# Geometric Sequences

## Writing Formulas for Geometric Sequences

A **geometric sequence** is a list of numbers, where the next term of the sequence is found by multiplying the term by a constant, called the common ratio.

The general form of the geometric sequence formula is: \(a_n=a_1r^{(n-1)}\), where \(r\) is the common ratio, \(a_1\) is the first term, and \(n\) is the placement of the term in the sequence.

Here is a geometric sequence: \(1,3,9,27,81,…\)

To find the formula for this geometric sequence, start by determining the common ratio, which is \(3\), since the terms are increasing by a factor of \(3\). Then identify the first term, \(a_1\), which is \(1\). Therefore, the formula for this geometric sequence is \(a_n=1·3^{(n-1)}\).

Use the formula for the geometric sequence to find the \(8^{th}\) term:

\(a_8=1·3^{(8-1)}\)

\(a_8=2{,}187\)

**Example:**

Here is a geometric sequence: \(2,10,50,250,1{,}250,…\)

Find the formula for the geometric sequence, then find the \(7^{th}\) term.

Common ratio is \(5\), first term is \(2\), therefore the formula for the geometric sequence is, \(a_n=2·5^{(n-1)}\).

\(a_7=2·5^{(7-1)}=31{,}250\)

## Finding Terms in Geometric Sequences

### What is a geometric sequence?

A geometric sequence is an ordered set of numbers in which each term is a fixed multiple of the number that comes before it. Geometric sequences use multiplication to find each subsequent term. Each term gets multiplied by a common ratio, resulting in the next term in the sequence. In the geometric sequence shown below, the common ratio is 2. In other words, each term is multiplied by 2. The resulting product is the next term in the sequence.

\(1,2,4,8,16,32,64…\)

### How do you find a term in a geometric sequence when given a formula?

The formula \(a_n=a_1r^{(n-1)}\) is used to identify any number in a given geometric sequence. In this formula, \(n\) stands for the number in the sequence that needs to be identified. \(a_1\) stands for the first term in the sequence, and \(r\) stands for the common ratio. Consider how this formula applies to the following geometric sequence:

\(2,10,50,250…\)

In this geometric sequence, the common ratio, or \(r\), equals \(5\). As the sequence progresses, each term is multiplied by \(5\). Use the formula given to identify the \(6^{th}\) term in this sequence.

Since the variable \(n\) stands for the number in the sequence that needs to be identified, replace \(n\) with \(6\) in the formula. The variable \(a_1\) stands for the first term in the sequence, which is \(2\). The variable r represents the common ratio, so replace \(r\) with \(5\) in the formula:

\(n=6\)

\(a_1=2\)

\(r=5\)

From here, rewrite the formula and substitute the variables with the numbers they represent:

\(a_6=(2)(5)^{(6-1)}\) | Now that the variables are replaced with their corresponding values, solve the equation using the Order of Operations. |

\(a_6=(2)(5)^{(5)}\) | First, simplify the exponents. Since \(6-1=5\), we can rewrite the equation using the exponent \(5\). |

\(a_6=(10)(3{,}125)\) | Next, simplify the exponent \(5^5\). \(5^5=5×5×5×5×5\), which equals \(3{,}125\). Rewrite the equation using \(3{,}125\). |

\(a_6=31{,}250\) | From here, multiply \(10×3{,}125\) to get the final answer. Since \(10×3{,}125=31{,}250\), the \(6^{th}\) term in the geometric sequence is \(31{,}250\). |

### How do you find a term in a geometric sequence without a formula?

A missing term in a geometric sequence can be found without using a formula. Analyze the pattern and consider the common ratio to identify the missing term. Consider the following geometric sequence:

\(10,30,90,270,\) _____ \(,2{,}430…\)

First, analyze the pattern to identify the common ratio. Start with the first term, \(10\). \(10\) multiplied by what number results in a product of \(30\)? The correct answer is \(3\), so the common ratio is \(3\). Check the rest of the terms using this common ratio. \(30×3=90\) and \(90×3=270\).

Next, multiply \(270\) by \(3\) to get the missing term in the geometric sequence. Since \(270×3=810\), the missing term is \(810\).

\(10,30,90,270,\mathbf{810},2{,}430…\)

Check your work by multiplying \(810\) by \(3\). If correct, the product should be the subsequent term in the sequence, \(2{,}450\). Since \(810×3=2{,}430\), the answer is correct.

