# Geometric Series

### Learning Objectives

- Write the first n terms of a geometric sequence
- Solve an annuity problem using a geometric series

**geometric series**. Recall that a

**geometric sequence**is a sequence in which the ratio of any two consecutive terms is the

**common ratio**, [latex]r[/latex]. We can write the sum of the first [latex]n[/latex] terms of a geometric series as

[latex]{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}[/latex].

Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first [latex]n[/latex] terms of a geometric series. We will begin by multiplying both sides of the equation by [latex]r[/latex].[latex]r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}[/latex]

Next, we subtract this equation from the original equation.[latex]\begin{array}{l}\\ \frac{\begin{array}{l}\text{ }{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}\hfill \\ -r{S}_{n}=-\left(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\right)\hfill \end{array}}{\left(1-r\right){S}_{n}={a}_{1}-{r}^{n}{a}_{1}}\end{array}[/latex]

Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for [latex]{S}_{n}[/latex], divide both sides by[latex]\left(1-r\right)[/latex]. [latex]{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\text{ r}\ne \text{1}[/latex]

### A General Note: Formula for the Sum of the First *n* Terms of a Geometric Series

A **geometric series**is the sum of the terms in a geometric sequence. The formula for the sum of the first [latex]n[/latex] terms of a geometric sequence is represented as

[latex]{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\text{ r}\ne \text{1}[/latex]

### How To: Given a geometric series, find the sum of the first *n* terms.

- Identify [latex]{a}_{1},r,\text{and}n[/latex].
- Substitute values for [latex]{a}_{1},r[/latex], and [latex]n[/latex] into the formula [latex]{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}[/latex].
- Simplify to find [latex]{S}_{n}[/latex].

### Example: Finding the First *n* Terms of a Geometric Series

Use the formula to find the indicated partial sum of each geometric series.
- [latex]{S}_{11}[/latex] for the series [latex]\text{ 8 + -4 + 2 + }\dots [/latex]
- [latex]\underset{6}{\overset{k=1}{{\sum }^{\text{ }}}}3\cdot {2}^{k}[/latex]

Answer:

- [latex]{a}_{1}=8[/latex], and we are given that [latex]n=11[/latex].We can find [latex]r[/latex] by dividing the second term of the series by the first. [latex-display]r=\frac{-4}{8}=-\frac{1}{2}[/latex-display] Substitute values for [latex]{a}_{1}, r, \text{and} n[/latex] into the formula and simplify. [latex]\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{11}=\frac{8\left(1-{\left(-\frac{1}{2}\right)}^{11}\right)}{1-\left(-\frac{1}{2}\right)}\approx 5.336\hfill \end{array}[/latex]
- Find [latex]{a}_{1}[/latex] by substituting [latex]k=1[/latex] into the given explicit formula. [latex-display]{a}_{1}=3\cdot {2}^{1}=6[/latex-display] We can see from the given explicit formula that [latex]r=2[/latex]. The upper limit of summation is 6, so [latex]n=6[/latex].Substitute values for [latex]{a}_{1},r[/latex], and [latex]n[/latex] into the formula, and simplify. [latex]\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{6}=\frac{6\left(1-{2}^{6}\right)}{1 - 2}=378\hfill \end{array}[/latex]

### Try It

Use the formula to find the indicated partial sum of each geometric series. [latex-display]{S}_{20}[/latex] for the series [latex]\text{ 1,000 + 500 + 250 + }\dots [/latex-display]Answer: [latex]\approx 2,000.00[/latex]

## Annuities

At the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An**annuity**is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the initial deposit. The account paid 6%

**annual interest**, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added. We can find the value of the annuity right after the last deposit by using a geometric series with [latex]{a}_{1}=50[/latex] and [latex]r=100.5%=1.005[/latex]. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned. We can find the value of the annuity after [latex]n[/latex] deposits using the formula for the sum of the first [latex]n[/latex] terms of a geometric series. In 6 years, there are 72 months, so [latex]n=72[/latex]. We can substitute [latex]{a}_{1}=50, r=1.005, \text{and} n=72[/latex] into the formula, and simplify to find the value of the annuity after 6 years.

[latex]{S}_{72}=\frac{50\left(1-{1.005}^{72}\right)}{1 - 1.005}\approx 4\text{,}320.44[/latex]

After the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of [latex]72\left(50\right) = $3,600[/latex]. This means that because of the annuity, the couple earned $720.44 interest in their college fund.### How To: Given an initial deposit and an interest rate, find the value of an annuity.

- Determine [latex]{a}_{1}[/latex], the value of the initial deposit.
- Determine [latex]n[/latex], the number of deposits.
- Determine [latex]r[/latex].
- Divide the annual interest rate by the number of times per year that interest is compounded.
- Add 1 to this amount to find [latex]r[/latex].

- Substitute values for [latex]{a}_{1}\text{,}r,\text{and}n[/latex] into the formula for the sum of the first [latex]n[/latex] terms of a geometric series, [latex]{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}[/latex].
- Simplify to find [latex]{S}_{n}[/latex], the value of the annuity after [latex]n[/latex] deposits.

### Example: Solving an Annuity Problem

A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit?Answer: The value of the initial deposit is $100, so [latex]{a}_{1}=100[/latex]. A total of 120 monthly deposits are made in the 10 years, so [latex]n=120[/latex]. To find [latex]r[/latex], divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.

[latex]r=1+\frac{0.09}{12}=1.0075[/latex]

Substitute [latex]{a}_{1}=100\text{,}r=1.0075\text{,}\text{and}n=120[/latex] into the formula for the sum of the first [latex]n[/latex] terms of a geometric series, and simplify to find the value of the annuity.[latex]{S}_{120}=\frac{100\left(1-{1.0075}^{120}\right)}{1 - 1.0075}\approx 19\text{,}351.43[/latex]

So the account has $19,351.43 after the last deposit is made.### Try It

At the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years?Answer: $92,408.18

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