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Geometric sequences, just like arithmetic sequences, can be described using explicit rules. For geometric sequences, the first term, a1, is also used to find the rule. Instead of the common difference of the arithmetic sequences, the common ratio is used for the geometric ones.

Every geometric sequence can be described by a function known as the explicit rule which receives as input the position of a term n, and returns as output the term's value $a_{n}.$ The explicit rule for a geometric sequence has the following general form.

Here, a1 is the first term of the sequence and r is the common ratio.

Note that every geometric sequence has a common ratio r. Therefore, it is possible to obtain every term of the sequence by multiplying enough times the first term a1 by this common ratio. Hence, knowing a1 and r is enough to describe the whole geometric sequence.

By making a table it is easier to identify a pattern and write a general expression. Note that r0 is equal to 1, and that r can be written as r1.

n | $a_{n}$ | Using a1 and r |
---|---|---|

1 | a1 | a1⋅r0 |

2 | a2 | a1⋅r1 |

3 | a3 | a1⋅r2 |

4 | a4 | a1⋅r3 |

It can be seen that the exponent of the common ratio is always 1 less than the value of the position n. This allows to write the explicit rule as the formula given above.

The first four terms of a geometric sequence are

96,48,24,and 12.

Find the explicit rule describing the geometric sequence. Then, use the rule to find the eighth term of the sequence. Show Solution

To write the explicit rule for the sequence, we first have to find the common ratio, r. To do so, we can divide any term in the sequence by the term that precedes it. Let's use the second and first term.
Substituting r=0.5 and a1=96 into the general rule for geometric sequences gives the desired rule.
Now, we can find the eighth term in the sequence by substituting n=8 into the rule above.
The eighth term in the sequence is 0.75.

$a_{n}=96⋅0.5_{n−1}$

Substitute

n=8

$a_{8}=96⋅0.5_{8−1}$

SubTerm

Subtract term

$a_{8}=96⋅0.5_{7}$

UseCalc

Use a calculator

a8=0.75

Pelle's good friend, Lisa, decides to play a trick on Pelle. While he is away, she rearranges his pellets so that they are grouped in a geometric sequence instead of an arithmetic one. The first group has 2 pellets, the second has 6, the third has 18, and so on. Find a rule describing this sequence. After finishing the seventh group, Lisa counted 3273 remaining pellets. Use the rule to figure out whether there are enough to make an eighth group.

Show Solution

To begin, we'll write the explicit rule describing this particular geometric sequence. It is given that a1=2. To find the common ratio, r, we can divide the second term by the first.
To write the rule, we can substitute a1=2 and r=3 into the general rule for geometric sequences.
To find if there are enough pellets to finish the eighth group, we must know the eighth term in the sequence. We'll substitute n=8 into the rule.
The eighth term, $a_{8}=4374,$ is greater than the number of remaining pellets, 3273. Thus, there are not enough pellets for Lisa to make the eighth group.

$a_{n}=2⋅3_{n−1}$

Substitute

n=8

$a_{8}=2⋅3_{8−1}$

SubTerm

Subtract term

$a_{8}=2⋅3_{7}$

CalcPow

Calculate power

$a_{8}=2⋅2187$

Multiply

Multiply

$a_{8}=4374$

For a geometric sequence, it is known that the common ratio is positive, and that

$a_{2}=4anda_{4}=64.$

Find the explicit rule for the sequence and give its first six terms. Show Solution

The terms we've been given are not consecutive. Therefore, we can't directly find r. However, the terms a2 and a4 are 2 positions apart, so the ratio between them must be r2.

This gives the equation$a_{2}a_{4} =r_{2}$

SubstituteII

$a_{4}=64$, $a_{2}=4$

$464 =r_{2}$

CalcQuot

Calculate quotient

16=r2

RearrangeEqn

Rearrange equation

r2=16

SqrtEqn

$LHS =RHS $

$r=±16 $

CalcRoot

Calculate root

r=±4

$r>0$

r=4

The desired explicit rule is $a_{n}=4_{n−1}.$ We already know the terms a1,a2, and a4. Let's use the rule to find the remaining three.

The terms a5 and a6 are evaluated similarly.

n | $4_{n−1}$ | $a_{n}$ |
---|---|---|

3 | $4_{3−1}$ | 16 |

5 | $4_{5−1}$ | 256 |

6 | $4_{6−1}$ | 1024 |

1,4,16,64,256,and 1024.

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