9.3 Geometric Sequences And Series

Geometric Sequences

A geometric sequenceA sequence of numbers where each successive number is the product of the previous number and some constant r., or geometric progressionUsed when referring to a geometric sequence., is a sequence of numbers where each successive number is the product of the previous number and some constant r.

an=ran−1     Geometic Sequence

And because anan−1=r, the constant factor r is called the common ratioThe constant r that is obtained from dividing any two successive terms of a geometric sequence; anan−1=r.. For example, the following is a geometric sequence,

9,27,81,243,729…

Here a1=9 and the ratio between any two successive terms is 3. We can construct the general term an=3an−1 where,

a1=9a2=3a1=3(9)=27a3=3a2=3(27)=81a4=3a3=3(81)=243a5=3a4=3(243)=729⋮

In general, given the first term a1 and the common ratio r of a geometric sequence we can write the following:

a2=ra1a3=ra2=r(a1r)=a1r2a4=ra3=r(a1r2)=a1r3a5=ra3=r(a1r3)=a1r4⋮

From this we see that any geometric sequence can be written in terms of its first element, its common ratio, and the index as follows:

an=a1rn−1       Geometric Sequence

In fact, any general term that is exponential in n is a geometric sequence.

Example 1

Find an equation for the general term of the given geometric sequence and use it to calculate its 10th term: 3,6,12,24,48…

Solution:

Begin by finding the common ratio,

r=63=2

Note that the ratio between any two successive terms is 2. The sequence is indeed a geometric progression where a1=3 and r=2.

an=a1rn−1=3(2)n−1

Therefore, we can write the general term an=3(2)n−1 and the 10th term can be calculated as follows:

a10=3(2)10−1=3(2)9=1,536

Answer: an=3(2)n−1; a10=1,536

The terms between given terms of a geometric sequence are called geometric meansThe terms between given terms of a geometric sequence..

Example 2

Find all terms between a1=−5 and a4=−135 of a geometric sequence. In other words, find all geometric means between the 1st and 4th terms.

Solution:

Begin by finding the common ratio r. In this case, we are given the first and fourth terms:

an=a1rn−1  Use n = 4.a4=a1r4−1a4=a1r3

Substitute a1=−5 and a4=−135 into the above equation and then solve for r.

−135=−5r327=r33=r

Next use the first term a1=−5 and the common ratio r=3 to find an equation for the nth term of the sequence.

an=a1rn−1an=−5(3)n−1

Now we can use an=−5(3)n−1 where n is a positive integer to determine the missing terms.

a1=−5(3)1−1=−5⋅30=−5a2=−5(3)2−1=−5⋅31=−15a3=−5(3)3−1=−5⋅32=−45     }    geometic meansa4=−5(3)4−1=−5⋅33=−135

Answer: −15, −45,

The first term of a geometric sequence may not be given.

Example 3

Find the general term of a geometric sequence where a2=−2 and a5=2125.

Solution:

To determine a formula for the general term we need a1 and r. A nonlinear system with these as variables can be formed using the given information and an=a1rn−1:

 {a2=a1r2−1a5=a1r5−1  ⇒     {−2=a1r2125=a1r4  Use a2=−2.  Use a5=2125.

Solve for a1 in the first equation,

{  −2=a1r ⇒ −2r=a12125=a1r4

Substitute a1=−2r into the second equation and solve for r.

2125=a1r42125=(−2r)r42125=−2r3−1125=r3−15=r

Back substitute to find a1:

a1=−2r=−2(−15)=10

Therefore, a1=10 and r=−15.

Answer: an=10(−15)n−1

Try this! Find an equation for the general term of the given geometric sequence and use it to calculate its 6th term: 2,43,89,…

Answer: an=2(23)n−1; a6=64243

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