Pre-Calc 11 · Lesson 1

Arithmetic Sequences.

An arithmetic sequence is a staircase of numbers that climbs by the same step every time, so once you know where you start and how big each step is, you can leap straight to any term without climbing all the stairs.

2
4
6
8
10

By the end, you can

  • 1I can recognize an arithmetic sequence and find its first term t₁ and common difference d.
  • 2I can derive and use the general term tₙ = t₁ + (n-1)d to find any term directly.
  • 3I can solve for an unknown (t₁, d, n, or tₙ) when the others are given, including setting up two equations from two known terms.
  • 4I can explain why an arithmetic sequence behaves like a linear function, connecting d to slope and the rule to a line.
  • 5I can model and solve real-world problems (pay schedules, populations, stacked rows, multiples) using arithmetic sequences.
01 / CONCEPTS

The core ideas.

What makes a sequence arithmetic

A sequence is just an ordered list of numbers called terms, written t₁, t₂, t₃, and so on. A sequence is arithmetic when the difference between every pair of consecutive terms is the same constant, called the common difference d. You find d by subtracting any term from the one right after it: d = tₙ - tₙ₋₁. The difference can be positive (increasing), negative (decreasing), or zero (constant).

d=tn−tn−1d = t_n - t_{n-1}

Why it works

The word 'common' means shared by all the gaps. If even one gap is different, the list is not arithmetic. This is exactly the test you run: walk down the list, subtract each term from the next, and check that you always land on the same number. That constancy is the entire personality of the sequence.

Deriving the general term

Instead of adding d over and over to reach a far-off term, you can jump straight there. Start at t₁. To reach t₂ you add d once, to reach t₃ you add d twice, to reach t₄ you add d three times. The pattern: to reach the nth term you add d exactly (n-1) times. That gives the general term tₙ = t₁ + (n-1)d.

tn=t1+(n−1)dt_n = t_1 + (n-1)d

Why it works

Notice it is (n-1)d, not nd. The first term already exists for free before you take any steps, so the number of steps is always one less than the term number. Counting fence posts and gaps: 5 posts have only 4 gaps between them. Term number is posts; the multiplier of d is gaps.

Arithmetic sequence as a linear function

Expand the general term: tₙ = t₁ + (n-1)d = dn + (t₁ - d). This has the form tₙ = (slope)n + (intercept). So if you plot term number n on the x-axis and term value tₙ on the y-axis, the points fall exactly on a straight line. The common difference d is the slope, and t₁ - d is the y-intercept (the value the line would hit at n=0).

tn=dn+(t1−d)t_n = dn + (t_1 - d)

Why it works

Adding the same amount d each time is precisely what a line does: constant rate of change. The reason the points are dots and not a full line is that n must be a natural number, you cannot have term 2.5. So an arithmetic sequence is the 'discrete' version of a linear function, lined up like beads on a straight wire.

Solving for any unknown

The general term links four quantities: t₁, d, n, and tₙ. Given any three, you can solve for the fourth by substituting into tₙ = t₁ + (n-1)d and isolating. When two different terms are known (say t₃ and t₈) but neither t₁ nor d is, write one equation per term and subtract them to eliminate t₁ and solve for d first.

tn=t1+(n−1)dt_n = t_1 + (n-1)d

Why it works

Think of the formula as a four-slot machine where filling any three slots forces the fourth. Subtracting two term-equations works because both contain the same t₁, so the subtraction cancels it cleanly. The difference in their term numbers times d equals the difference in their values: this is just slope = rise over run in disguise.

Reading direction from the sign of d

The sign of the common difference tells you which way the sequence travels. If d is positive the sequence increases; if d is negative it decreases; if d equals zero every term is identical. This lets you predict behavior before computing a single term and lets you check that your d has the right sign for the situation.

d>0 increasing,  d<0 decreasing,  d=0 constantd>0 \text{ increasing},\; d<0 \text{ decreasing},\; d=0 \text{ constant}

Why it works

In the cereal-box stack, rows shrink from bottom to top, so going up the rows gives d = -2. The negative sign is not a mistake, it encodes 'losing two boxes per row.' Always ask 'which direction am I counting?' because the same physical pattern can have d = +2 or d = -2 depending on which end you call term 1.

02 / WORKED EXAMPLES

From easy to tricky.

EX 1easy

Is the sequence -4, -7, -10, -13, -16, ... arithmetic? If so, state t₁ and d, and write the next three terms.

