Model a Pattern | Math Coach Amy

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Key Ideas Simple Interest Money Seating and Auditoriums Stacking and Building Predict Number

SL 1.2

$u_n = u_1 + (n-1)d$

$S_n = \dfrac{n}{2}\bigl(2u_1 + (n-1)d\bigr)$

$S_n = \dfrac{n}{2}(u_1 + u_n)$

What to Remember

A steady change is an arithmetic sequence

Simple interest, money saved, a row of seats, or a stack can share one common difference.

The starting balance, the first payment, the first row, or the bottom layer is u₁.

$u_n = u_1 + (n-1)d$

A model uses one common difference

When the gaps are close but not equal, keep the first value and the last value, and use one common difference between them.

$d = \dfrac{u_k - u_1}{k-1}$

Then predict the later term with the nth-term formula.

Easy 1

Simple Interest

The balance at the start of year 1 is the first term. The yearly interest is the common difference.

Example

An account has $400 at the start of year 1. It earns $20 of simple interest each year. Find the balance at the start of year 6.

$u_n = u_1 + (n-1)d$

$u_{6} = \text{the term to find}$

$u_1 = 400$

$n = 6$

$d = 20$

$u_{6} = 400 + (6-1) \times 20$

$u_{6} = 400 + 5 \times 20$

$u_{6} = 400 + 100$

$u_{6} = 500$

Answer: the balance = $$500$

An account has $250 at the start of year 1. It earns $15 of simple interest each year. Find the balance at the start of year 8.

Show Answer

$u_n = u_1 + (n-1)d$

$u_{8} = \text{the term to find}$

$u_1 = 250$

$n = 8$

$d = 15$

$u_{8} = 250 + (8-1) \times 15$

$u_{8} = 250 + 7 \times 15$

$u_{8} = 250 + 105$

$u_{8} = 355$

Answer: the balance = $$355$

An account has $800 at the start of year 1. It earns $40 of simple interest each year. Find the balance at the start of year 10.

Show Answer

$u_n = u_1 + (n-1)d$

$u_{10} = \text{the term to find}$

$u_1 = 800$

$n = 10$

$d = 40$

$u_{10} = 800 + (10-1) \times 40$

$u_{10} = 800 + 9 \times 40$

$u_{10} = 800 + 360$

$u_{10} = 1160$

Answer: the balance = $$1160$

An account has $120 at the start of year 1. It earns $8 of simple interest each year. Find the balance at the start of year 5.

Show Answer

$u_n = u_1 + (n-1)d$

$u_{5} = \text{the term to find}$

$u_1 = 120$

$n = 5$

$d = 8$

$u_{5} = 120 + (5-1) \times 8$

$u_{5} = 120 + 4 \times 8$

$u_{5} = 120 + 32$

$u_{5} = 152$

Answer: the balance = $$152$

An account has $1000 at the start of year 1. It earns $25 of simple interest each year. Find the balance at the start of year 12.

Show Answer

$u_n = u_1 + (n-1)d$

$u_{12} = \text{the term to find}$

$u_1 = 1000$

$n = 12$

$d = 25$

$u_{12} = 1000 + (12-1) \times 25$

$u_{12} = 1000 + 11 \times 25$

$u_{12} = 1000 + 275$

$u_{12} = 1275$

Answer: the balance = $$1275$

An account has $60 at the start of year 1. It earns $12 of simple interest each year. Find the balance at the start of year 9.

Show Answer

$u_n = u_1 + (n-1)d$

$u_{9} = \text{the term to find}$

$u_1 = 60$

$n = 9$

$d = 12$

$u_{9} = 60 + (9-1) \times 12$

$u_{9} = 60 + 8 \times 12$

$u_{9} = 60 + 96$

$u_{9} = 156$

Answer: the balance = $$156$

Easy 2

Money

The first week or the first month is term 1.

Example

Jordan saves $30 in week 1. Each later week, Jordan saves $5 more than the week before. How much does Jordan save in week 10?

