Pre-Calc 11 · Lesson 2

Arithmetic Series.

Fold an arithmetic series in half like Gauss did, and a long chain of additions collapses into a single multiplication: average term times how many terms.

2
+6
+12
+20

By the end, you can

  • 1I can add up the terms of an arithmetic sequence using Sₙ = (n/2)(t₁ + tₙ) or Sₙ = (n/2)[2t₁ + (n-1)d].
  • 2I can choose the right sum formula depending on whether I know the last term or the common difference.
  • 3I can find an unknown quantity (t₁, d, n, or Sₙ) when the others are given, including by solving systems of equations from two given sums.
  • 4I can translate a real-world counting situation (chimes, stacked cans, falling distances) into an arithmetic series and compute its total.
  • 5I can explain WHY the sum formula works using Gauss's pairing trick and the rectangle-of-disks picture.
01 / CONCEPTS

The core ideas.

What an arithmetic series is

An arithmetic series is the SUM of the terms of an arithmetic sequence. The sequence 2, 4, 6, 8 has terms; the series 2 + 4 + 6 + 8 is what you get when you add them. We write Sₙ for the sum of the first n terms, read 'S sub n'. So for 2 + 4 + 6 + 8 + ..., S₄ = 20.

Sn=t1+t2+t3+⋯+tnS_n = t_1 + t_2 + t_3 + \dots + t_n

Why it works

A sequence is a list; a series is a total. The whole point of this section is that because the terms grow by a fixed step d, their total has hidden structure you can exploit, instead of grinding through the addition term by term.

Gauss's pairing trick

Write the series forwards, then write it again backwards underneath, and add the two rows column by column. Every column gives the SAME total, t₁ + tₙ. There are n columns, so the doubled sum is n(t₁ + tₙ), and the real sum is half of that.

2Sn=n(t1+tn)2S_n = n(t_1 + t_n)

Why it works

As you move right along the top row each term goes UP by d, and the bottom row goes DOWN by d, so every column is balanced and constant. For 1 to 100, each of the 100 columns sums to 101, giving 2S = 10100, so S = 5050. That is exactly how 10-year-old Gauss beat his teacher.

The sum formula (last-term version)

Because every paired column equals t₁ + tₙ and there are n of them, dividing by 2 gives the clean formula. Read it as 'number of terms times the average of the first and last term.' Use this version when you already know the last term tₙ.

Sn=n2(t1+tn)S_n = \frac{n}{2}(t_1 + t_n)

Why it works

In an arithmetic sequence the average of all the terms equals the average of just the first and last, because the terms are evenly spaced and symmetric about the middle. So total = count times average = n times (t₁ + tₙ)/2. This is why the formula feels like the area of a rectangle of disks: two triangles joined make an n by (t₁ + tₙ) rectangle.

The sum formula (common-difference version)

If you do NOT know the last term but you know the common difference d, substitute tₙ = t₁ + (n-1)d into the last-term formula. This packages everything in terms of t₁, d, and n so you never need a separate step to find tₙ.

Sn=n2[2t1+(n−1)d]S_n = \frac{n}{2}\bigl[2t_1 + (n-1)d\bigr]

Why it works

It is the SAME formula in disguise. Replace tₙ with what it must equal, t₁ + (n-1)d, and t₁ + tₙ becomes 2t₁ + (n-1)d. Both formulas always give the same answer; you just pick whichever matches the information you were handed.

Working backward to a missing quantity

The sum formula links t₁, d, n, and Sₙ. Given any three of these (or two sums), you can solve for the fourth. Often this means plugging in and solving a linear equation, or, when two different sums are given, solving a system of two equations.

Sn=n2[2t1+(n−1)d]S_n = \frac{n}{2}\bigl[2t_1 + (n-1)d\bigr]

Why it works

The formula is a single equation with four slots. Algebra does not care which slot is the unknown. When a problem gives you S₂ and S₄, each becomes one equation in t₁ and d, and subtracting them eliminates t₁ so d pops right out, just like solving any linear system.