Hello, and welcome to this video on geometric sequences!

Today we’ll explore how to find a term in a geometric sequence using a formula. We’ll also learn how to find the sum of a specific number of terms in a geometric sequence.

Before we dive in, let’s review a few things. First, a **sequence** is an ordered set of numbers with a pattern. The pattern helps us predict what each term might be. For example, 2, 4, 6, 8, and 10 are a sequence.

A **term** is an individual expression, or number, in the sequence.

A **geometric sequence** is a special type of sequence. Each term is a fixed multiple of the number that comes before it.

For example, let’s say my first number is 2, and I multiply 2 by 5 to get 10. Then I multiply 10 by 5 to get 50. I can multiply 50 by 5 to get 250, and so on.

\(2, 10, 50, 250…\)

In geometric sequences, we use multiplication to find each subsequent term. The number we multiply by is called the **common ratio**. Each term gets multiplied by the common ratio, resulting in the next term in the sequence. In this geometric sequence, the common ratio is 5.

To find any number in a given sequence, we can use the following formula:

\(a_{n}=a_{1}r^{(n-1)}\)

In this formula, \(n\) stands for the number in the sequence we are asked to find. \(a_{1}\) stands for the first term in the sequence, and \(r\) stands for the common ratio. Let’s look at how we can use this formula. Given the geometric sequence:

\(10, 30, 90, 270…\)

we can see that the common ratio, or \(r\), equals 3 because each time we’re multiplying by 3.

\(10\times 3=30\)

\(30\times 3=90\)

\(90\times 3=270\)

And so on and so forth.

Let’s use our formula to find the 6th term in this sequence.

Since we need to identify the 6th term, we can replace the variable \(n\) with 6. Remember that \(a_{1}\) always stands for the first term in our sequence. In this case, the first term is 10. Let’s rewrite the formula and replace the variables with the numbers that they represent.

So we’re looking for the 6th term, so:

\(a_{6}=10(3)^{6-1}\)

Now that we’ve replaced the variables with numbers, we can solve the equation using the order of operations. First, we need to simplify the exponents. Since 6 – 1 = 5, we can rewrite the equation using one exponent.

\(a_{6}=10(3)^{5}\)

Next, we can solve \(3^{5}\), or \(3\times 3\times 3\times 3\times 3\), which equals 243.

\(a_{6}=10(243)\)

Finally, we can multiply \(10\times 243\) to get the final answer. The 6th term in the geometric sequence is 2,430.

A **geometric series** is the sum of all terms in a geometric sequence. Let’s consider the last sequence we looked at, which was 10, 30, 90, 270, and so on.

As a geometric series, this is written as \(10+30+90+270+…\)

To find the sum of a specific number of terms in a geometric sequence, we can use this formula:

\(s_{n}=\frac{a_{1}(1-r^{n})}{1-r}\)

In this formula, \(n\) stands for the number of terms added together. Like the first formula we learned about, \(a_{1}\) stands for the first term in the sequence, and \(r\) stands for the common ratio.

Let’s walk through an example together. Using the same geometric sequence as our last example, let’s find the sum of the first 8 terms:

\(10, 30, 90, 270…\)

Using this sequence and the formula \(s_{n}=\frac{a_{1}(1-r^{n})}{1-r}\), let’s consider what each variable can be replaced with.

Remember that \(n\) stands for the number of terms. Since we are looking for the sum of the first 8 terms, we can substitute 8 for \(n\). \(a_{1}\) stands for the first term in the sequence, which is 10. Recall that \(r\) stands for the common ratio. The common ratio is 3, so we can substitute \(r\) with 3 in our formula.

Now, let’s rewrite the equation and replace the variables with the numbers they represent:

\(s_{8}=\frac{10(1-3^{8})}{1-3}\)

Now that we replaced the variables with numbers, we can solve the equation using the order of operations. First, we need to simplify the exponents. Since \(3^{8}=6,561\), we can rewrite the equation using this product.

\(s_{8}=\frac{10(1-6,561)}{1-3}\)

Next, we can simplify the parentheses. Since \(1-6,561=-6,560\), we can rewrite the equation again.

\(s_{8}=\frac{10(-6,560)}{1-3}\)

Now we can simplify the numerator and denominator.