$u_n = u_1 + (n-1)d$

$u_{10} = \text{the term to find}$

$u_1 = 30$

$n = 10$

$d = 5$

$u_{10} = 30 + (10-1) \times 5$

$u_{10} = 30 + 9 \times 5$

$u_{10} = 30 + 45$

$u_{10} = 75$

Answer: the amount saved = $$75$

Maya saves $20 in week 1. Each later week, Maya saves $4 more than the week before. How much does Maya save in week 8?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{8} = \text{the term to find}$

$u_1 = 20$

$n = 8$

$d = 4$

$u_{8} = 20 + (8-1) \times 4$

$u_{8} = 20 + 7 \times 4$

$u_{8} = 20 + 28$

$u_{8} = 48$

Answer: the amount saved = $$48$

An allowance is $12 in week 1. Each later week, the allowance is $3 more than the week before. How much is the allowance in week 15?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{15} = \text{the term to find}$

$u_1 = 12$

$n = 15$

$d = 3$

$u_{15} = 12 + (15-1) \times 3$

$u_{15} = 12 + 14 \times 3$

$u_{15} = 12 + 42$

$u_{15} = 54$

Answer: the allowance = $$54$

A ticket costs $18 in week 1. Each later week, the ticket costs $2 more than the week before. What is the price in week 9?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{9} = \text{the term to find}$

$u_1 = 18$

$n = 9$

$d = 2$

$u_{9} = 18 + (9-1) \times 2$

$u_{9} = 18 + 8 \times 2$

$u_{9} = 18 + 16$

$u_{9} = 34$

Answer: the price = $$34$

Priya saves $45 in month 1. Each later month, Priya saves $10 more than the month before. How much does Priya save in month 6?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{6} = \text{the term to find}$

$u_1 = 45$

$n = 6$

$d = 10$

$u_{6} = 45 + (6-1) \times 10$

$u_{6} = 45 + 5 \times 10$

$u_{6} = 45 + 50$

$u_{6} = 95$

Answer: the amount saved = $$95$

Leo earns $50 in week 1. Each later week, Leo earns $8 more than the week before. How much does Leo earn in week 11?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{11} = \text{the term to find}$

$u_1 = 50$

$n = 11$

$d = 8$

$u_{11} = 50 + (11-1) \times 8$

$u_{11} = 50 + 10 \times 8$

$u_{11} = 50 + 80$

$u_{11} = 130$

Answer: the amount earned = $$130$

Easy 3

Seating and Auditoriums

Row 1 is the first term.

Example

Row 1 of an auditorium has 18 seats. Each row behind it has 2 more seats than the row in front of it. How many seats are in row 14?

$u_n = u_1 + (n-1)d$

$u_{14} = \text{the term to find}$

$u_1 = 18$

$n = 14$

$d = 2$

$u_{14} = 18 + (14-1) \times 2$

$u_{14} = 18 + 13 \times 2$

$u_{14} = 18 + 26$

$u_{14} = 44$

Answer: 44 seats

Row 1 of an auditorium has 22 seats. Each row behind it has 3 more seats than the row in front of it. How many seats are in row 10?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{10} = \text{the term to find}$

$u_1 = 22$

$n = 10$

$d = 3$

$u_{10} = 22 + (10-1) \times 3$

$u_{10} = 22 + 9 \times 3$

$u_{10} = 22 + 27$

$u_{10} = 49$

Answer: 49 seats

Row 1 of an auditorium has 16 seats. Each row behind it has 4 more seats than the row in front of it. How many seats are in row 12?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{12} = \text{the term to find}$

$u_1 = 16$

$n = 12$

$d = 4$

$u_{12} = 16 + (12-1) \times 4$

$u_{12} = 16 + 11 \times 4$

$u_{12} = 16 + 44$

$u_{12} = 60$

Answer: 60 seats

Row 1 of an auditorium has 30 seats. Each row behind it has 1 fewer seats than the row in front of it. How many seats are in row 9?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{9} = \text{the term to find}$

$u_1 = 30$

$n = 9$

$d = -1$

$u_{9} = 30 + (9-1) \times \left(-1\right)$

$u_{9} = 30 + 8 \times \left(-1\right)$

$u_{9} = 30 - 8$

$u_{9} = 22$

Answer: 22 seats

Row 1 of an auditorium has 12 seats. Each row behind it has 2 more seats than the row in front of it. How many seats are in row 20?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{20} = \text{the term to find}$