From real situation to series

Many counting problems hide an arithmetic series: a clock chiming 1, 2, 3, ... times each hour; cans stacked 1, 2, 3, ... per row; a falling object covering 5, 15, 25, ... metres each second. Identify t₁ and d from the pattern, decide n, then apply a sum formula.

Why it works

The skill is translation, not new math. Ask: what is the first count, by how much does each step change, and how many steps are there? Once you name t₁, d, and n, the abstract formula does the heavy lifting and turns a tedious tally into one calculation.

02 / WORKED EXAMPLES

From easy to tricky.

EX 1easy

Find the sum of the arithmetic series 5 + 8 + 11 + ... + 53.

1

Identify t₁ and d

t₁ = 5, and the common difference is d = 8 - 5 = 3.

2

Find how many terms (n)

Use tₙ = t₁ + (n-1)d with tₙ = 53: 53 = 5 + (n-1)(3), so 48 = 3(n-1), giving n - 1 = 16 and n = 17.

3

Apply the last-term formula

Sₙ = (n/2)(t₁ + tₙ) = (17/2)(5 + 53) = (17/2)(58).

4

Compute

(17)(58)/2 = 986/2 = 493.

EX 2easy

For the series with t₁ = 7, tₙ = 79, and n = 8, find the sum S₈.

1

Choose the formula

We know first term, last term, and the count, so use Sₙ = (n/2)(t₁ + tₙ).

2

Substitute

S₈ = (8/2)(7 + 79) = 4(86).

3

Compute

4 times 86 = 344.

EX 3medium

A particular firefly flashes 2 times in the first minute, 4 times in the second, 6 in the third, and so on. (a) How many flashes occur in the 30th minute? (b) What is the total number of flashes in 30 minutes? [MHR Example 1]

1

Set up the sequence

Flashes per minute form 2, 4, 6, ... so t₁ = 2, d = 2, and we want n = 30.

2

(a) The 30th term

t₃₀ = t₁ + (n-1)d = 2 + (29)(2) = 2 + 58 = 60 flashes in the 30th minute alone.

3

(b) The total, using Sₙ = (n/2)(t₁ + tₙ)

S₃₀ = (30/2)(2 + 60) = 15(62) = 930.

4

Interpret

Notice 60 is ONE term and 930 is the running total. They are different questions.

EX 4medium

For the arithmetic series with t₁ = 8 and tₙ = 68, the sum is Sₙ = 608. Find the number of terms n. [MHR Q5a]

1

Pick the formula with tₙ

Sₙ = (n/2)(t₁ + tₙ), since we know both end terms and the sum.

2

Substitute

608 = (n/2)(8 + 68) = (n/2)(76) = 38n.

3

Solve for n

n = 608 / 38 = 16.

EX 5hard

The sum of the first two terms of an arithmetic series is 13 and the sum of the first four terms is 46. Find the first six terms and the sum to six terms. [MHR Example 2]

1

Write each sum with the d-formula

S₂ = (2/2)[2t₁ + (2-1)d] = 2t₁ + d = 13. S₄ = (4/2)[2t₁ + (4-1)d] = 2(2t₁ + 3d) = 4t₁ + 6d = 46, i.e. 2t₁ + 3d = 23.

2

Solve the system

Subtract (2t₁ + d = 13) from (2t₁ + 3d = 23): 2d = 10, so d = 5.

3

Back-substitute

2t₁ + 5 = 13, so 2t₁ = 8 and t₁ = 4.

4

List the first six terms

4, 9, 14, 19, 24, 29 (adding d = 5 each time).

5

Sum to six terms

S₆ = (6/2)(4 + 29) = 3(33) = 99.

EX 6hard

The second and fifth terms of an arithmetic series are 40 and 121. Determine the sum of the first 25 terms. [MHR Q10]

1

Translate the given terms

t₂ = t₁ + d = 40 and t₅ = t₁ + 4d = 121.

2

Solve for d

Subtract: (t₁ + 4d) - (t₁ + d) = 121 - 40, so 3d = 81 and d = 27.

3

Solve for t₁

t₁ + 27 = 40, so t₁ = 13.

4

Apply the d-formula for n = 25

S₂₅ = (25/2)[2(13) + (25-1)(27)] = (25/2)[26 + 648] = (25/2)(674).