\(s_{8}=\frac{-65,600}{-2}\)

And to get our final answer, we just divide. \(-65,600\div -2=32,800\). So the sum of our 8 terms is 32,800.

Geometric sequences are patterns, and patterns are all around us. Knowing how they work will help you identify and use them in the real world.

For example, let’s say that you want to share a YouTube video with friends. You might start by messaging 2 friends about this. In your message, you might ask each friend to share the video with 4 other friends. That would mean 8 people were messaged, since \(2\times 4=8\). If each of those people shares with 4 more people, then how many people would have seen the video? 32, because \(8\times 4=32\).

As you can see, we have a geometric pattern. 2, 8, 32, and so on. The common ratio is 4 since each person is asked to share the video with 4 friends. Let’s find out how many people received the YouTube video after 8 rounds of messaging, including the original message shared with 2 friends.

To solve this problem, we need to know the sum of the first 8 terms in the sequence. Remember which formula can help us find the sum. Our sum formula is:

\(s_{n}=\frac{a_{1}(1-r^{n})}{1-r}\)

Using information from the problem, let’s replace variables with the numbers they represent. \(n=8\) because we are finding the sum of the first 8 terms in the sequence. \(a_{1}=2\) since the first number in the sequence is 2. And \(r=4\) because the common ratio is 4. So we’re gonna substitute in our variables.

\(s_{8}=\frac{2(1-(4)^{8})}{1-4}\)

From here, we can solve the equation using the order of operations. First, we need to simplify the exponent inside the parentheses. Since \(4^{8}=65,536\), we can rewrite the equation without an exponent.

\(s_{8}=\frac{2(1-65,536)}{1-4}\)

From here, we can solve the expression in parentheses. \(1-65,536=-65,535\), so we can rewrite the equation again.

\(s_{8}=\frac{2(-65,535)}{1-4}\)

Next, we can simplify the numerator and denominator. \(2\times -65,535=-131,070\), and \(1-4=-3\).

\(s_{8}=\frac{-131,070}{-3}\)

Finally, we divide to solve. \(-131,070\div -3=43,690\).

\(s_{8}=43,690\)

Therefore, 43,690 people received the YouTube video after 8 rounds of messaging.

Let’s try another problem together. Let’s say that Anne’s club is selling cookies online. They sell 4 boxes on day 1. Their goal for each day is to double the amount sold on the previous day. At this rate, how many boxes will they sell on day 10?

In this problem, we need to find the 10th term in the geometric sequence described.

Let’s start by identifying what we know from the problem. Since 4 boxes are sold on Day 1, that’s the first number in our sequence. If their goal is to double, that means that each subsequent term needs to be multiplied by 2.

We can use the first formula we learned to solve this problem.

\(a_{n}=a_{1}r^{n-1}\)

Let’s replace the variables with their corresponding numbers. \(n=10\), since we need to find the 10th term in this sequence. \(a_{1}=4\), which is the first number in the sequence. And \(r=2\) because their goal is to double the previous day’s sales.

\(a_{10}=4(2)^{10-1}\)

Now that the variables are replaced, we can solve the equation. First, we simplify our exponents. Since \(10-1=9\), we can rewrite the equation using the exponent 9.

\(a_{10}=4(2)^{9}\)

Next, we can rewrite the exponent term. \(2^{9}=512\), so we can rewrite the equation using 512.

\(a_{10}=4(512)\)

Finally, we multiply to solve. \(4\times 512=2,048\).

\(a_{10}=2,048\)

At their current rate, Anne’s club will sell 2,048 boxes of cookies on day 10.

I have one last problem for you to try on your own. It combines everything we’ve learned in this video, so it’s a little more challenging, but I know you can handle it.

Malik opened a bank account in January with $10. Each month, he plans to double the amount deposited in the previous month. Based on this information, how much money will Malik deposit in September? And assuming he doesn’t spend any of the deposited money, how much will Malik have at the end of the year?

This problem asks 2 questions. Let’s start with the first: How much money does Malik plan to deposit in September? See if you can try this part on your own. Pause the video here. When you’re done, press play and check your work.