$u_1 = 12$

$n = 20$

$d = 2$

$u_{20} = 12 + (20-1) \times 2$

$u_{20} = 12 + 19 \times 2$

$u_{20} = 12 + 38$

$u_{20} = 50$

Answer: 50 seats

Row 1 of an auditorium has 25 seats. Each row behind it has 5 more seats than the row in front of it. How many seats are in row 8?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{8} = \text{the term to find}$

$u_1 = 25$

$n = 8$

$d = 5$

$u_{8} = 25 + (8-1) \times 5$

$u_{8} = 25 + 7 \times 5$

$u_{8} = 25 + 35$

$u_{8} = 60$

Answer: 60 seats

Easy 4

Stacking and Building

The bottom layer is the first term.

Example

The bottom layer of a can display has 15 pieces. Each higher layer has 1 fewer pieces. How many pieces are in layer 6?

$u_n = u_1 + (n-1)d$

$u_{6} = \text{the term to find}$

$u_1 = 15$

$n = 6$

$d = -1$

$u_{6} = 15 + (6-1) \times \left(-1\right)$

$u_{6} = 15 + 5 \times \left(-1\right)$

$u_{6} = 15 - 5$

$u_{6} = 10$

Answer: 10 pieces

The bottom layer of a brick wall has 20 pieces. Each higher layer has 2 fewer pieces. How many pieces are in layer 7?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{7} = \text{the term to find}$

$u_1 = 20$

$n = 7$

$d = -2$

$u_{7} = 20 + (7-1) \times \left(-2\right)$

$u_{7} = 20 + 6 \times \left(-2\right)$

$u_{7} = 20 - 12$

$u_{7} = 8$

Answer: 8 pieces

The bottom layer of a log wall has 12 pieces. Each higher layer has 1 fewer pieces. How many pieces are in layer 5?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{5} = \text{the term to find}$

$u_1 = 12$

$n = 5$

$d = -1$

$u_{5} = 12 + (5-1) \times \left(-1\right)$

$u_{5} = 12 + 4 \times \left(-1\right)$

$u_{5} = 12 - 4$

$u_{5} = 8$

Answer: 8 pieces

The bottom layer of a box display has 24 pieces. Each higher layer has 3 fewer pieces. How many pieces are in layer 6?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{6} = \text{the term to find}$

$u_1 = 24$

$n = 6$

$d = -3$

$u_{6} = 24 + (6-1) \times \left(-3\right)$

$u_{6} = 24 + 5 \times \left(-3\right)$

$u_{6} = 24 - 15$

$u_{6} = 9$

Answer: 9 pieces

The bottom layer of a cup pyramid has 18 pieces. Each higher layer has 2 fewer pieces. How many pieces are in layer 8?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{8} = \text{the term to find}$

$u_1 = 18$

$n = 8$

$d = -2$

$u_{8} = 18 + (8-1) \times \left(-2\right)$

$u_{8} = 18 + 7 \times \left(-2\right)$

$u_{8} = 18 - 14$

$u_{8} = 4$

Answer: 4 pieces

The bottom layer of a display has 16 pieces. Each higher layer has 1 fewer pieces. How many pieces are in layer 10?

Show Answer

$u_n = u_1 + (n-1)d$

$u_{10} = \text{the term to find}$

$u_1 = 16$

$n = 10$

$d = -1$

$u_{10} = 16 + (10-1) \times \left(-1\right)$

$u_{10} = 16 + 9 \times \left(-1\right)$

$u_{10} = 16 - 9$

$u_{10} = 7$

Answer: 7 pieces

Medium

Predict Number

The gaps are close, but not equal. Use one common difference from the first value to the last value.

Example

Five practice scores are 20, 33, 44, 57, and 68. Model the scores as an arithmetic sequence and predict the 9th score.

The differences are 13, 11, 13, and 11, so one difference does not fit every step.

Use the first value as $u_1$ and the last value as $u_{5}$.