5

Compute

25 times 674 = 16850; divide by 2 to get 8425.

EX 7challenge

The sum of the first n terms of an arithmetic series is Sₙ = 2n² + 5n. (a) Find the first three terms. (b) Confirm that S₁₀ from the arithmetic sum formula matches the given rule. [MHR Q18]

1

Get terms from sums

S₁ = t₁ = 2(1)² + 5(1) = 7. S₂ = 2(4) + 10 = 18, so t₂ = S₂ - S₁ = 11. S₃ = 2(9) + 15 = 33, so t₃ = S₃ - S₂ = 15.

2

Identify t₁ and d

The terms 7, 11, 15 give t₁ = 7 and d = 4 (a genuine arithmetic series).

3

(b) Arithmetic formula for S₁₀

S₁₀ = (10/2)[2(7) + (10-1)(4)] = 5[14 + 36] = 5(50) = 250.

4

Check against the given rule

S₁₀ = 2(10)² + 5(10) = 200 + 50 = 250. They agree.

03 / COMMON TRAPS

Where students slip.

Trap

Confusing the term tₙ with the sum Sₙ. Students compute t₃₀ = 60 for the firefly and report 60 as the total number of flashes.

Fix

tₙ is the size of a single term (the 30th minute alone has 60 flashes). Sₙ adds up ALL the terms (the total over 30 minutes is 930). Always ask: am I being asked for ONE term or the RUNNING TOTAL?

Trap

Using n = tₙ, that is, plugging the last value into the formula where the count of terms belongs. For 5 + 8 + 11 + ... + 53 they set n = 53.

Fix

n is HOW MANY terms there are, not the value of the last term. First find n from tₙ = t₁ + (n-1)d. Here 53 = 5 + (n-1)3 gives n = 17, and only then S₁₇ = (17/2)(5 + 53) = 493.

Trap

Forgetting the factor of 1/2 (the 'fold in half' step), so they report n(t₁ + tₙ) instead of (n/2)(t₁ + tₙ), doubling the true sum.

Fix

Gauss's trick gives 2Sₙ = n(t₁ + tₙ). That 2 is the whole reason for the trick: you wrote the series twice. You must divide by 2 to undo the doubling and get Sₙ.

Trap

Mixing the two formulas, for example writing Sₙ = (n/2)[2t₁ + (n-1)d] but plugging in tₙ where d should go, or using (n/2)(t₁ + tₙ) with d in place of tₙ.

Fix

Pick ONE formula based on what you know. Know the last term? Use (n/2)(t₁ + tₙ). Know the common difference? Use (n/2)[2t₁ + (n-1)d]. Do not blend ingredients from both recipes.

04 / QUIZ

Test yourself.

The Test

32 problems across three tiers. Auto-graded on submit. Hints available before you submit, full solutions after.

0/32Answered
Core

Core (13)

C1

Determine the sum of the arithmetic series 5 + 8 + 11 + ... + 53.

Need a hint?
  • Find d, then use tₙ = t₁ + (n-1)d to find n.
  • 53 = 5 + (n-1)(3) gives n = 17. Then use Sₙ = (n/2)(t₁ + tₙ).
C2

Determine the sum of the arithmetic series 7 + 14 + 21 + ... + 98.

Need a hint?
  • d = 7. Find n from 98 = 7 + (n-1)(7).
  • n = 14. Use Sₙ = (n/2)(t₁ + tₙ).
C3

Determine the sum of the arithmetic series 8 + 3 + (-2) + ... + (-102).

Need a hint?
  • The difference is negative: d = 3 - 8 = -5.
  • -102 = 8 + (n-1)(-5) gives n = 23. Then Sₙ = (n/2)(t₁ + tₙ).
C4

For the arithmetic series 1 + 3 + 5 + ..., determine t₁, d, and S₈.

Need a hint?
  • d = 3 - 1 = 2. You want the sum of 8 terms.
  • Use Sₙ = (n/2)[2t₁ + (n-1)d] with n = 8.
C5

For the arithmetic series 40 + 35 + 30 + ..., determine t₁, d, and S₁₁.