We can use the sequence formula to identify Malik’s deposit amount in September. Since he makes monthly deposits, this would be his 9th deposit. So:

\(a_{n}=a_{1}r^{n-1}\)

That’s the formula we’re gonna use to solve this problem. And \(n=9\) since September is the 9th month of the year. \(r=2\) because the sequence doubles each month. And \(a_{1}=10\) because Malik’s first deposit was 10.

\(a_{9}=10(2)^{9-1}\)

Now that we’ve replaced our variables, we’re ready to solve the equation. First, we’ll simplify our exponents. Since \(9-1=8\), we can rewrite the equation using the exponent 8.

\(a_{9}=10(2)^{8}\)

Next, we can rewrite the exponent term. \(2^{8}=256\), so we can rewrite the equation using 256.

\(a_{9}=10(256)\)

Finally, we multiply to solve. \(10\times 256=2,560\).

\(a_{9}=2,560\)

Malik plans to deposit $2,560 in September.

\(a_{n}=a_{1}r^{(n-1)}\)

\(a_{9}=(10)(2)^{(9-1)}\)

\(a_{9}=(10)(2)^{(8)}\)

\(a_{9}=(10)(256)\)

\(a_{9}=2,560\)

Let’s move onto the second question: How much money will Malik have at the end of the year? Pause the video and try to solve this on your own. When you’re done, press play and check your work.

We can use the series formula to find the total amount of money Malik plans to deposit.

\(s_{n}=\frac{a_{1}(1-r^{n})}{1-r}\)

First, we can replace our variables with numbers from the problem. Since we are looking for the total amount deposited after 12 months, \(n=12\). We know that \(a_{1}=10\) because Malik’s first deposit was $10. And we know that \(r=2\) because our sequence doubles each month.

\(s_{12}=\frac{10(1-2^{12})}{1-2}\)

Next, we can simplify the exponent. Since \(2^{12}=4,096\), we can rewrite the equation.

\(s_{12}=\frac{10(1-4,096)}{1-2}\)

Now we can solve the expression in parentheses. \(1-4,096=-4,095\), so we can rewrite the equation again.

\(s_{12}=\frac{10(-4,095)}{1-2}\)

Our next step is to simplify the numerator and denominator. \(10\times -4,095=-40,950\) and \(1-2=-1\).

\(s_{12}=\frac{-40,950}{-1}\)

Finally, we can divide to solve.

\(s_{12}=40,950\)

Malik will deposit $40,950 by the end of the year.

I hope this video on geometric sequences was helpful. Thanks for watching, and happy studying!

## Geometric Sequence Practice Questions

**Question #1:**

Find the fifth term in the following geometric sequence:

\(5,30,180,…\)

**Answer:**

Solving these types of problems can typically be done easiest by first determining the formula for \(a_n\), and then plugging in the appropriate \(n\)-value. Since a geometric sequence is defined such that \(a_n=a_1r^{n-1}\), determine what \(a_1\) and \(r\) equal, then plug them in.

The first term of the sequence is \(5\), so \(a_1=5\). The next term is \(30\), and \(r=\frac{a_2}{a_1}=\frac{30}{5}=6\). Plugging these into the geometric sequence formula, you get \(a_n=5\times6^{n-1}\).

Now, since the problem is asking for the fifth term, plug \(n=5\) into this formula and simplify.

\(a_5=5\times6^{5-1}\)

\(=5\times6^4\)

\(=5\times1{,}296\)

\(=6{,}480\)

**Question #2:**

Determine the common ratio \(r\) in the following geometric sequence and write the formula for \(a_n\).

\(13,143,1{,}573,17{,}303, …\)

\(r=9\); \(a_n=9\times13^{n-1}\)

\(r=11\); \(a_n=13\times11^{n-1}\)

\(r=13\); \(a_n=13\times11^{n-1}\)

\(r=11\); \(a_n=11\times13^{n-1}\)

**Answer:**

The common ratio \(r\) can be found by dividing one term by the preceding term. \(r=\frac{a_{n+1}}{a_n}\). Usually the simplest way to determine \(r\) is by dividing the first two terms. In this case, \(r=\frac{a_2}{a_1}=\frac{143}{13}=11\).

Once \(r\) is found, you must write the formula for \(a_n\); this means you’ll also need to know \(a_1\). By looking at the first term of the sequence above, it is clear that \(a_1\) is \(13\). So the formula for the \(n^{th}\) term is \(a_n=13\times11^{n-1}\).

