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Writing and Using Explicit Rules for Geometric Sequences

Geometric sequences, just like arithmetic sequences, can be described using explicit rules. For geometric sequences, the first term, a1,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.

Explicit Rule of Geometric Sequences

All geometric sequences have a common ratio, r.r. Using the common ratio, together with the value of the first term of the sequence, a1,a_1, an explicit rule describing the sequence can be found. By expressing the terms in a geometric sequence using a1a_1 and r,r, a pattern emerges. Note that r0r^0 is equal to 11, and that rr can be written as r1.r^1.

nn ana_n Using a1a_1 and rr
11 a1a_1 a1r0a_1 \cdot r^0
22 a2a_2 a1r1a_1 \cdot r^1
33 a3a_3 a1r2a_1 \cdot r^2
44 a4a_4 a1r3a_1 \cdot r^3

When nn increases by 1,1, the exponent on rr increases by 11 as well. Due to this, and that the exponent is 00 when nn is 1,1, the exponent is always 11 less than n.n. Expressing this in a general form gives the explicit rule.

an=a1rn1a_n = a_1 \cdot r^{n - 1}

Thus, knowing a1a_1 and rr is enough to write the explicit rule of a geometric sequence. Notice that the common ratio could be a negative number, leading to alternating positive and negative terms in the sequence.

The first four terms of a geometric sequence are 96,48,24,and 12. 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, rr. 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=4896=0.5 r = \dfrac{48}{96} = 0.5 Substituting r=0.5r=0.5 and a1=96a_1=96 into the general rule for geometric sequences gives the desired rule. an=a1rn1an=960.5n1 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=8n=8 into the rule above.

a8=960.581a_{\color{#0000FF}{8}}=96\cdot 0.5^{{\color{#0000FF}{8}}-1}
The eighth term in the sequence is

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 22 pellets, the second has 6,6, the third has 18,18, and so on. Find a rule describing this sequence. After finishing the seventh group, Lisa counted 32733273 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.a_1=2. To find the common ratio, r,r, we can divide the second term by the first. r=62=3 r = \dfrac{6}{2} = 3 To write the rule, we can substitute a1=2a_1=2 and r=3r=3 into the general rule for geometric sequences. an=a1rn1an=23n1 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=8n = 8 into the rule.

an=23n1a_{n} = 2 \cdot 3^{n-1}
a8=2381a_{{\color{#0000FF}{8}}} = 2 \cdot 3^{{\color{#0000FF}{8}}-1}
a8=237a_{8} = 2 \cdot 3^{7}
a8=22187a_{8} = 2\cdot 2187
a8=4374a_{8} = 4374
The eighth term, a8=4374,a_{8}=4374, is greater than the number of remaining pellets, 3273.3273. Thus, there are not enough pellets for Lisa to make the eighth group.

For a geometric sequence, it is known that the common ratio is positive, and that a2=4anda4=64. 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.r. However, the terms a2a_2 and a4a_4 are 22 positions apart, so the ratio between them must be r2.r^2.

This gives the equation a4a2=r2, \dfrac{a_4}{a_2} = r^2, which we can solve for r.r.

a4a2=r2\dfrac{a_4}{a_2} = r^2
644=r2\dfrac{{\color{#0000FF}{64}}}{{\color{#009600}{4}}} = r^2
16=r216 = r^2
r2=16r^2 = 16
r=±16r = \pm \sqrt{16}
r=±4r = \pm 4
r>0 r \gt 0
r=4r = 4

Now that we know the common ratio, we have to find a1a_1 as well, to be able to write the explicit rule. Knowing one term, a subsequent one can by found by multiplying by r.r. Therefore, a previous term is instead found by dividing by r.r. Using a2a_2 and rr this way, we can find a1.a_1. a1=a2ra1=44=1 a_1 = \dfrac{a_2}{r} \quad \Rightarrow \quad a_1 = \dfrac{4}{4} = 1 With a1a_1 and r,r, we have enough information to state the explicit rule.

an=a1rn1a_n = a_1 \cdot r^{n - 1}
an=14n1a_n = {\color{#0000FF}{1}} \cdot {\color{#009600}{4}}^{n - 1}
an=4n1a_n = 4^{n - 1}

The desired explicit rule is an=4n1.a_n = 4^{n - 1}. We already know the terms a1,a2,a_1, a_2, and a4.a_4. Let's use the rule to find the remaining three.

an=4n1a_n = 4^{n - 1}
a3=431a_{\color{#0000FF}{3}} = 4^{{\color{#0000FF}{3}} - 1}
a3=42a_3 = 4^2
a3=16a_3 = 16

The terms a5a_5 and a6a_6 are evaluated similarly.

nn 4n14^{n - 1} ana_n
3{\color{#0000FF}{3}} 4314^{{\color{#0000FF}{3}} - 1} 1616
5{\color{#0000FF}{5}} 4514^{{\color{#0000FF}{5}} - 1} 256256
6{\color{#0000FF}{6}} 4614^{{\color{#0000FF}{6}} - 1} 10241024

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

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