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{{ printedBook.courseTrack.name }} {{ printedBook.name }} A geometric sequence is a sequence where the ratio $r$ of any term to its preceding term is a constant other than $0.$ This ratio is called the common ratio. In the following geometric sequence, the first term is $3,$ and the common ratio is $2.$

Each term of a geometric sequence is multiplied by the common ratio $r$ to get the next term. Like any other sequence, the first term of a geometric sequence is denoted by $a_{1},$ the second $a_{2},$ and so on.

Therefore, geometric sequences have the following form.

$a_{1},a_{1}r,a_{1}r_{2},a_{1}r_{3},a_{1}r_{4},… $Consider the following geometric sequence. $2,6,18,54,162,…$ Determine the common ratio and find the next three terms.

Show Solution

In geometric sequences, the terms increase or decrease by a common ratio. Since we know that this sequence is geometric, it's enough to find the ratio between two consecutive terms. The ratio for the others must then be the same. Let's take the first two: $2and6.$ If we let $r$ be the common ratio we get the equation $2⋅r=6⇔r=3.$ The common ratio, $r,$ is $3.$ To find the next terms, we multiply by $3,$ three times. $162⋅3486⋅31458⋅3 =486=1458=4374 $ In summary, the common ratio is $3,$ and the next three terms are $486,$ $1458,$ and $4374.$

All geometric sequences have a common ratio, $r.$ Using the common ratio, together with the value of the first term of the sequence, $a_{1},$ an explicit rule describing the sequence can be found. By expressing the terms in a geometric sequence using $a_{1}$ and $r,$ a pattern emerges. Note that $r_{0}$ is equal to $1$, and that $r$ can be written as $r_{1}.$

$n$ | $a_{n}$ | Using $a_{1}$ and $r$ |
---|---|---|

$1$ | $a_{1}$ | $a_{1}⋅r_{0}$ |

$2$ | $a_{2}$ | $a_{1}⋅r_{1}$ |

$3$ | $a_{3}$ | $a_{1}⋅r_{2}$ |

$4$ | $a_{4}$ | $a_{1}⋅r_{3}$ |

When $n$ increases by $1,$ the exponent on $r$ increases by $1$ as well. Due to this, and that the exponent is $0$ when $n$ is $1,$ the exponent is always $1$ less than $n.$ Expressing this in a general form gives the explicit rule.

$a_{n}=a_{1}⋅r_{n−1}$

The first four terms of a geometric sequence are $96,48,24,and12.$ 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. $r=9648 =0.5$ Substituting $r=0.5$ and $a_{1}=96$ into the general rule for geometric sequences gives the desired rule. $a_{n}=a_{1}⋅r_{n−1}⇒a_{n}=96⋅0.5_{n−1}$ Now, we can find the eighth term in the sequence by substituting $n=8$ into the rule above.

$a_{n}=96⋅0.5_{n−1}$

Substitute

$n=8$

$a_{8}=96⋅0.5_{8−1}$

SubTerm

Subtract term

$a_{8}=96⋅0.5_{7}$

UseCalc

Use a calculator

$a_{8}=0.75$

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 $a_{2}$ and $a_{4}$ are $2$ positions apart, so the ratio between them must be $r_{2}.$

This gives the equation $a_{2}a_{4} =r_{2},$ which we can solve for $r.$

$a_{2}a_{4} =r_{2}$

SubstituteII

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

$464 =r_{2}$

CalcQuot

Calculate quotient

$16=r_{2}$

RearrangeEqn

Rearrange equation

$r_{2}=16$

SqrtEqn

$LHS =RHS $

$r=±16 $

CalcRoot

Calculate root

$r=±4$

$r>0$

$r=4$

Now that we know the common ratio, we have to find $a_{1}$ as well, to be able to write the explicit rule. Knowing one term, a subsequent one can by found by multiplying by $r.$ Therefore, a previous term is instead found by dividing by $r.$ Using $a_{2}$ and $r$ this way, we can find $a_{1}.$ $a_{1}=ra_{2} ⇒a_{1}=44 =1$ With $a_{1}$ and $r,$ we have enough information to state the explicit rule.

The desired explicit rule is $a_{n}=4_{n−1}.$ We already know the terms $a_{1},a_{2},$ and $a_{4}.$ Let's use the rule to find the remaining three.

$a_{n}=4_{n−1}$

Substitute

$n=3$

$a_{3}=4_{3−1}$

SubTerm

Subtract term

$a_{3}=4_{2}$

CalcPow

Calculate power

$a_{3}=16$

The terms $a_{5}$ and $a_{6}$ are evaluated similarly.

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

$3$ | $4_{3−1}$ | $16$ |

$5$ | $4_{5−1}$ | $256$ |

$6$ | $4_{6−1}$ | $1024$ |

Thus, the first six terms of the sequence are $1,4,16,64,256,and1024.$

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 $a_{1}=2.$ To find the common ratio, $r,$ we can divide the second term by the first. $r=26 =3$ To write the rule, we can substitute $a_{1}=2$ and $r=3$ into the general rule for geometric sequences. $a_{n}=a_{1}⋅r_{n−1}⇒a_{n}=2⋅3_{n−1}$ 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.

$a_{n}=2⋅3_{n−1}$

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

$a_{8}=4374$

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