Need a hint?
  • d = 35 - 40 = -5.
  • S₁₁ = (11/2)[2(40) + (11-1)(-5)].
C6

Determine the sum Sₙ for the arithmetic sequence with t₁ = 7, tₙ = 79, n = 8.

Need a hint?
  • You know both end terms and n, so use Sₙ = (n/2)(t₁ + tₙ).
C7

Determine the sum Sₙ for the arithmetic sequence with t₁ = 58, tₙ = -7, n = 26.

Need a hint?
  • Use Sₙ = (n/2)(t₁ + tₙ); the terms are decreasing but the formula is identical.
C8

Determine the sum Sₙ for the arithmetic sequence with t₁ = 12, d = 8, n = 9.

Need a hint?
  • You have d (not tₙ), so use Sₙ = (n/2)[2t₁ + (n-1)d].
C9

For an arithmetic series, t₁ = 8, tₙ = 68, and Sₙ = 608. Determine n.

Need a hint?
  • Use Sₙ = (n/2)(t₁ + tₙ) since both end terms are known.
  • 608 = (n/2)(76) = 38n.
C10

For an arithmetic series, t₁ = -6, tₙ = 21, and Sₙ = 75. Determine n.

Need a hint?
  • Sₙ = (n/2)(t₁ + tₙ).
  • 75 = (n/2)(-6 + 21) = (n/2)(15).
C11

For the series 5 + 10 + 15 + ..., find t₁₀ and S₁₀.

Need a hint?
  • d = 5. Use tₙ = t₁ + (n-1)d for t₁₀.
  • Then S₁₀ = (10/2)(t₁ + t₁₀).
C12

For the series 10 + 7 + 4 + ..., find t₁₀ and S₁₀.

Need a hint?
  • d = -3.
  • t₁₀ = 10 + 9(-3); then use S₁₀ = (10/2)(t₁ + t₁₀).
C13

In the firefly series 2 + 4 + 6 + ..., which expression correctly gives the TOTAL number of flashes over 30 minutes?

Need a hint?
  • A total means a SUM, Sₙ, not a single term tₙ.
  • t₃₀ = 60, so use Sₙ = (n/2)(t₁ + tₙ).
Apply

Apply (13)

A1

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

Need a hint?
  • The multiples are 4, 8, 12, ..., up to the largest multiple of 4 below 999.
  • Largest is 996. t₁ = 4, d = 4. Find n from 996 = 4 + (n-1)4, then sum.
A2

What is the sum of the multiples of 6 between 6 and 999?

Need a hint?
  • Multiples of 6: 6, 12, 18, ..., up to the largest one below 999.
  • Largest is 996. t₁ = 6, d = 6. Find n, then Sₙ = (n/2)(t₁ + tₙ).
A3

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

Need a hint?
  • In 12 hours the chimes are 1 + 2 + 3 + ... + 12. A day has two such cycles.
  • Sum 1 to 12 with Sₙ = (n/2)(t₁ + tₙ), then double.
A4

A pilot flies circuits of an airfield, flying three more circuits each day than the day before. On the fifth day she flew 14 circuits. (a) How many circuits on the first day? (b) Total by the end of the fifth day?

Need a hint?
  • d = 3 and t₅ = 14. Use t₅ = t₁ + 4d to find t₁.
  • Once t₁ = 2, sum five terms with S₅ = (5/2)(t₁ + t₅).
A5

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

Need a hint?
  • The rows form 1 + 2 + 3 + ... + 18.
  • Use Sₙ = (n/2)(t₁ + tₙ) with t₁ = 1, tₙ = 18, n = 18.
A6

Determine the first term t₁ of an arithmetic series given d = 6, Sₙ = 574, and n = 14.

Need a hint?
  • Use Sₙ = (n/2)[2t₁ + (n-1)d] and solve for t₁.
  • 574 = 7[2t₁ + 13(6)] = 7[2t₁ + 78].
A7

Determine the first term t₁ of an arithmetic series given d = 0.5, Sₙ = 218.5, and n = 23.

