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Geometric sequences, just like arithmetic sequences, can be described using explicit rules. For geometric sequences, the first term, $a_1,$ 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.

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 $r$ |
---|---|---|

$1$ | $a_1$ | $a_1 \cdot r^0$ |

$2$ | $a_2$ | $a_1 \cdot r^1$ |

$3$ | $a_3$ | $a_1 \cdot r^2$ |

$4$ | $a_4$ | $a_1 \cdot r^3$ |

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.

$a_n = a_1 \cdot r^{n - 1}$

The first four terms of a geometric sequence are $96, 48, 24, \text{and } 12.$ 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. $r = \dfrac{48}{96} = 0.5$ Substituting $r=0.5$ and $a_1=96$ into the general rule for geometric sequences gives the desired rule. $a_n=a_1 \cdot r^{n-1} \quad \Rightarrow \quad a_n = 96 \cdot 0.5^{n - 1}$ Now, we can find the eighth term in the sequence by substituting $n=8$ into the rule above.

$a_n=96\cdot0.5^{n-1}$

Substitute$n={\color{#0000FF}{8}}$

$a_{\color{#0000FF}{8}}=96\cdot 0.5^{{\color{#0000FF}{8}}-1}$

SubTermSubtract term

$a_8=96\cdot0.5^{7}$

UseCalcUse a calculator

$a_8=0.75$

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 $a_1=2.$ To find the common ratio, $r,$ we can divide the second term by the first. $r = \dfrac{6}{2} = 3$ To write the rule, we can substitute $a_1=2$ and $r=3$ into the general rule for geometric sequences. $a_{n} = a_{1} \cdot r^{n-1} \quad \Rightarrow \quad a_n = 2 \cdot 3^{n-1}$ 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.

$a_{n} = 2 \cdot 3^{n-1}$

Substitute$n={\color{#0000FF}{8}}$

$a_{{\color{#0000FF}{8}}} = 2 \cdot 3^{{\color{#0000FF}{8}}-1}$

SubTermSubtract term

$a_{8} = 2 \cdot 3^{7}$

CalcPowCalculate power

$a_{8} = 2\cdot 2187$

MultiplyMultiply

$a_{8} = 4374$

For a geometric sequence, it is known that the common ratio is positive, and that $a_2 = 4 \quad \text{and} \quad a_4 = 64.$ 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 $\dfrac{a_4}{a_2} = r^2,$ which we can solve for $r.$

$\dfrac{a_4}{a_2} = r^2$

$\dfrac{{\color{#0000FF}{64}}}{{\color{#009600}{4}}} = r^2$

CalcQuotCalculate quotient

$16 = r^2$

RearrangeEqnRearrange equation

$r^2 = 16$

SqrtEqn$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$r = \pm \sqrt{16}$

CalcRootCalculate root

$r = \pm 4$

$r \gt 0$

$r = 4$

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 $a_1.$ $a_1 = \dfrac{a_2}{r} \quad \Rightarrow \quad a_1 = \dfrac{4}{4} = 1$ With $a_1$ and $r,$ we have enough information to state the explicit rule.

$a_n = a_1 \cdot r^{n - 1}$

$a_n = {\color{#0000FF}{1}} \cdot {\color{#009600}{4}}^{n - 1}$

MultiplyMultiply

$a_n = 4^{n - 1}$

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$n={\color{#0000FF}{3}}$

$a_{\color{#0000FF}{3}} = 4^{{\color{#0000FF}{3}} - 1}$

SubTermSubtract term

$a_3 = 4^2$

CalcPowCalculate power

$a_3 = 16$

The terms $a_5$ and $a_6$ are evaluated similarly.

$n$ | $4^{n - 1}$ | $a_n$ |
---|---|---|

${\color{#0000FF}{3}}$ | $4^{{\color{#0000FF}{3}} - 1}$ | $16$ |

${\color{#0000FF}{5}}$ | $4^{{\color{#0000FF}{5}} - 1}$ | $256$ |

${\color{#0000FF}{6}}$ | $4^{{\color{#0000FF}{6}} - 1}$ | $1024$ |

Thus, the first six terms of the sequence are $1, 4, 16, 64, 256, \text{and } 1024.$

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