# Geometric Sequences.

## Presentation on theme: "Geometric Sequences."— Presentation transcript:

Geometric Sequences

Definition of a geometric
sequence. An geometric sequence is a sequence in which each term after the first is found by multiplying the previous term by a constant called the common ratio, r.

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

Example 1. Find the next two terms
in 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 of
3, 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 the
terms of a geometric sequence are formed.

Let’s look at example 1. 3 12 48 192 numerical symbols a1 a2 a3 a4 an In terms of r and the previous term a2 = r • a1 a1 = a1 a3 = r • a2 a4 = r • a3

In terms of r and the first term
numerical a2 = 3(41) a1 = 3(40) a3 = 3(42) a4 = 3(43) symbols a2 = a1(r1) a1 = a1(r0) 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 an = an-1 • r or an = a1 • rn-1

Example 2. Write the first six terms of a geometric sequence in which a1 = 3 and r = 2 Method 1 use an = an-1(r) a3 = 6•2 = 12 a1 = 3 a2 = 3•2 = 6 a5 = 24•2 = 48 a4 = 12•2 = 24 a6 = 48•2 = 96

Example 2. Write the first six terms of a geometric sequence in which a1 = 3 and r = 2 Method 2 use an = a1(rn-1) a1 = 3•21-1 = 3 a2 = 3•22-1 = 6 a4 = 3•24-1 = 24 a3 = 3•23-1 = 12 a5 = 3•25-1 = 48 a6 = 3•26-1 = 96

Example 3. Find the ninth term of a geometric sequence in which a3 = 63 and r = -3. Method 1 use the common ratio and the given term. a4 = a3(-3) = 63(-3) = -189 a5 = a4(-3) = (-189)(-3) = 567

Example 3. Ninth term with a3 = 63
and r = -3 Method 1 use the common ratio and the given term. a4 = a3(-3) = 63(-3) = -189 a5 = a4(-3) = (-189)(-3) = 567 a6 = a5(-3) = (567)(-3) = -1701 a7 = a6(-3) = (-1701)(-3) = 5103

Example 3. Ninth term with a3 = 63
and r = -3 Method 1 use the common ratio and the given term. a6 = a5(-3) = (567)(-3) = -1701 a7 = a6(-3) = (-1701)(-3) = 5103 a8 = a7(-3) = (5103)(-3) = a9 = a8(-3) = (-15309)(-3) = 45927

Example 3. Ninth term with a3 = 63
and r = -3 Method 2 find a1 a3 = a1(r3-1) an = a1(rn-1) 63 = a1(-3)(2) 63 = a1(9) a1 = 93/9 = 7 a9 = a1(r(9-1)) = 7(-3)8 = 45927

The terms between any two
nonconsecutive terms of a geometric sequence are called the geometric means. In the sequence 3, 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 a5 = 3.4(r4) an = a1(rn-1) 2125 = 3.4(r4) 625 = (r4) r = 5

Example 4. Three geometric means
between 3.4 and 2125 r = 5 3.4 is a1 2125 is a5 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 means
between 3.4 and 2125 r = 5 3.4 is a1 2125 is a5 Check both solutions If r = -5 a2 = 3.4(-5) = -17 a4 = 85(-5) = -425 a3 = -17(-5) = 85 a5 = -425(-5) = 2125 Both solutions check

Example 4. Three geometric means
between 3.4 and 2125 r = 5 3.4 is a1 2125 is a5 Both solutions check There are two sets of geometric means between 3.4 and 2125. 17, 85, and 425 and -17, 85, and -425