Need a hint?
  • Sₙ = (n/2)[2t₁ + (n-1)d].
  • 218.5 = (23/2)[2t₁ + 22(0.5)]. Multiply both sides by 2/23.
A8

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

Need a hint?
  • The sixth term equals S₆ - S₅.
  • t₆ = 123 - 85 = 38, so t₁ + 5d = 38. Also S₅ = (5/2)(2t₁ + 4d) = 85 gives t₁ + 2d = 17.
A9

The second and fifth terms of an arithmetic series are 40 and 121. Determine the sum of the first 25 terms.

Need a hint?
  • t₂ = t₁ + d = 40 and t₅ = t₁ + 4d = 121. Subtract to find d.
  • 3d = 81 gives d = 27, t₁ = 13. Then S₂₅ = (25/2)[2t₁ + 24d].
A10

Two students compute the same series sum: Pierre uses Sₙ = (n/2)(t₁ + tₙ) and Jeanette uses Sₙ = (n/2)[2t₁ + (n-1)d]. Both get 912. Why do both work?

Need a hint?
  • Substitute tₙ = t₁ + (n-1)d into t₁ + tₙ.
A11

True or false: doubling EVERY term in an arithmetic series doubles the sum of the series. Justify.

Need a hint?
  • If each term becomes 2tₖ, what happens to t₁ + t₂ + ... + tₙ?
A12

True or false: keeping the first term constant but doubling the NUMBER of terms doubles the sum. Justify with the series 2 + 4 + 6 + ... .

Need a hint?
  • Compare Sₙ and S₂n for, say, 2 + 4 + 6 + ... and check if S₂n = 2 Sₙ.
A13

A pilot flies 3 more circuits each day; on day 5 she flew 14. Write a simplified expression for the total circuits by the end of the nth day.

Need a hint?
  • From earlier, t₁ = 2 and d = 3.
  • Plug into Sₙ = (n/2)[2t₁ + (n-1)d] and simplify.
Extension

Extension (6)

E1

A falling object travels 5 m in the first second, 15 m in the second, 25 m in the third, and so on. Using the sum formula, show the total distance after n seconds equals d(n) = 5n², and verify for n = 100.

Need a hint?
  • The per-second distances form 5, 15, 25, ... with t₁ = 5, d = 10.
  • Compute Sₙ = (n/2)[2(5) + (n-1)(10)] and simplify.
E2

In a room everyone shakes hands with everyone else exactly once. (a) How many handshakes occur with 30 people? (b) Explain why this is the series 1 + 2 + ... + 29.

Need a hint?
  • Person number k can shake a 'new' hand with each earlier person; the counts build up 1, 2, 3, ...
  • For 30 people the new handshakes total 1 + 2 + ... + 29. Sum with Sₙ = (n/2)(t₁ + tₙ).
E3

Triangular numbers come from 1, (1+2), (1+2+3), .... Using the sum formula, the nth triangular number equals ____, and the 10th triangular number is ____.

Need a hint?
  • The nth triangular number is the series 1 + 2 + ... + n.
  • Apply Sₙ = (n/2)(t₁ + tₙ) with t₁ = 1, tₙ = n.
E4

The 15th term of an arithmetic sequence is 43 and the sum of the first 15 terms is 120. Determine the first three terms.

Need a hint?
  • S₁₅ = (15/2)(t₁ + t₁₅) gives an equation in t₁ since t₁₅ = 43.
  • Solve for t₁, then use t₁₅ = t₁ + 14d to find d.
E5

The first three terms of an arithmetic sequence are x, (2x - 5), 8.6. (a) Find t₁ and d. (b) Find t₂₀.

Need a hint?
  • Consecutive terms have equal differences: (2x - 5) - x = 8.6 - (2x - 5).
  • Solve that equation for x to get t₁, then d = (2x-5) - x.
E6

The sum of the first n terms of an arithmetic series is Sₙ = 2n² + 5n. Find the first three terms and S₁₀ two ways to confirm they agree.

Need a hint?
  • t₁ = S₁, and each later term is tₖ = Sₖ - Sₖ₋₁.
  • Once you have t₁ = 7 and d = 4, compute S₁₀ with the sum formula and also from 2(10)² + 5(10).