**Question #3:**

Determine the sixth term, \(a_6\), in the geometric series given by the terms in the geometric sequence below:

\(3, 21, 147, 1\text{,}029, …\)

**Answer:**

To find the sixth term in the sequence, it’s essential to use the formula for the nth term of a geometric sequence:

\(a_n=a_1\times r^{(n-1)}\)

First, determine the values of \(a_1\)and \(r\). The first term is 3. The common ratio \(r\) can be found by dividing any term of the sequence by the preceding term.

Plugging \(a_1\)and \(r\) into the formula, and setting \(n=6\), we get the following:

\(a_6=3\times 7^{(6-1)}\)

\(a_6=3\times 7^5\)

\(a_6=3\times 16,807\)

\(a_6=50,421\)

Therefore, the sixth term of the geometric sequence is 50,421.

**Question #4:**

Claudia is training for a marathon. On her first day of training, she runs for one minute. Each day she will increase her running time by 50%. How much time will she have run cumulatively after seven days? Round to three decimal places.

21.818 minutes

23.541 minutes

29.940 minutes

32.172 minutes

**Answer:**

The word “cumulatively” is a signal that this is a geometric series problem. The formula for the \(n^{th}\) term of a geometric series is \(s_n=\frac{a_1(1-r^n)}{1-r}\), so first determine what \(a_1\) and \(r\) are.

The problem statement says that Claudia runs for one minute on her first day of training, so \(a_1=1\). It also says that she increases her running time by \(50\%\) each day. Don’t be fooled though—this does not mean that \(r=0.50\). If that were the case, then Claudia would only be running *half* of her previous time each day. Instead, she is increasing her time every day, making \(r\) equal to \(1.5\).

In this case, \(s_n=\frac{1(1-1.5^n)}{1-1.5}\). So Claudia’s cumulative running time after seven days will equal \(s_7\).

\(s_7=\frac{1(1-1.5^7)}{1-1.5}\)

\(=\frac{1-1.5^7}{-0.5}\)

\(=\frac{1-17.086}{-0.5}\)

\(=\frac{-16.086}{-0.5}\)

\(=32.172\text{ minutes}\)

**Question #5:**

Every year on his birthday, Wilbur donates canned food to his local food bank. He started this tradition at age \(20\) by donating two cans of food. Each year, Wilbur gives triple the amount of the previous year. How many cans of food will Wilbur have donated in total after his \(30^{th}\) birthday?

**Answer:**

Before the formula for the geometric series can be used, first determine what \(a_1\) and \(r\) are equal to. The first time Wilbur donated food was at age \(20\), when he gave two cans of food. This means \(a_1=2\). Each year, he triples his donation amount for the year. This indicates that \(r=3\).

The geometric series formula is then \(s_n=\frac{2(1-3^n)}{1-3}\). Now determine what \(n\)-value should be plugged in. It is tempting to say that \(n=10\) after Wilbur’s \(30^{th}\) birthday, but remember that his \(20^{th}\) birthday is included as the first iteration of the geometric series. Because of this, plug in \(n=11\).

\(s_{11}=\frac{2(1-3^{11})}{1-3}\)

\(=\frac{2(1-177{,}147)}{-2}\)

\(=\frac{2(-177{,}146)}{-2}\)

\(=177{,}146\)

After his \(30^{th}\) birthday and \(11\) years of donations, Wilbur will have given \(177{,}146\) cans of food.

**Question #6:**

Here is the formula for a geometric sequence: \(a_n= 4·5^{(n-1)}\). What is the \(4^{th}\) term of the geometric sequence?

**Answer:**

To find the \(4^{th}\) term of the geometric sequence, we substitute \(4\) for \(n\) in the formula and evaluate to find the value of the term, which is \(500\), since the first term is \(4\) and the common ratio is \(5\), according to the formula.

**Question #7:**

Here is a geometric sequence: \(1,7,49,343,2{,}401,…\)

What is the \(9^{th}\) term of the geometric sequence?

**Answer:**

The first term of the geometric sequence is \(1\) and the common ratio is \(7\); therefore, the formula for the geometric sequence is \(a_n=1·7^{(n-1)}\). Therefore, \(a_9=1·7^{(9-1)}=5{,}764{,}801\).

**Question #8:**

Here is a geometric sequence: \(3,12,48,192,768,…\)

What is the formula for this geometric sequence?

