Algebra Examples

160

$160$ ,

80

$80$ ,

40

$40$

Step 1

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by

\frac{1}{2}

$\frac{1}{2}$ gives the next term. In other words,

a_{n} = a_{1} r^{n - 1}

$a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence:

r = \frac{1}{2}

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

Step 2

This is the form of a geometric sequence.

a_{n} = a_{1} r^{n - 1}

$a_{n} = a_{1} r^{n - 1}$

Step 3

Substitute in the values of

a_{1} = 160

$a_{1} = 160$ and

r = \frac{1}{2}

$r = \frac{1}{2}$ .

a_{n} = 160 {(\frac{1}{2})}^{n - 1}

$a_{n} = 160 {(\frac{1}{2})}^{n - 1}$

Step 4

Apply the product rule to

\frac{1}{2}

$\frac{1}{2}$ .

a_{n} = 160 \frac{1^{n - 1}}{2^{n - 1}}

$a_{n} = 160 \frac{1^{n - 1}}{2^{n - 1}}$

Step 5

One to any power is one.

a_{n} = 160 \frac{1}{2^{n - 1}}

$a_{n} = 160 \frac{1}{2^{n - 1}}$

Step 6

Combine

160

$160$ and

\frac{1}{2^{n - 1}}

$\frac{1}{2^{n - 1}}$ .

a_{n} = \frac{160}{2^{n - 1}}

$a_{n} = \frac{160}{2^{n - 1}}$

Step 7

Substitute in the value of

n

$n$ to find the

n

$n$ th term.

a_{4} = \frac{160}{2^{(4) - 1}}

$a_{4} = \frac{160}{2^{(4) - 1}}$

Step 8

Simplify the denominator.

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Step 8.1

Subtract

1

$1$ from

4

$4$ .

a_{4} = \frac{160}{2^{3}}

$a_{4} = \frac{160}{2^{3}}$

Step 8.2

Raise

2

$2$ to the power of

3

$3$ .

a_{4} = \frac{160}{8}

$a_{4} = \frac{160}{8}$

a_{4} = \frac{160}{8}

$a_{4} = \frac{160}{8}$

Step 9

Divide

160

$160$ by

8

$8$ .

a_{4} = 20

$a_{4} = 20$