17. Find n from 53 = 5 + (n − 1)(3) before summing.

Trap

Forgetting that you need n, and trying to sum straight from the first and last values.

Fix

Sₙ = n/2 (t₁ + tₙ) still contains n. If only t₁ and tₙ are given, first find n with the general term, then substitute everything into the sum formula.

Trap

Off-by-one when counting terms in 'multiples between A and B' problems.

Fix

Count carefully with n = (last − first)/d + 1. For multiples of 4 from 4 to 996: n = (996 − 4)/4 + 1 = 249, not 248. The '+1' counts the first term itself.

★ / EASY TO MASTER

Ten worked examples, simple to deeper.

No grades, no pressure. Just ten examples worked out slowly, building your confidence one small step at a time.

05 / QUIZ

Test yourself.

The Test

28 problems across three tiers. Auto-graded on submit. Hints and a full walkthrough on every question. Your work is saved on this device — sign in to keep it across devices.

0/28Answered

How to answerType just the value. Formatting won't trip you up — we ignore spaces, capital letters, commas, and subscript style, so t₁ = 5, d = 3 and t1=5,d=3 both count. For a list of numbers, separate them with commas (e.g. 3, 10, 17).

Core

Core (10)

Find the sum 7 + 14 + 21 + … + 98.

Need a hint?

t₁ = 7, d = 7, tₙ = 98. First find n.
98 = 7 + (n − 1)(7) gives n = 14. Then S = 14/2 (7 + 98).

Find the sum 5 + 8 + 11 + … + 53.

Need a hint?

tₙ = 53 gives n = 17; then S₁₇ = 17/2 (5 + 53).

An arithmetic series has t₁ = 7, tₙ = 79, and n = 8. Find Sₙ.

Need a hint?

Use Sₙ = n/2 (t₁ + tₙ).

An arithmetic series has t₁ = 12, d = 8, and n = 9. Find S₉.

Need a hint?

tₙ unknown, so use Sₙ = n/2 [2t₁ + (n − 1)d].

Add the first 100 counting numbers: 1 + 2 + 3 + … + 100.

Need a hint?

Pair 1 + 100, 2 + 99, … : 50 pairs of 101.

Which expression is an arithmetic series (a sum), not a sequence?

Need a hint?

A series uses plus signs; it adds the terms.

Find S₁₀ for the series 5 + 10 + 15 + …

Need a hint?

t₁ = 5, d = 5, n = 10. Use the second form.

Find S₁₀ for the series 10 + 7 + 4 + …

Need a hint?

d = −3 here.

An arithmetic series has t₁ = −12, tₙ = 51, and n = 10. Find Sₙ.

Need a hint?

Sₙ = n/2 (t₁ + tₙ).

C10

Find the sum 1 + 3 + 5 + … + (S₈), i.e. the first 8 terms of 1 + 3 + 5 + …

Need a hint?

t₁ = 1, d = 2, n = 8.

Apply

Apply (12)

For the arithmetic series with t₁ = 8, tₙ = 68, and Sₙ = 608, find the number of terms n.

Need a hint?

Use Sₙ = n/2 (t₁ + tₙ).
608 = n/2 (76) = 38n.

Find the first term t₁ for the series with d = 6, Sₙ = 574, and n = 14.

Need a hint?

574 = 14/2 [2t₁ + 13(6)].

Find the first term t₁ for the series with d = −3, Sₙ = 279, and n = 18.

Need a hint?

279 = 18/2 [2t₁ + 17(−3)].

For the arithmetic series with t₁ = −6, tₙ = 21, and Sₙ = 75, find n.

Need a hint?

75 = n/2 (−6 + 21) = n/2 (15).

Determine the sum of all the multiples of 4 between 1 and 999.

Need a hint?

The multiples are 4, 8, …, 996. Find how many there are.
n = (996 − 4)/4 + 1 = 249, then S = 249/2 (4 + 996).

A grandfather clock chimes the number of the hour (e.g., 4 chimes at 4:00). How many times does it chime in a 24-hour period?

Need a hint?

In 12 hours it chimes 1 + 2 + … + 12. There are two such cycles in 24 hours.
1 + 2 + … + 12 = 78, then double it.

A pilot flies 3 more circuits each day than the day before. On the 5th day she flew 14 circuits. (a) How many on the first day? (b) How many in total by the end of the 5th day?

Need a hint?

t₅ = 14, d = 3. Find t₁, then S₅.
14 = t₁ + 4(3) ⇒ t₁ = 2; S₅ = 5/2 (2 + 14).

The 2nd and 5th terms of an arithmetic series are 40 and 121. Find the sum of the first 25 terms.

Need a hint?

t₅ − t₂ = 3d, so find d, then t₁.
3d = 121 − 40 = 81 ⇒ d = 27; t₁ = 40 − 27 = 13.

Find S₁₀ for the series (−10) + (−14) + (−18) + …

Need a hint?

t₁ = −10, d = −4.

A10

For the series 40 + 35 + 30 + …, find the sum of the first 11 terms (S₁₁).

Need a hint?

t₁ = 40, d = −5, n = 11.

A11

Find the sum 8 + 3 + (−2) + … + (−102).

Need a hint?

t₁ = 8, d = −5, tₙ = −102. Find n first.
−102 = 8 + (n − 1)(−5) ⇒ n = 23.

A12

How many terms of the series 5 + 10 + 15 + … are needed for the sum to reach 720?

Need a hint?

Set Sₙ = 720 with t₁ = 5, d = 5: 720 = n/2 [10 + 5(n − 1)].
1440 = n(5n + 5) ⇒ 5n² + 5n − 1440 = 0 ⇒ n² + n − 288 = 0.

Extension

Extension (6)

The sum of the first 5 terms of an arithmetic series is 85, and the sum of the first 6 terms is 123. What are the first four terms?

Need a hint?

t₆ = S₆ − S₅. Also S₅ = 5/2 (2t₁ + 4d).
t₆ = 38 gives t₁ + 5d = 38; S₅ gives t₁ + 2d = 17. Subtract.

Galileo found a falling object travels 5 m in the 1st second, 15 m in the 2nd, 25 m in the 3rd, and so on. Using the series formula, find a formula for the total distance d(n) after n seconds.

Need a hint?

Distances 5, 15, 25, … are arithmetic: t₁ = 5, d = 10. Sum the first n.
d(n) = n/2 [2(5) + (n − 1)(10)] = n/2 (10n).

Determine the sum of all the multiples of 6 between 6 and 999.

Need a hint?

The multiples are 6, 12, …, 996.
n = (996 − 6)/6 + 1 = 166, then S = 166/2 (6 + 996).

An arithmetic series has t₁ = 1/2, d = 1, and n = 7. Find S₇.

Need a hint?

S₇ = 7/2 [2(1/2) + 6(1)].

For the series 2.5 + 3 + 3.5 + …, find S₁₀.

Need a hint?

t₁ = 2.5, d = 0.5, n = 10.

A theatre has 20 seats in the front row and 2 more seats in each row behind it. If there are 15 rows, how many seats are in the theatre?

Need a hint?

Seats per row: 20, 22, 24, … with t₁ = 20, d = 2, n = 15.
S₁₅ = 15/2 [2(20) + 14(2)].