1

Test consecutive differences

-7-(-4)=-3, -10-(-7)=-3, -13-(-10)=-3. The gap is always -3.

2

Conclude and name the parts

The difference is constant, so it is arithmetic with t₁=-4 and d=-3.

3

Extend by adding d

-16+(-3)=-19, -19+(-3)=-22, -22+(-3)=-25.

EX 2easy

An events leader is paid $12 for the first hour, $19 for two hours, $26 for three hours, and so on. Write the general term, then find the pay for 6 hours.

1

Identify t₁ and d

t₁=12, d=19-12=7.

2

Substitute into the general term

tₙ=t₁+(n-1)d=12+(n-1)7=12+7n-7=7n+5.

3

Evaluate at n=6

t₆=7(6)+5=42+5=47.

EX 3medium

A musk-ox population starts at 9250 and increases by about 1650 each year. How many years until it reaches 100000?

1

List known quantities

t₁=9250, d=1650, tₙ=100 000.

2

Substitute into the formula

100 000=9250+(n-1)1650.

3

Expand and simplify

100 000=9250+1650n-1650=1650n+7600.

4

Solve for n

92 400=1650n ⇒ n=56.

EX 4medium

Find the position of 170 in the sequence -4, 2, 8, ... .

1

Find t₁ and d

t₁=-4, d=2-(-4)=6.

2

Set tₙ equal to 170

170=-4+(n-1)(6).

3

Solve for n

170=-4+6n-6=6n-10 ⇒ 180=6n ⇒ n=30.

EX 5hard

In a stack of cereal boxes the rows form an arithmetic sequence. There are 16 boxes in the 3rd row from the bottom and 6 boxes in the 8th row from the bottom. Find t₁ (bottom row), the general term, and the total number of rows if the top row has 2 boxes.

1

Write two equations

t₃: 16=t₁+2d. t₈: 6=t₁+7d.

2

Subtract to eliminate t₁

16-6=(t₁+2d)-(t₁+7d) ⇒ 10=-5d ⇒ d=-2.

3

Back-substitute for t₁

16=t₁+2(-2)=t₁-4 ⇒ t₁=20.

4

Build the general term

tₙ=20+(n-1)(-2)=-2n+22.

5

Find n when the top row has 2 boxes

2=-2n+22 ⇒ -20=-2n ⇒ n=10.

EX 6hard

The terms 5x+2, 7x-4, and 10x+6 are consecutive terms of an arithmetic sequence. Find x and state the three terms.

1

Use equal consecutive differences

(7x-4)-(5x+2)=(10x+6)-(7x-4).

2

Simplify each side

2x-6=3x+10.

3

Solve for x

-6-10=3x-2x ⇒ -16=x.

4

Substitute x=-16 into each term

5(-16)+2=-78, 7(-16)-4=-116, 10(-16)+6=-154.

5

Verify constant difference

-116-(-78)=-38 and -154-(-116)=-38. ✓

EX 7challenge

The side lengths of a quadrilateral form an arithmetic sequence. The longest side is 24 cm and the perimeter is 60 cm. Find the four side lengths, explaining your reasoning.

1

Name the four sides

Let the sides be a, a+d, a+2d, a+3d, with the longest being a+3d=24.

2

Use the perimeter

a+(a+d)+(a+2d)+(a+3d)=4a+6d=60, so 2a+3d=30.

3

Substitute a=24-3d

2(24-3d)+3d=30 ⇒ 48-3d=30 ⇒ d=6.

4

Find a and list sides

a=24-3(6)=6, giving 6, 12, 18, 24.

5

Check

Sum=6+12+18+24=60 ✓, longest=24 ✓, common difference=6 ✓.

03 / COMMON TRAPS

Where students slip.

Trap

Using nd instead of (n-1)d, so students compute the 10th term as t₁ + 10d.

Fix

You add the common difference one fewer time than the term number, because t₁ is already there before any step. The 10th term is t₁ + 9d. Test it on a small case: t₂ should equal t₁ + d, and the formula gives t₁ + (2-1)d = t₁ + d. Correct.

Trap

Assuming any sequence with a recognizable pattern is arithmetic, for example calling 2, 4, 8, 16 arithmetic.

Fix

Arithmetic requires a constant difference, not just any pattern. In 2, 4, 8, 16 the gaps are 2, 4, 8, which are not equal, so it is not arithmetic (it is geometric, multiplying by 2). Always subtract consecutive terms and confirm every gap matches before using the arithmetic formula.

