Model a Pattern | Math Coach Amy

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Key Ideas Percent Change Money Growth Bounce Predict Number

SL 1.3

$u_n = u_1 r^{n-1}$

$S_n = \dfrac{u_1(r^n - 1)}{r - 1},\ r \neq 1$

$S_n = \dfrac{u_1(1 - r^n)}{1 - r},\ r \neq 1$

What to Remember

A percent change is the common ratio

An increase of p% multiplies by 1 + p/100. A decrease of p% multiplies by 1 − p/100.

The starting amount, or the first bounce, is u₁.

$u_n = u_1 r^{n-1}$

A model uses one common ratio

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

$u_k = u_1 r^{k-1}$

Easy 1

Percent Change

The first amount is term 1.

Example

The first amount is 80. The amount increases by 50% each time. Find the 4th amount.

The amount increases by 50%, so multiply by 150%.

$r = 1 + \dfrac{50}{100}$

$r = \dfrac{3}{2}$

$u_n = u_1 r^{n-1}$

$u_{4} = 80 \times \left(\dfrac{3}{2}\right)^{3}$

$\left(\dfrac{3}{2}\right)^{3} = \dfrac{27}{8}$

$u_{4} = 80 \times \dfrac{27}{8}$

$u_{4} = 270$

Answer: the amount = $270$

The first amount is 64. The amount increases by 50% each time. Find the 5th amount.

Show Answer

The amount increases by 50%, so multiply by 150%.

$r = 1 + \dfrac{50}{100}$

$r = \dfrac{3}{2}$

$u_n = u_1 r^{n-1}$

$u_{5} = 64 \times \left(\dfrac{3}{2}\right)^{4}$

$\left(\dfrac{3}{2}\right)^{4} = \dfrac{81}{16}$

$u_{5} = 64 \times \dfrac{81}{16}$

$u_{5} = 324$

Answer: the amount = $324$

The first amount is 200. The amount doubles each time. Find the 4th amount.

Show Answer

The amount doubles. Doubling is an increase of 100%, so r = 2.

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{4} = 200 \times 2^{3}$

$2^{3} = 8$

$u_{4} = 200 \times 8$

$u_{4} = 1600$

Answer: the amount = $1600$

The first amount is 160. The amount decreases by 50% each time. Find the 4th amount.

Show Answer

The amount decreases by 50%, so multiply by 50%.

$r = 1 - \dfrac{50}{100}$

$r = \dfrac{1}{2}$

$u_n = u_1 r^{n-1}$

$u_{4} = 160 \times \left(\dfrac{1}{2}\right)^{3}$

$\left(\dfrac{1}{2}\right)^{3} = \dfrac{1}{8}$

$u_{4} = 160 \times \dfrac{1}{8}$

$u_{4} = 20$

Answer: the amount = $20$

The first amount is 250. The amount increases by 20% each time. Find the 3rd amount.

Show Answer

The amount increases by 20%, so multiply by 120%.

$r = 1 + \dfrac{20}{100}$

$r = \dfrac{6}{5}$

$u_n = u_1 r^{n-1}$

$u_{3} = 250 \times \left(\dfrac{6}{5}\right)^{2}$

$\left(\dfrac{6}{5}\right)^{2} = \dfrac{36}{25}$

$u_{3} = 250 \times \dfrac{36}{25}$

$u_{3} = 360$

Answer: the amount = $360$

The first amount is 81. The amount doubles each time. Find the 5th amount.

Show Answer

The amount doubles. Doubling is an increase of 100%, so r = 2.

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{5} = 81 \times 2^{4}$

$2^{4} = 16$

$u_{5} = 81 \times 16$

$u_{5} = 1296$

Answer: the amount = $1296$

Easy 2

Money

The balance at the start of year 1 is the first term.

Example

An account has $200 at the start of year 1. It increases by 50% each year. Find the balance at the start of year 4.

The amount increases by 50%, so multiply by 150%.

$r = 1 + \dfrac{50}{100}$

$r = \dfrac{3}{2}$

$u_n = u_1 r^{n-1}$

$u_{4} = 200 \times \left(\dfrac{3}{2}\right)^{3}$

$\left(\dfrac{3}{2}\right)^{3} = \dfrac{27}{8}$

$u_{4} = 200 \times \dfrac{27}{8}$

$u_{4} = 675$

Answer: the amount = $675$

An account has $400 at the start of year 1. It increases by 50% each year. Find the balance at the start of year 3.