**Answer:**

We will start by identifying the first term of the geometric sequence, which is \(3\), then we find the common ratio, which is \(4\), since the terms are increasing by a factor of \(4\). Therefore, the formula for this geometric sequence is \(a_n= 3·4^{(n-1)}\).

**Question #9:**

What is the \(5^{th}\) term of a geometric sequence where the first term is \(8\) and the common ratio is \(3\)?

**Answer:**

Since the first term of the geometric sequence is \(8\) and the common ratio is \(3\), the formula for this geometric sequence is \(a_n=8·3^{(n-1)}\); therefore, the \(5^{th}\) term can be found using the formula, substitute \(5\) for \(n\) and evaluate, \(a_5=8·3^{(5-1)}=648\).

**Question #10:**

Here is a geometric sequence: \(2,12,72,432,2592,…\)

What is the \(10^{th}\) term of the geometric sequence?

**Answer:**

Start by identifying the first term of the geometric sequence, which is \(2\). The common ratio is \(6\) since the terms are increasing by a factor of \(6\). Therefore, the formula for this geometric sequence is \(a_n= 2·6^{(n-1)}\). Using this formula, calculate the value of the \(10^{th}\) term of the geometric sequence, \(a_{10}= 2·6^{(10-1)}=20{,}155{,}392\).

**Question #11:**

Find the missing term in the geometric sequence without using a formula.

\(3,15,75,375,\) _____ \(,9{,}375…\)

**Answer:**

To identify the missing term without a formula, analyze the pattern to find the common ratio. Start with the first term, \(3\). \(3\) multiplied by what number results in a product of the next term, \(15\)? The correct answer is \(5\), so the common ratio is \(5\). Check the rest of the terms in the sequence: \(15×5=75\) and \(75×5=375\). Next, multiply \(375\) by \(5\) to find the missing term in the sequence. Since \(375×5=1{,}875\), the missing term in the geometric sequence is \(1{,}875\). Check your work by multiplying \(1{,}875\) by \(5\). If correct, the product should be the subsequent term in the sequence, \(9{,}375\). Since \(1{,}875×5=9,375\), answer \(B\) is correct.

\(3,15,75,375,\mathbf{1{,}875},9{,}375…\)

**Question #12:**

Use the formula \(a_n=a_1r^{(n-1)}\) to identify the \(8^{th}\) term in the sequence shown:

\(1,6,36,216,1{,}296…\)

**Answer:**

First, identify the values of the variables used in the equation. Since n stands for the number in the sequence that needs to be identified, replace \(n\) with \(8\) in the formula. The variable \(a_1\) stands for the first term in the sequence, which is \(1\). The variable \(r\) represents the common ratio. In this case, the common ratio is \(6\). Each term is multiplied by \(6\), resulting in the subsequent term in the sequence. Replace \(r\) with \(6\) in the formula.

\(n=8\)

\(a_1=1\)

\(r=6\)

From here, rewrite the formula \(a_n=a_1r^{(n-1)}\) and substitute the variables with the numbers they represent:

\(a_8=(1)(6)^{(8-1)}\) | Now that the variables are replaced with their corresponding values, solve the equation using the Order of Operations. |

\(a_8=(1)(6)^{(7)}\) | First, simplify the exponents. Since \(8-1=7\), we can rewrite the equation using the exponent \(7\). |

\(a_8=(1)(279{,}936)\) | Next, simplify the exponent \(6^7\). \(6^7=6×6×6×6×6×6×6\), which equals \(279{,}936\). Rewrite the equation using \(279{,}936\). |

\(a_8=279{,}936\) | From here, multiply \(1×279{,}936\) to get the final answer. Since \(1×279{,}936=279{,}936\), the \(6^{th}\) term in the geometric sequence is \(279{,}936\). Therefore, \(A\) is the correct answer. |

**Question #13:**

Use the formula \(a_n=a_1r^{(n-1)}\) to identify the \(15^{th}\) term in the sequence shown:

\(7,21,63,189,567…\)

**Answer:**

First, identify the values of the variables used in the equation. Since n stands for the number in the sequence that needs to be identified, replace \(n\) with \(15\) in the formula. The variable \(a_1\) stands for the first term in the sequence, which is \(7\). The variable \(r\) represents the common ratio. In this case, the common ratio is \(3\). Each term is multiplied by \(3\), resulting in the subsequent term in the sequence. Replace \(r\) with \(3\) in the formula.

