Geometric Sequences

Definition of a geometric sequence.

An geometric sequence is a sequence in which each term afterthe first is found by multiplyingthe previous term by a constantcalled the common ratio, r.

You can name the terms of a geometric sequence using a1, a2, a3,and so onIf we define the nth term as an thenthen previous term is an-1.And by the definition of a geometricsequence an = r(an-1) now solve for r

r =an

an-1

Example 1. Find the next two termsin the geometric sequence 3, 12, 48, ...

First find the common ratio.

Let 3 = an-1 and let 12 = an.

r = an

an-1

=12 3

= 4

The common ratio is 4.

Example 1. Geometric sequence of3, 12, 48, … find a4 and a5.The common ratio is 4.

a4 = r(a3) = 4(48) = 192

a5 = r(a4) = 4(192) = 768

The next two terms are 192 and 768

There is a pattern in the way theterms of a geometric sequenceare formed.

Let’s look at example 1.

numerical

symbols

3 12 48 192

a1 a2 a3 a4 an

In terms of r and the previous term

a1 = a1a2 = r • a1

a3 = r • a2 a4 = r • a3

In terms of r and the first termnumerical

a1 = 3(40) a2 = 3(41)

a3 = 3(42) a4 = 3(43)

symbols

a1 = a1(r0) a2 = a1(r1)

a3 = a1(r2) a4 = a1(r3)

Formula for the nth term of a geometric sequence.

The nth term of a geometric sequence with first term a1 and common ratio r is given by

or

an = an-1 • r

an = a1 • rn-1

Example 2.Write the first six terms of a Write the first six terms of a geometric sequence in which geometric sequence in which aa11 = 3 and r = 2 = 3 and r = 2

Method 1 use aMethod 1 use ann = a = an-1n-1(r)(r)aa11 = 3 = 3 aa22 = 3•2 = 6 = 3•2 = 6 aa33 = 6•2 = 12 = 6•2 = 12

aa44 = 12•2 = 24 = 12•2 = 24 aa55 = 24•2 = 48 = 24•2 = 48

aa66 = 48•2 = 96 = 48•2 = 96

Example 2.Write the first six terms of a Write the first six terms of a geometric sequence in which geometric sequence in which aa11 = 3 and r = 2 = 3 and r = 2

Method 2 use aMethod 2 use ann = a = a11(r(rn-1n-1))aa11 = 3•2 = 3•21-11-1 = 3 = 3 aa22 = 3•2 = 3•22-12-1 = 6 = 6

aa33 = 3•2 = 3•23-13-1 = 12 = 12 aa44 = 3•2 = 3•24-14-1 = 24 = 24

aa55 = 3•2 = 3•25-15-1 = 48 = 48 aa66 = 3•2 = 3•26-16-1 = 96 = 96

Example 3. Find the ninth term of a geometricFind the ninth term of a geometricsequence in which asequence in which a33 = 63 and = 63 and

r = -3.r = -3.Method 1 use the common ratio Method 1 use the common ratio and the given term.and the given term.aa44 = a = a33(-3) = 63(-3) = -189(-3) = 63(-3) = -189

aa55 = a = a44(-3) = (-189)(-3) = 567(-3) = (-189)(-3) = 567

Example 3. Ninth term with a3 = 63and r = -3Method 1 use the common ratio Method 1 use the common ratio and the given term.and the given term.aa44 = a = a33(-3) = 63(-3) = -189(-3) = 63(-3) = -189

aa55 = a = a44(-3) = (-189)(-3) = 567(-3) = (-189)(-3) = 567

aa66 = a = a55(-3) = (567)(-3) = -1701(-3) = (567)(-3) = -1701

aa77 = a = a66(-3) = (-1701)(-3) = 5103(-3) = (-1701)(-3) = 5103

Example 3. Ninth term with a3 = 63and r = -3Method 1 use the common ratio Method 1 use the common ratio and the given term.and the given term.aa66 = a = a55(-3) = (567)(-3) = -1701(-3) = (567)(-3) = -1701

aa77 = a = a66(-3) = (-1701)(-3) = 5103(-3) = (-1701)(-3) = 5103

aa88 = a = a77(-3) = (5103)(-3) = -15309(-3) = (5103)(-3) = -15309

aa99 = a = a88(-3) = (-15309)(-3) = 45927(-3) = (-15309)(-3) = 45927

Example 3. Ninth term with a3 = 63and r = -3Method 2 find aMethod 2 find a11

aann = a = a11(r(rn-1n-1)) aa33 = a = a11(r(r3-13-1))

63 = a63 = a11(-3)(-3)(2)(2) 63 = a63 = a11(9)(9)

aa11 = 93/9 = 7 = 93/9 = 7

aa99 = a = a11(r(r(9-1)(9-1))) = 7(-3)= 7(-3)8 8 = 45927= 45927

The terms between any two nonconsecutive terms of a geometric sequence are called thegeometric means.

In the sequence3, 12, 48, 192, 769, ...

12, 48, and 192 are the three geometric means between 3 and 769

Example 4.

Find the three geometric means between 3.4 and 2125.

Use the nth term formula to find r.

3.4, ____, ____, ____, 2125

3.4 is a1 2125 is a5

Example 4.Find the three geometric means between 3.4 and 2125.Use the nth term formula to find r.3.4, ____, ____, ____, 2125

3.4 is a1 2125 is a5

an = a1(rn-1) a5 = 3.4(r4)

2125 = 3.4(r4) 625 = (r4) r = 5

Example 4. Three geometric meansbetween 3.4 and 2125

3.4 is a1 2125 is a5r = 5

Check both solutions

If r = 5 a2 = 3.4(5) = 17

a3 = 17(5) = 85 a4 = 85(5) = 425

a5 = 425(5) = 2125

Example 4. Three geometric meansbetween 3.4 and 2125

3.4 is a1 2125 is a5r = 5

Check both solutions

If r = -5 a2 = 3.4(-5) = -17

a3 = -17(-5) = 85 a4 = 85(-5) = -425

a5 = -425(-5) = 2125Both solutions check

Example 4. Three geometric meansbetween 3.4 and 2125

3.4 is a1 2125 is a5r = 5

Both solutions check

There are two sets of geometric means between 3.4 and 2125.17, 85, and 425 and

-17, 85, and -425