Show Answer

The amount increases by 50%, so multiply by 150%.

$r = 1 + \dfrac{50}{100}$

$r = \dfrac{3}{2}$

$u_n = u_1 r^{n-1}$

$u_{3} = 400 \times \left(\dfrac{3}{2}\right)^{2}$

$\left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$

$u_{3} = 400 \times \dfrac{9}{4}$

$u_{3} = 900$

Answer: the amount = $900$

An account has $100 at the start of year 1. It doubles each year. Find the balance at the start of year 6.

Show Answer

The amount doubles. Doubling is an increase of 100%, so r = 2.

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{6} = 100 \times 2^{5}$

$2^{5} = 32$

$u_{6} = 100 \times 32$

$u_{6} = 3200$

Answer: the amount = $3200$

An account has $500 at the start of year 1. It increases by 20% each year. Find the balance at the start of year 3.

Show Answer

The amount increases by 20%, so multiply by 120%.

$r = 1 + \dfrac{20}{100}$

$r = \dfrac{6}{5}$

$u_n = u_1 r^{n-1}$

$u_{3} = 500 \times \left(\dfrac{6}{5}\right)^{2}$

$\left(\dfrac{6}{5}\right)^{2} = \dfrac{36}{25}$

$u_{3} = 500 \times \dfrac{36}{25}$

$u_{3} = 720$

Answer: the amount = $720$

An account has $80 at the start of year 1. It increases by 50% each year. Find the balance at the start of year 5.

Show Answer

The amount increases by 50%, so multiply by 150%.

$r = 1 + \dfrac{50}{100}$

$r = \dfrac{3}{2}$

$u_n = u_1 r^{n-1}$

$u_{5} = 80 \times \left(\dfrac{3}{2}\right)^{4}$

$\left(\dfrac{3}{2}\right)^{4} = \dfrac{81}{16}$

$u_{5} = 80 \times \dfrac{81}{16}$

$u_{5} = 405$

Answer: the amount = $405$

An account has $320 at the start of year 1. It decreases by 50% each year. Find the balance at the start of year 4.

Show Answer

The amount decreases by 50%, so multiply by 50%.

$r = 1 - \dfrac{50}{100}$

$r = \dfrac{1}{2}$

$u_n = u_1 r^{n-1}$

$u_{4} = 320 \times \left(\dfrac{1}{2}\right)^{3}$

$\left(\dfrac{1}{2}\right)^{3} = \dfrac{1}{8}$

$u_{4} = 320 \times \dfrac{1}{8}$

$u_{4} = 40$

Answer: the amount = $40$

Easy 3

Growth

The population at the start is term 1.

Example

A population is 500 at the start of year 1 and doubles each year. Find the population at the start of year 5.

The amount doubles. Doubling is an increase of 100%, so r = 2.

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{5} = 500 \times 2^{4}$

$2^{4} = 16$

$u_{5} = 500 \times 16$

$u_{5} = 8000$

Answer: the amount = $8000$

A bacteria count is 40 at the start of hour 1 and doubles each hour. Find the count at the start of hour 6.

Show Answer

The amount doubles. Doubling is an increase of 100%, so r = 2.

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{6} = 40 \times 2^{5}$

$2^{5} = 32$

$u_{6} = 40 \times 32$

$u_{6} = 1280$

Answer: the amount = $1280$

A town’s population is 120 at the start of year 1 and increases by 50% each year. Find the population at the start of year 4.

Show Answer

The amount increases by 50%, so multiply by 150%.

$r = 1 + \dfrac{50}{100}$

$r = \dfrac{3}{2}$

$u_n = u_1 r^{n-1}$

$u_{4} = 120 \times \left(\dfrac{3}{2}\right)^{3}$

$\left(\dfrac{3}{2}\right)^{3} = \dfrac{27}{8}$

$u_{4} = 120 \times \dfrac{27}{8}$

$u_{4} = 405$

Answer: the amount = $405$

A colony is 1000 at the start of day 1 and increases by 20% each day. Find the size at the start of day 3.

Show Answer

The amount increases by 20%, so multiply by 120%.

