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a 1 = 2 a 2 = ( 2 4 ) = 8 a 3 = ( 8 4 ) = 32 a 4 = ( 32 4 ) 128

The first four terms are { –2 –8 –32 –128 } .

Given the first term and the common factor, find the first four terms of a geometric sequence.

  1. Multiply the initial term, a 1 , by the common ratio to find the next term, a 2 .
  2. Repeat the process, using a n = a 2 to find a 3 and then a 3 to find a 4, until all four terms have been identified.
  3. Write the terms separated by commons within brackets.

Writing the terms of a geometric sequence

List the first four terms of the geometric sequence with a 1 = 5 and r = –2.

Multiply a 1 by 2 to find a 2 . Repeat the process, using a 2 to find a 3 , and so on.

a 1 = 5 a 2 = 2 a 1 = 10 a 3 = 2 a 2 = 20 a 4 = 2 a 3 = 40

The first four terms are { 5 , –10 , 20 , –40 } .

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List the first five terms of the geometric sequence with a 1 = 18 and r = 1 3 .

{ 18 , 6 , 2 , 2 3 , 2 9 }

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Using recursive formulas for geometric sequences

A recursive formula    allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.

Recursive formula for a geometric sequence

The recursive formula for a geometric sequence with common ratio r and first term a 1 is

a n = r a n 1 , n 2

Given the first several terms of a geometric sequence, write its recursive formula.

  1. State the initial term.
  2. Find the common ratio by dividing any term by the preceding term.
  3. Substitute the common ratio into the recursive formula for a geometric sequence.

Using recursive formulas for geometric sequences

Write a recursive formula for the following geometric sequence.

{ 6 9 13.5 20.25 ... }

The first term is given as 6. The common ratio can be found by dividing the second term by the first term.

r = 9 6 = 1.5

Substitute the common ratio into the recursive formula for geometric sequences and define a 1 .

a n = r a n 1 a n = 1.5 a n 1  for  n 2 a 1 = 6
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Do we have to divide the second term by the first term to find the common ratio?

No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio.

Write a recursive formula for the following geometric sequence.

{ 2 4 3 8 9 16 27 ... }

a 1 = 2 a n = 2 3 a n 1  for  n 2

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Using explicit formulas for geometric sequences

Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.

a n = a 1 r n 1

Let’s take a look at the sequence { 18 36 72 144 288 ... } . This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is

a n = 18 · 2 n 1

The graph of the sequence is shown in [link] .

Graph of the geometric sequence.

Explicit formula for a geometric sequence

The n th term of a geometric sequence is given by the explicit formula    :

a n = a 1 r n 1

Writing terms of geometric sequences using the explicit formula

Given a geometric sequence with a 1 = 3 and a 4 = 24 , find a 2 .

The sequence can be written in terms of the initial term and the common ratio r .

3 , 3 r , 3 r 2 , 3 r 3 , ...

Find the common ratio using the given fourth term.

a n = a 1 r n 1 a 4 = 3 r 3 Write the fourth term of sequence in terms of  α 1 and  r 24 = 3 r 3 Substitute  24  for a 4 8 = r 3 Divide r = 2 Solve for the common ratio

Find the second term by multiplying the first term by the common ratio.

a 2 = 2 a 1 = 2 ( 3 ) = 6
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Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
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I rally confuse this number And equations too I need exactly help
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But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
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Sherica
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Uday
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salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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Practice Key Terms 2

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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