$u_k = u_1 + (k-1)d$

$68 = 20 + (5-1)d$

$68 = 20 + 4d$

$68 - 20 = 4d$

$48 = 4d$

$d = \dfrac{48}{4}$

$d = 12$

$u_n = u_1 + (n-1)d$

$u_{9} = 20 + (9-1) \times 12$

$u_{9} = 20 + 8 \times 12$

$u_{9} = 20 + 96$

$u_{9} = 116$

Answer: the model gives $116$

Five game scores are 18, 29, 41, 50, and 62. Model the scores as an arithmetic sequence and predict the score on the 10th game.

Show Answer

The differences are 11, 12, 9, and 12, so one difference does not fit every step.

Use the first value as $u_1$ and the last value as $u_{5}$.

$u_k = u_1 + (k-1)d$

$62 = 18 + (5-1)d$

$62 = 18 + 4d$

$62 - 18 = 4d$

$44 = 4d$

$d = \dfrac{44}{4}$

$d = 11$

$u_n = u_1 + (n-1)d$

$u_{10} = 18 + (10-1) \times 11$

$u_{10} = 18 + 9 \times 11$

$u_{10} = 18 + 99$

$u_{10} = 117$

Answer: the model gives $117$

Free-throw counts over five practices are 8, 15, 21, 30, and 36. Model the counts as an arithmetic sequence and predict the count at practice 10.

Show Answer

The differences are 7, 6, 9, and 6, so one difference does not fit every step.

Use the first value as $u_1$ and the last value as $u_{5}$.

$u_k = u_1 + (k-1)d$

$36 = 8 + (5-1)d$

$36 = 8 + 4d$

$36 - 8 = 4d$

$28 = 4d$

$d = \dfrac{28}{4}$

$d = 7$

$u_n = u_1 + (n-1)d$

$u_{10} = 8 + (10-1) \times 7$

$u_{10} = 8 + 9 \times 7$

$u_{10} = 8 + 63$

$u_{10} = 71$

Answer: the model gives $71$

Tickets sold over five days are 30, 46, 61, 78, and 94. Model the sales as an arithmetic sequence and predict the sales on day 10.

Show Answer

The differences are 16, 15, 17, and 16, so one difference does not fit every step.

Use the first value as $u_1$ and the last value as $u_{5}$.

$u_k = u_1 + (k-1)d$

$94 = 30 + (5-1)d$

$94 = 30 + 4d$

$94 - 30 = 4d$

$64 = 4d$

$d = \dfrac{64}{4}$

$d = 16$

$u_n = u_1 + (n-1)d$

$u_{10} = 30 + (10-1) \times 16$

$u_{10} = 30 + 9 \times 16$

$u_{10} = 30 + 144$

$u_{10} = 174$

Answer: the model gives $174$

Five daily totals are 10, 22, 31, 43, and 54. Model the totals as an arithmetic sequence and predict the 12th total.

Show Answer

The differences are 12, 9, 12, and 11, so one difference does not fit every step.

Use the first value as $u_1$ and the last value as $u_{5}$.

$u_k = u_1 + (k-1)d$

$54 = 10 + (5-1)d$

$54 = 10 + 4d$

$54 - 10 = 4d$

$44 = 4d$

$d = \dfrac{44}{4}$

$d = 11$

$u_n = u_1 + (n-1)d$

$u_{12} = 10 + (12-1) \times 11$

$u_{12} = 10 + 11 \times 11$

$u_{12} = 10 + 121$

$u_{12} = 131$

Answer: the model gives $131$

A count records 4, 13, 21, and 34 on the first four days. Model the counts as an arithmetic sequence and predict the number on day 6.

Show Answer

The differences are 9, 8, and 13, so one difference does not fit every step.

Use the first value as $u_1$ and the last value as $u_{4}$.

$u_k = u_1 + (k-1)d$

$34 = 4 + (4-1)d$

$34 = 4 + 3d$

$34 - 4 = 3d$

$30 = 3d$

$d = \dfrac{30}{3}$

$d = 10$

$u_n = u_1 + (n-1)d$

$u_{6} = 4 + (6-1) \times 10$

$u_{6} = 4 + 5 \times 10$

$u_{6} = 4 + 50$

$u_{6} = 54$

Answer: the model gives $54$

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