$r = 1 + \dfrac{20}{100}$

$r = \dfrac{6}{5}$

$u_n = u_1 r^{n-1}$

$u_{3} = 1000 \times \left(\dfrac{6}{5}\right)^{2}$

$\left(\dfrac{6}{5}\right)^{2} = \dfrac{36}{25}$

$u_{3} = 1000 \times \dfrac{36}{25}$

$u_{3} = 1440$

Answer: the amount = $1440$

A herd is 64 at the start of year 1 and decreases by 50% each year. Find the herd at the start of year 4.

Show Answer

The amount decreases by 50%, so multiply by 50%.

$r = 1 - \dfrac{50}{100}$

$r = \dfrac{1}{2}$

$u_n = u_1 r^{n-1}$

$u_{4} = 64 \times \left(\dfrac{1}{2}\right)^{3}$

$\left(\dfrac{1}{2}\right)^{3} = \dfrac{1}{8}$

$u_{4} = 64 \times \dfrac{1}{8}$

$u_{4} = 8$

Answer: the amount = $8$

A cell count is 25 at the start of hour 1 and doubles each hour. Find the count at the start of hour 5.

Show Answer

The amount doubles. Doubling is an increase of 100%, so r = 2.

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{5} = 25 \times 2^{4}$

$2^{4} = 16$

$u_{5} = 25 \times 16$

$u_{5} = 400$

Answer: the amount = $400$

Easy 4

Bounce

The first bounce is term 1.

Example

A tennis ball’s first bounce reaches 80 cm. Each later bounce is 50% of the height of the previous bounce. Find the height of the 5th bounce.

Each later bounce is 50% of the previous bounce.

$r = \dfrac{50}{100}$

$r = \dfrac{1}{2}$

The first bounce is term 1.

$u_1 = 80$

$u_n = u_1 r^{n-1}$

$u_{5} = 80 \times \left(\dfrac{1}{2}\right)^{4}$

$\left(\dfrac{1}{2}\right)^{4} = \dfrac{1}{16}$

$u_{5} = 80 \times \dfrac{1}{16}$

$u_{5} = 5$

Answer: the height = $5$ cm

A golf ball’s first bounce reaches 64 cm. Each later bounce is 50% of the height of the previous bounce. Find the height of the 6th bounce.

Show Answer

Each later bounce is 50% of the previous bounce.

$r = \dfrac{50}{100}$

$r = \dfrac{1}{2}$

The first bounce is term 1.

$u_1 = 64$

$u_n = u_1 r^{n-1}$

$u_{6} = 64 \times \left(\dfrac{1}{2}\right)^{5}$

$\left(\dfrac{1}{2}\right)^{5} = \dfrac{1}{32}$

$u_{6} = 64 \times \dfrac{1}{32}$

$u_{6} = 2$

Answer: the height = $2$ cm

A basketball’s first bounce reaches 160 cm. Each later bounce is 50% of the height of the previous bounce. Find the height of the 5th bounce.

Show Answer

Each later bounce is 50% of the previous bounce.

$r = \dfrac{50}{100}$

$r = \dfrac{1}{2}$

The first bounce is term 1.

$u_1 = 160$

$u_n = u_1 r^{n-1}$

$u_{5} = 160 \times \left(\dfrac{1}{2}\right)^{4}$

$\left(\dfrac{1}{2}\right)^{4} = \dfrac{1}{16}$

$u_{5} = 160 \times \dfrac{1}{16}$

$u_{5} = 10$

Answer: the height = $10$ cm

A ping-pong ball’s first bounce reaches 96 cm. Each later bounce is 50% of the height of the previous bounce. Find the height of the 4th bounce.

Show Answer

Each later bounce is 50% of the previous bounce.

$r = \dfrac{50}{100}$

$r = \dfrac{1}{2}$

The first bounce is term 1.

$u_1 = 96$

$u_n = u_1 r^{n-1}$

$u_{4} = 96 \times \left(\dfrac{1}{2}\right)^{3}$

$\left(\dfrac{1}{2}\right)^{3} = \dfrac{1}{8}$

$u_{4} = 96 \times \dfrac{1}{8}$

$u_{4} = 12$

Answer: the height = $12$ cm

A superball’s first bounce reaches 128 cm. Each later bounce is 50% of the height of the previous bounce. Find the height of the 6th bounce.

