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Geometric Sequence Solutions

A sequence is an ordered list of numbers that contains a pattern. 
A geometric sequence is an ordered list of numbers in which each term after $a_{1}$ is found by multiplying the previous term by the common ratio, r. 
The common ratio is the number that each term is multiplied by to find the next term in the sequence .
A finite list can be written with variables as: $a_{1}, a_{1} r, a_{1} r^{2}, a_{1} r^{3}, . . ., a_{1} r^{n - 1}$  
An infinite list can be written with variables as: $a_{1}, a_{1} r, a_{1} r^{2}, a_{1} r^{3}, . . .$
 To find successive terms in a geometric sequence when $a_{1}$ and r are known: $a_{n + 1} = a_{n} \cdot r$  
To find the $n^{t h}$ term in a geometric sequence, use the formula: $a_{n} = (a_{1}) \cdot r^{n - 1}$ , where n is any natural number.
 To find r, divide successive terms in a geometric sequence: $r = \frac{a_{n + 1}}{a_{n}}$

Example 1

For the geometric sequence, ${128, 32, 8, 2, . . .}$ , complete the following:

Determine the common ratio.

Implement

Worked solution content here

$r = \frac{a_{n + 1}}{a_{n}} = \frac{a_{4}}{a_{3}} = \frac{2}{8} = \frac{1}{4}$

Explain

Choose any two consecutive terms

Substitute terms into: $r = \frac{a_{n + 1}}{a_{n}}$

Simplify

Determine the next term in the geometric sequence.

Implement

Next term = 5th

$a_{n} = 2, r = \frac{1}{4} \begin{matrix} \begin{array}{rcl} a_{n + 1} & = & a_{n} \cdot r \\ a_{5} & = & 2 (\frac{1}{4}) = \frac{2}{4} \\ a_{5} & = & \frac{1}{2} \end{array} \end{matrix}$

Explain

Define terms

Substitute terms into: $a_{n + 1} = a_{n} \cdot r$

Solve

Find the ninth term in the sequence.

Implement

$n = 9, a_{1} = 128, r = \frac{1}{4} \begin{matrix} a_{n} = (a_{1}) \cdot r^{n - 1} \\ a_{9} = (128) {(\frac{1}{4})}^{9 - 1} \\ a_{9} = 128 {(\frac{1}{4})}^{8} = \frac{1}{512} \end{matrix}$

Explain

Define terms

Substitute terms into: $a_{n} = (a_{1}) \cdot r^{n - 1}$

Solve

Example 2

Find the first term, $a_{1}$ , whose fourth and fifth terms are –54 and 162, respectively.

Plan

Solve for r

Substitute known values, $a_{4} and r$ , into $a_{n} = (a_{1}) \cdot r^{n - 1}$

Solve for $a_{1}$

Implement

$a_{4} = - 54, a_{5} = 162 r = \frac{a_{n + 1}}{a_{n}} = \frac{a_{5}}{a_{4}} = \frac{162}{- 54} = - 3$

$\begin{matrix} a_{n} = (a_{1}) \cdot r^{n - 1} \\ a_{4} = a_{1} {(- 3)}^{4 - 1} \\ - 54 = a_{1} {(- 3)}^{3} \\ - 54 = a_{1} (- 27) \\ a_{1} = \frac{- 54}{- 27} = 2 \end{matrix}$