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# Geometric Sequences Help

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## Introduction to Geometric Sequences

In an arithmetic sequence, the difference of any two consecutive terms is the same, and in a geometric sequence, the quotient of any two consecutive terms is the same. A term in a geometric sequence can be found by multiplying the previous term by a fixed number. For example, the next term in the sequence 1,2,4,8,16, ... is 2(16)=32, and the term after that is 2(32)=64. This fixed number is called the common ratio . We can define the n th term of a geometric sequence recursively by a n = ra n −1 . The general formula is a n = a 1 r n −1 .

#### Examples

• Determine if the sequence 5, 15, 45, 135,405, ... is geometric.
• We need to see if the ratio of each consecutive pair of numbers is the same.

The ratio is the same number, so the sequence is geometric.

• Determine if the sequence is geometric.

The ratio is the same number, so the sequence is geometric.

• Determine if the sequence 2430, 729, 240.57, 80.10981, ... is geometric.

The ratios are different, so this is not a geometric sequence.

• Find the first four terms and the tenth term of the sequence .

## Finding the n th Term of a Geometric Sequence

We can find the n th term of a geometric sequence by either knowing one term and the common ratio or by knowing two terms. This is similar to what we did to find the n th term of an arithmetic sequence.

#### Examples

Find the n th term of the geometric sequence.

• The common ratio is 3 and the fourth term is 54.
• a 4 = 54 and r = 3, becomes . This gives us a 1 = 2. The n th term is .

• The third term is 320, and the fifth term is 204.8.
• a 3 = 320 and a 5 = 204.8 give us the system of equations 320 = a 1 r 3−1 and 204.8 = a 1 r 5−1 . Elimination by addition will not work for the systems in this section, so we will use substitution. Solving for a 1 in a 1 r 2 = 320 gives us a 1 = 320/ r 2. Substituting this in a 1 r 4 = 204.8 gives us the following.

There are two geometric sequences whose third term is 320 and whose fifth term is 204.8, one has a common ratio of 0.8 and the other, −0.8. a 1 for both the sequences is the same.

The n th term for one sequence is a n = 500(0.8) n −1 , and the other is a n = 500(−0.8) n −1 .

• The third term is 20, and the sixth term is 81.92.
• From a 3 = 20 and a 6 = 81.92 we have the system of equations 20 = a 1 r 3−1 and 81.92 = a 1 r 6−1 . We will solve for a 1 in 20 = a 1 r 2 . Now we will substitute a 1 = 20/ r 2 for a 1 in 81.92 = a 1 r 5 .

The n th term is a n = 7.8125(1.6) n −1 .

We can add the first n terms of a geometric sequence using the following formula (except for r = 1).

#### Examples

• Find the sum of the first five terms of the geometric sequence whose n th term is a n = 3(2) n −1 , a 1 = 3 and r = 2.

• Compute
• Find the sum of the first five terms of the geometric sequence whose fourth term is 1.3824 and whose seventh term is 2.3887872.
• We need to find a 1 and r . The terms a 4 = 1.3824 and a 7 = 2.3887872 give us the system of equations 1.3824 = a 1 r 3 and 2.3887872 = a 1 r 6 . We will solve for a 1 in the first equation and substitute this for a 1 in the second equation.

We have enough information to compute S 5 .

• We are adding the first six terms of the geometric sequence whose n th term is a n = 6.4(1.5) n −1 .

• This problem is tricky because the sum begins with k = 0 instead of k = 1. These terms are the first eight terms of the geometric sequence Now we can see that n = 8, a 1 = .

• We have a 1 = 54 and . We need n for .

Because 3 7 = 2187, n − 1 = 7, so n = 8.

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