\(n=15\)

\(a_1=7\)

\(r=3\)

From here, rewrite the formula \(a_n = a_1r^{(n-1)}\) and substitute the variables with the numbers they represent:

\(a_{15}=(7)(3)^{(15-1)}\) | Now that the variables are replaced with their corresponding values, solve the equation using the Order of Operations. |

\(a_{15}=(7)(3)^{(14)}\) | First, simplify the exponents. Since \(15-1=14\), we can rewrite the equation using the exponent \(14\). |

\(a_{15}=(7)(4{,}782{,}969)\) | Next, simplify the exponent \(3^{14}\). \(3^{14}=3×3×3×3×3×3×3×3\) \(×3×3×3×3×3×3\), which equals \(4{,}782{,}969\). Rewrite the equation using \(4{,}782{,}969\). |

\(a_{15}=33{,}480{,}783\) | From here, multiply \(7×4{,}782{,}969\) to get the final answer. Since \(7×4{,}782{,}969=33{,}480{,}783\), the \(15^{th}\) term in the geometric sequence is \(33{,}480{,}783\). Therefore, \(D\) is the correct answer. |

**Question #14:**

Kelsey opened a bank account in January and deposited \($5\). Each month, she plans to double the amount deposited in the previous month. Based on this information, how much money will Kelsey deposit in October?

**Answer:**

To figure out how much money Kelsey plans to deposit, use the sequences formula. First, identify the values of the variables used in the equation. October will be Kelsey’s tenth deposit, so we are finding the tenth term in the geometric sequence. Since n stands for the number in the sequence that needs to be identified, replace \(n\) with \(10\) in the formula. The variable \(a_1\) stands for the first term in the sequence. Kelsey’s first deposit was in January when she opened her bank account with \($5\). Therefore, the first term in the sequence is \(5\). Each month, Kelsey plans to double the amount of money she deposits. Each term gets doubled, or multiplied by \(2\), resulting in the subsequent term in the sequence. Based on this information, the common ratio in this sequence is \(2\). Therefore, replace \(r\) with \(2\) in the formula.

\(n=10\)

\(a_1=5\)

\(r=2\)

From here, rewrite the formula \(a_n = a_1r^{(n-1)}\) and substitute the variables with the numbers they represent:

\(a_{10}=(5)(2)^{(10-1)}\) | |

\(a_{10}=(5)(2)^{(9)}\) | First, simplify the exponents. Since \(10-1=9\), we can rewrite the equation using the exponent \(9\). |

\(a_{15}=(512)\) | Next, simplify the exponent \(3^{9}\). \(2^{2}=2×2×2×2×2×2×2×2×2\), which equals \(512\). Rewrite the equation using \(512\). |

\(a_{10}=2{,}560\) | From here, multiply \(5×512\) to get the final answer. Since \(5×512=2{,}560\), the \(10^{th}\) term in the geometric sequence is \(2{,}560\). Kelsey plans to deposit \($2{,}560\) to her bank account in October. Therefore, \(D\) is the correct answer. |

**Question #15:**

Ken is researching the reproduction of a species of bacteria. Based on his research, the bacteria triples in amount every 24 hours. Ken keeps a log of the bacteria count in his laboratory, but the bacteria count on day 4 of his research is missing. Assuming the bacteria tripled each day as predicted, identify the bacteria count on day 4 without using the geometric formula:

\(8,24,72,\) _____ \(,648…\)

**Answer:**

To identify the missing term without a formula, analyze the pattern to find the common ratio. Start with the first term, \(8\). \(8\) multiplied by what number results in a product of the next term, \(24\)? The correct answer is \(3\), so the common ratio is \(3\). Check the rest of the terms in the sequence: \(24×3=72\). Next, multiply \(72\) by \(3\) to get the missing term in the sequence. Since \(72×3=216\), the missing term in the geometric sequence is \(216\). Check your work by multiplying \(216\) by \(3\). If correct, the product should be the subsequent term in the sequence, \(648\). Since \(216×3=648\), answer \(C\) is correct. The bacteria count on day \(4\) of Ken’s experiment was \(216\).

\(8,24,72,\mathbf{216},648…\)