Show Answer

Each later bounce is 50% of the previous bounce.

$r = \dfrac{50}{100}$

$r = \dfrac{1}{2}$

The first bounce is term 1.

$u_1 = 128$

$u_n = u_1 r^{n-1}$

$u_{6} = 128 \times \left(\dfrac{1}{2}\right)^{5}$

$\left(\dfrac{1}{2}\right)^{5} = \dfrac{1}{32}$

$u_{6} = 128 \times \dfrac{1}{32}$

$u_{6} = 4$

Answer: the height = $4$ cm

A rubber ball’s first bounce reaches 250 cm. Each later bounce is 80% of the height of the previous bounce. Find the height of the 4th bounce.

Show Answer

Each later bounce is 80% of the previous bounce.

$r = \dfrac{80}{100}$

$r = \dfrac{4}{5}$

The first bounce is term 1.

$u_1 = 250$

$u_n = u_1 r^{n-1}$

$u_{4} = 250 \times \left(\dfrac{4}{5}\right)^{3}$

$\left(\dfrac{4}{5}\right)^{3} = \dfrac{64}{125}$

$u_{4} = 250 \times \dfrac{64}{125}$

$u_{4} = 128$

Answer: the height = $128$ cm

Medium

Predict Number

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

Example

Four readings are 8, 20, 40, and 64. Model them as a geometric sequence and predict the 6th reading.

The ratios from one reading to the next are not equal.

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

$u_{4} = u_1 r^{3}$

$64 = 8 \times r^{3}$

$r^{3} = \dfrac{64}{8}$

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{6} = 8 \times 2^{5}$

$2^{5} = 32$

$u_{6} = 8 \times 32$

$u_{6} = 256$

Answer: the model gives $256$

Four readings are 5, 12, 20, and 40. Model them as a geometric sequence and predict the 6th reading.

Show Answer

The ratios from one reading to the next are not equal.

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

$u_{4} = u_1 r^{3}$

$40 = 5 \times r^{3}$

$r^{3} = \dfrac{40}{5}$

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{6} = 5 \times 2^{5}$

$2^{5} = 32$

$u_{6} = 5 \times 32$

$u_{6} = 160$

Answer: the model gives $160$

Four readings are 3, 8, 14, and 24. Model them as a geometric sequence and predict the 6th reading.

Show Answer

The ratios from one reading to the next are not equal.

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

$u_{4} = u_1 r^{3}$

$24 = 3 \times r^{3}$

$r^{3} = \dfrac{24}{3}$

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{6} = 3 \times 2^{5}$

$2^{5} = 32$

$u_{6} = 3 \times 32$

$u_{6} = 96$

Answer: the model gives $96$

Four readings are 4, 10, 18, and 32. Model them as a geometric sequence and predict the 7th reading.

Show Answer

The ratios from one reading to the next are not equal.

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

$u_{4} = u_1 r^{3}$

$32 = 4 \times r^{3}$

$r^{3} = \dfrac{32}{4}$

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{7} = 4 \times 2^{6}$

$2^{6} = 64$

$u_{7} = 4 \times 64$

$u_{7} = 256$

Answer: the model gives $256$

Five readings are 2, 6, 10, 16, and 32. Model them as a geometric sequence and predict the 8th reading.

Show Answer

The ratios from one reading to the next are not equal.

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

$u_{5} = u_1 r^{4}$

$32 = 2 \times r^{4}$

$r^{4} = \dfrac{32}{2}$

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{8} = 2 \times 2^{7}$

$2^{7} = 128$

$u_{8} = 2 \times 128$

$u_{8} = 256$

Answer: the model gives $256$

Four readings are 9, 20, 36, and 72. Model them as a geometric sequence and predict the 6th reading.

Show Answer

The ratios from one reading to the next are not equal.

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

$u_{4} = u_1 r^{3}$

$72 = 9 \times r^{3}$

$r^{3} = \dfrac{72}{9}$

$r = 2$

$u_n = u_1 r^{n-1}$

$u_{6} = 9 \times 2^{5}$

$2^{5} = 32$

$u_{6} = 9 \times 32$

$u_{6} = 288$

Answer: the model gives $288$

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