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
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 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 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.
We have enough information to compute S _{5} .
Because 3 ^{7} = 2187, n − 1 = 7, so n = 8.
Finite, Infinite, and Partial Sums
When the common ratio is small enough (− 1 < r < 1 and r ≠ 0), the sum of all terms in a geometric sequence is a number. In the finite sum , r ^{n} is very small when the ratio is a fraction, so 1 − r ^{n} is very close to 1. Using this fact and calculus techniques (usually learned in a later calculus course), it can be shown that the sum of all terms of this kind of geometric sequence is
The only difference between the infinite sum formula and the partial sum formula is that 1 − r ^{n} is replaced by 1. If n is large enough, there is very little difference between the partial sum and the entire sum. We will compare the sum of the first 20 terms of the sequence whose n th term is with the sum of all terms.
Examples
Geometric Sequences Practice Problems
Practice
 What term comes after 18 in the sequence
 Find the first four terms and the tenth term of the geometric sequence whose n th term is a _{n} = −2(4) ^{n −1} .
 Determine if the sequence 900, 90, 9, 0.9, 0.09, ... is geometric.
 Determine if the sequence 9, 99, 999, 9999, ... is geometric.
 Find the n th term of the geometric sequence(s) whose first term is 9 and whose fifth term is .
 Find the n th term of the geometric sequence whose common ratio is −3 and whose sixth term is −1701.
 Find the n th term of the geometric sequence whose third term is 1 and whose sixth term is .
 Compute the sum.
Solutions

3(18) = 54

a _{1} = −2, a _{2} = −8, a _{3} = −32, a _{4} = −128 and a _{10} = −524, 288

The sequence is geometric because the following ratios are the same.

The sequence is not geometric because the ratios are not the same.

Because the fifth term of the sequence is , we have the equation = 9 r ^{5−1} . Once we have solved this equation for r , we will be done.
There are two sequences. The n th term for one of them is and the other is .

The sixth term is −1701 and r = −3, which gives us the equation − 1701 = a _{1} (−3) ^{6−1} .
−1701 = a _{1} (−3) ^{5}
−1701 = −243 a _{1}
7 = a _{1}
The n th term is a _{n} = 7(−3) ^{n −1}.

The third term is 1 and the sixth term is , which gives us the system of equations . Solving 1 = a _{1} r ^{2} for a _{1}, we get a _{1} = 1/ r ^{2}. We will substitute this in .
The n th term is .

. We know but we need n . We will solve for n .
Because 2 ^{6} = 64, n − 1 = 6, so n = 7. Now we can find the sum.

. This is all we need for the infinite sum formula.

a _{1} is not −4 because the sum begins at n = 0 instead of n = 1.
Now we can add all of the terms of the geometric sequence whose n th term is .
Geometric Sequences to Solve Savings and Lottery Problems
When regular payments are made to a savings account or to a lottery winner, the monthly balances act like terms in a geometric sequence. The common ratio is either 1 + i (for savings payments) or (1 + i ) ^{−1} (for lottery payments), where i is the interest rate per payment period. We learned in Chapter 9 that if we leave P dollars in an account, earning annual interest r , compounded n times per year, for t years, then this will grow to A dollars where A = P (1 + r / n ) ^{nt}. (This is why i replaces r / n .)
Savings Problem
We will see what happens to the balance of an account if $2000 is deposited on January 1 every year for 5 years, earning 10% per year, compounded annually. The first $2000 will earn interest for the entire 5 years, so it will grow to 2000 (1 + 0.10/1) ^{5} = 2000(1.10) ^{5} . The second $2000 will earn interest for 4 years, so it will grow to 2000(1.10) ^{4} . The third $2000 will earn interest for 3 years, so it will grow to 2000(1.10) ^{3} . The fourth $2000 will earn interest for 2 years, so it will grow to 2000(1.10) ^{2} . And the fifth $2000 will earn interest for 1 year, so it will grow to 2000(1.10) ^{1}. The balance after five years is
2000(1.10) ^{5} + 2000(1.10) ^{4} + 2000(1.10) ^{3} + 2000(1.10) ^{2} + 2000(1.10) ^{1}.
This is the sum of the first five terms of the geometric sequence whose n th term is a _{n} = 2000(1.10) ^{n}. If we want to use the partial sum formula, we need to rewrite the n th term in the form a _{n} = a _{1} r ^{n−1}. We will use exponent properties to change 2000(1.10) ^{n} to a _{1} (1.10) ^{n−1} . We will also use the fact that n = 1 + n − 1.
2000(1.10) ^{n} = 2000(1.10) ^{1+ n −1} = 2000(1.10) ^{1} (1.10) ^{n −1}
= [2000(1. 10)](1.10) ^{n −1} = 2200(1. 10) ^{n −1}
Now we can use the partial sum formula.
The balance in the account will be $13,431.22.
Lottery Problem
When a lottery winner wins a $1,000,000 jackpot, the money is likely to be paid out in $50,000 annual payments for 20 years. Some states allow the winner to take the cash value as a lump sum payment instead. The cash value is the present value of $1,000,000 to be paid in annual payments over 20 years. The formula for the present value of A dollars, due in t years, earning annual interest r, compounded n times per year is A (1 + r / n ) ^{− nt}. Assume that the money is expected to earn 5% per year. Then the cash value of the jackpot will need to be enough money so that at the beginning of the year (for a payment at the end of the year), they have 50,000(1.05) ^{−1} . For a payment at the end of two years, they need 50,000(1.05) ^{−2} ; at the end of three years, they need 50,000(1.05) ^{−3} , and so on until they reach the last payment after 20 years, 50,000(1.05) ^{−20} . In other words, the cash value of a $1,000,000 jackpot with a 20year payout (assuming 5% interest) is
50,000(1.05) ^{−1} + 50,000(1.05) ^{−2} + 50,000(1.05) ^{−3} + ... + 50,000(1.05) ^{−20} .
This is the sum of the first 20 terms of the geometric sequence whose n th term is a _{n} = 50,000(1.05) ^{− n}. We need to use exponent properties to rewrite the n th term in the form a _{n} = a _{1} r ^{n −1}. We will use the fact that −n = −n −1 + 1 and the exponent facts that x ^{m + n} = x ^{m} x ^{n} and x ^{mn} = ( x ^{m} ) ^{n} .
1.05 ^{−n} = 1.05 ^{− n +1−1} = 1.05 ^{−1} · 1.05 ^{− n +1} = 1.05 ^{−1} · 1.05 ^{−1( n −1)}
= 1.05 ^{−1} · (1.05 ^{−1} ) ^{n −1}
Now the n th term can be written as a _{n} = [50,000(1.05) ^{−1} ](1.05 ^{−1} ) ^{n −1} , where a _{1} = 50,000(1.05 ^{−1} ). Now we can use the partial sum formula.
The cash value is $623,110.51.
Practice problems for this concept can be found at: Sequence and Series Practice Test.