Geometric Sequences Help

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

1. What term comes after 18 in the sequence
2. Find the first four terms and the tenth term of the geometric sequence whose n th term is a _n = −2(4) ^{n −1} .
3. Determine if the sequence 900, 90, 9, 0.9, 0.09, ... is geometric.
4. Determine if the sequence 9, 99, 999, 9999, ... is geometric.
5. Find the n th term of the geometric sequence(s) whose first term is 9 and whose fifth term is .
6. Find the n th term of the geometric sequence whose common ratio is −3 and whose sixth term is −1701.
7. Find the n th term of the geometric sequence whose third term is 1 and whose sixth term is .
8. Compute the sum.

Solutions

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 20-year 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.