Trap

Getting the sign of d backward in a decreasing sequence, treating d as positive because 'difference' sounds like a positive amount.

Fix

Compute d as later term minus earlier term, keeping the sign. For 20, 18, 16, ... you get 18 - 20 = -2, so d = -2. A decreasing sequence must have a negative d, otherwise your formula will produce terms that grow instead of shrink.

Trap

Confusing the term value tₙ with the term position n, for example answering 'which term is 170' with the value 170 itself.

Fix

n is the position (1st, 2nd, 3rd...) and tₙ is the value sitting at that position. To find the position of a known value, substitute that value as tₙ and solve for n. For -4, 2, 8, ... with value 170: 170 = -4 + (n-1)(6) gives n = 30, so 170 is the 30th term.

04 / QUIZ

Test yourself.

The Test

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

0/30Answered
Core

Core (12)

C1

Which sequence is arithmetic?

Need a hint?
  • Subtract each term from the next and look for a constant gap.
  • For 16, 32, 48, ... the gaps are all 16; the others multiply or use squares.
C2

Write the first four terms of the arithmetic sequence with t₁ = 5 and d = 3.

Need a hint?
  • Add d = 3 to each term to get the next.
C3

Write the first four terms of the arithmetic sequence with t₁ = -1 and d = -4.

Need a hint?
  • Adding a negative d makes the terms decrease.
C4

For the sequence defined by tₙ = 3n + 8, find t₁, t₇, and t₁₄.

Need a hint?
  • Substitute the term number for n in 3n + 8.
C5

Find t₁ and d, then fill the missing terms: __, __, __, 19, 23.

Need a hint?
  • From 19 to 23 the difference is 4, so d = 4. Step backward by subtracting 4.
C6

What position is 170 in the sequence -4, 2, 8, ... ?

Need a hint?
  • Here t₁ = -4 and d = 6. Set tₙ = 170 and solve for n.
  • 170 = -4 + (n-1)(6).
C7

What position is 97 in the sequence -3, 1, 5, ... ?

Need a hint?
  • t₁ = -3, d = 4. Solve 97 = -3 + (n-1)(4).
C8

What position is -10 in the sequence 14, 12.5, 11, ... ?

Need a hint?
  • d = 12.5 - 14 = -1.5. Solve -10 = 14 + (n-1)(-1.5).
C9

An arithmetic sequence has first term 6 and fourth term 33. Find the second and third terms.

Need a hint?
  • There are 3 steps from t₁ to t₄, so d = (33-6)/3.
C10

Write the general term for the arithmetic sequence 3, 0, -3, -6, -9, ... in the form tₙ = dn + c.

Need a hint?
  • t₁ = 3 and d = -3. Substitute into tₙ = t₁ + (n-1)d and simplify.
C11

True or false: in the sequence 3, 7, 11, ... the value 34 appears as a term. Justify.

Need a hint?
  • t₁ = 3, d = 4, so tₙ = 4n - 1. Check if 34 = 4n - 1 gives a whole number n.
C12

Fill the missing terms: __, 4, __, __, 10.

Need a hint?
  • Position 2 is 4 and position 5 is 10. From t₂ to t₅ is 3 steps, so 3d = 10 - 4.
Apply

Apply (12)

A1

A child is 70 cm tall at age 3 and grows about 5 cm per year. Write a general term for height versus age (let n=1 at age 3) and find the expected height at age 10.

Need a hint?
  • t₁ = 70, d = 5. Age 10 corresponds to n = 8.
  • tₙ = 70 + (n-1)(5) = 5n + 65.
A2

A carpenter ant colony starts with 40 ants and grows by about 80 ants each month. How many months until it reaches 3000 ants?

Need a hint?
  • t₁ = 40, d = 80, tₙ = 3000. Solve for n.
A3

The 16th term of an arithmetic sequence is 110 and the common difference is 7. Find the first term.

Need a hint?
  • Use 110 = t₁ + (16-1)(7).
A4

The first term of an arithmetic sequence is 5y and the common difference is -3y. Write the equations for tₙ and t₁₅.

Need a hint?
  • Substitute t₁ = 5y and d = -3y into tₙ = t₁ + (n-1)d.
A5

The terms 5x+2, 7x-4, and 10x+6 are consecutive terms of an arithmetic sequence. Find x and the three terms.

