Exercise 51 Page 265

a The tank contains

8000

liters of water, and after each day half of the water in the tank is removed. Therefore, if we multiply the original amount of water by

\frac{1}{2}

we get what is left in the tank by the end of day

1 .

Initial value \to 8000 \cdot \frac{1}{2}

Similarly, by the end of day

2

we multiply the volume of water at the end of day

1

\frac{1}{2} .

end of day 2 8000 \cdot \frac{1}{2} end of day 1 \cdot \frac{1}{2}

We notice that the amount of water can be described by a geometric sequence where

8000

is the initial value

a_{0},

and

\frac{1}{2}

is the common ratio.

a_{n} = 8000 (\frac{1}{2})^{n}

To find the amount of water at the end of the

4^{th}

day, we substitute

n = 4

in our rule and simplify.

a_{n} = 8000 (\frac{1}{2})^{n}

$n = 4$

a_{4} = 8000 (\frac{1}{2})^{4}

▼

Simplify right-hand side

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

a_{4} = 8000 (\frac{1}{8})

$a \cdot \frac{1}{b} = \frac{a}{b}$

a_{4} = \frac{4 0 0 0}{8}

Calculate quotient

a_{4} = 500

At the end of the

4^{th}

day there will be

500

liters of water in the tank.

b To find the amount of water after the

8^{th}

day, we substitute

n = 8

in our rule from Part A.

a_{n} = 8000 (\frac{1}{2})^{n}

$n = 8$

a_{8} = 8000 (\frac{1}{2})^{8}

▼

Simplify right-hand side

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

a_{8} = 8000 (\frac{1}{2 5 6})

$a \cdot \frac{1}{b} = \frac{a}{b}$

a_{8} = \frac{4 0 0 0}{2 5 6}

Calculate quotient

a_{8} = 31.25

Therefore, at the end of the

8^{th}

day there will be

31.25

liters of water in the tank.

5.2.1 Generating and Investigating Sequences p.264-265

5.2.2 Generalizing Arithmetic Sequences p.269-270

5.2.3 Recursive Sequences p.273-274