Need a hint?
  • Set the two consecutive differences equal: (7x-4)-(5x+2) = (10x+6)-(7x-4).
  • 2x - 6 = 3x + 10.
A6

A furnace technician charges $65 per house call plus $42 per hour. Write the general term for the charge and find the cost of 10 hours.

Need a hint?
  • The 1-hour charge is 65 + 42 = 107, so t₁ = 107 and d = 42.
A7

Susan does 11 sit-ups on day 6 and 29 sit-ups on day 15, following an arithmetic sequence. Write the general term, then find the first day she does at least 100 sit-ups.

Need a hint?
  • From day 6 to day 15 is 9 steps: 9d = 29 - 11.
  • d = 2, then t₁ = 11 - 5(2) = 1, so tₙ = 2n - 1.
A8

Alkanes follow the formula where 1 carbon has 4 hydrogen atoms, 2 carbons have 6, 3 carbons have 8, and so on. Write the general term relating hydrogen atoms to carbon atoms, then find how many carbons support 202 hydrogen atoms (hectane).

Need a hint?
  • t₁ = 4, d = 2 (hydrogen increases by 2 per carbon).
  • tₙ = 4 + (n-1)(2) = 2n + 2. Set it equal to 202.
A9

Earth rotates 360 degrees per day, and 1 degree takes 4 minutes. Write an equation for time as a function of degrees, then find the time for an 80 degree rotation.

Need a hint?
  • t₁ = 4 (one degree) and d = 4 minutes per degree.
A10

Which arithmetic sequence contains the term 34?

Need a hint?
  • For each, set the rule equal to 34 and check whether n is a natural number.
  • 6 + (n-1)4 = 34 gives n = 8, a whole number.
A11

An arithmetic sequence has first term 42 and fourth term 27. Find the second and third terms.

Need a hint?
  • Three steps go from 42 to 27, so d = (27-42)/3.
A12

Golf tee-off times begin at 8:00 and are 8 minutes apart, with groups of four players (max 132 players). Taking 8:00 as time 0, write the general term for tee-off times and find when the last group tees off.

Need a hint?
  • 132 players in groups of 4 means 33 groups. t₁ = 0, d = 8.
  • tₙ = 0 + (n-1)(8) = 8n - 8; use n = 33.
Extension

Extension (6)

E1

A wall hanging is 22 in by 27 in (594 square inches total) completed over 27 days. The cumulative area finished by the end of each day forms an arithmetic sequence, with 48 square inches done by end of day 1. How much area is added each subsequent day?

Need a hint?
  • Cumulative area: t₁ = 48 and t₂₇ = 594. The daily increment is d.
  • 594 = 48 + (27-1)d.
E2

Water pressure is 14.7 psi at the surface and increases by 14.7 psi for every 30 ft of descent. Write the general term for pressure at each 30-ft interval (n=1 at surface, 14.7 psi) and find the pressure at 1000 ft.

Need a hint?
  • Surface is t₁ = 14.7 and each 30 ft adds d = 14.7.
  • At 1000 ft the pressure equals 14.7 + 14.7(1000/30).
E3

The side lengths of a quadrilateral form an arithmetic sequence. The longest side is 24 cm and the perimeter is 60 cm. Find all four side lengths.

Need a hint?
  • Let the sides be a, a+d, a+2d, a+3d with a+3d = 24.
  • The perimeter gives 4a + 6d = 60, i.e. 2a + 3d = 30. Substitute a = 24 - 3d.
E4

Diamond extraction forms an arithmetic sequence: 3.8 million carats in year 1 (2003) and 113.2 million carats in year 20. Find the common difference and explain what it represents.

Need a hint?
  • t₁ = 3.8, t₂₀ = 113.2. There are 19 steps from year 1 to year 20.
  • 113.2 = 3.8 + (20-1)d.
E5

A wheel-line irrigation system attaches wheel 1 at 50 m from the pivot, with each later wheel 20 m further out. Find the circumference of the circle traced by wheel 12. Use pi ≈ 3.14159.

Need a hint?
  • Radii form an arithmetic sequence: t₁ = 50, d = 20. Find t₁₂ first.
  • t₁₂ = 50 + (12-1)(20), then use C = 2πr.
E6

The numbers x, y, z are the first three terms of an arithmetic sequence. Show that z = 2y - x.

Need a hint?
  • Consecutive differences are equal: y - x = z - y.