problem solving involving geometric sequence in real life

Geometric Sequences Problems with Solutions

Geometric sequences are used in several branches of applied mathematics to engineering, sciences, computer sciences, biology, finance... Problems and exercises involving geometric sequences, along with answers are presented.

Review OF Geometric Sequences

The sequence shown below

Problems with Solutions

Problem 1 Find the terms a 2 , a 3 , a 4 and a 5 of a geometric sequence if a 1 = 10 and the common ratio r = - 1. Solution to Problem 1: Use the definition of a geometric sequence \( a_2 = a_1 \times r = 10 (-1) = - 10 \\ a_3 = a_2 \times r = - 10 (-1) = 10 \\ a_4 = a_3 \times r = 10 (-1) = - 10 \\ a_5 = a_4 \times r = - 10 (-1) = 10 \)

Find the 10 th term of a geometric sequence if a 1 = 45 and the common ration r = 0.2. Solution to Problem 2: Use the formula \[ a_n = a_1 \times r^{n-1} \] that gives the n th term to find a 10 as follows \( a_{10} = 45 \times 0.2^{10-1} = 2.304 \times 10^{-5} \)

Find a 20 of a geometric sequence if the first few terms of the sequence are given by

Given the terms a 10 = 3 / 512 and a 15 = 3 / 16384 of a geometric sequence, find the exact value of the term a 30 of the sequence. Solution to Problem 4: We first use the formula for the n th term to write a 10 and a 15 as follows \( a_{10} = a_1 \times r^{10-1} = a_1 r^9 = 3 / 512 \\ \\ a_{15} = a_1 \times r^{15-1} = a_1 r^{14} = 3 / 16384 \) We now divide the terms a 10 and a 15 to write \( a_{15} / a_{10} = a_1 \times r^{14} / (a_1 \times r^9) = (3 / 16384) / (3 / 512) \) Simplify expressions in the above equation to obtain. r 5 = 1 / 32 which gives r = 1/2 We now use a 10 to find a 1 as follows. \( a_{10} = 3 / 512 = a_1 (1/2)^9 \) Solve for a 1 to obtain. \( a_1 = 3 \) We now use the formula for the n th term to find a 30 as follows. \( a_{30} = 3(1/2)^{29} = 3 / 536870912 \)

Find the sum \[ S = \sum_{k=1}^{6} 3^{k - 1} \] Solution to Problem 5: We first rewrite the sum S as follows S = 1 + 3 + 9 + 27 + 81 + 243 = 364 Another method is to first note that the terms making the sum are those of a geometric sequence with a 1 = 1 and r = 3 using the formula s n = a 1 (1 - r n ) / (1 - r) with n = 6. s 6 = 1 (1 - 3 6 ) / (1 - 3) = 364

Find the sum \[ S = \sum_{i=1}^{10} 8 \times (1/4)^{i - 1} \] Solution to Problem 6: An examination of the terms included in the sum are 8 , 8× ((1/4) 1 , 8×((1/4) 2 , ... , 8×((1/4) 9 These are the terms of a geometric sequence with a 1 = 8 and r = 1/4 and therefore we can use the formula for the sum of the terms of a geometric sequence s 10 = a 1 (1 - r n ) / (1 - r) = 8 × (1 - (1/4) 10 ) / (1 - 1/4) = 10.67 (rounded to 2 decimal places)

Write the rational number 5.31313131... as the ratio of two integers. Solution to Problem 7: We first write the given rational number as an infinite sum as follows 5.313131... = 5 + 0.31 + 0.0031 + 0.000031 + .... The terms making 0.31 + 0.0031 + 0.000031 ... are those of a geometric sequence with a 1 = 0.31 and r = 0.01. Hence the use of the formula for an infinite sum of a geometric sequence S = a 1 / (1 - r) = 0.31 / (1 - 0.01) = 0.31 / 0.99 = 31 / 99 We now write 5.313131... as follows 5.313131... = 5 + 31/99 = 526 / 99

Exercises with Answers

Answer the following questions related to geometric sequences: a) Find a 20 given that a 3 = 1/2 and a 5 = 8 b) Find a 30 given that the first few terms of a geometric sequence are given by -2 , 1 , -1/2 , 1/4 ... c) Find r given that a 1 = 10 and a 20 = 10 -18 d) write the rational number 0.9717171... as a ratio of two positive integers.

a) a 20 = 2 18 b) a 30 = 1 / 2 28 c) r = 0.1 d) 0.9717171... = 481/495

More References and links

• Arithmetic Sequences Problems with Solutions
• math problems with detailed solutions
• Math Tutorials and Problems

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Course: Algebra 1   >   Unit 9

• Explicit & recursive formulas for geometric sequences
• Recursive formulas for geometric sequences
• Explicit formulas for geometric sequences
• Converting recursive & explicit forms of geometric sequences

Geometric sequences review

Parts and formulas of geometric sequence

Extending geometric sequences

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

Writing recursive formulas

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Examples of arithmetic and geometric sequences and series in daily life

In this part of the course I am just trying to show that we actually see a lot of sequences and series every day in our regular life. I already found some examples such as the house numbers when you drive down a street, the number of people you reach in those 'chain mails', the value of your block in the game 2048 , ...

I am still looking for some fun, everyday sequences or series that without much context give an idea of what those sequences are.

• mathematics-in-daily-life

• $\begingroup$ Related: matheducators.stackexchange.com/questions/1294/… $\endgroup$ –  kjetil b halvorsen Commented Aug 30, 2017 at 10:00

7 Answers 7

Here are a few more examples:

the amount on your savings account ;

the amount of money in your piggy bank if you deposit the same amount each week (a bank account with regular deposits leads you to arithmetico-geometric sequences) ;

the size of a population in exponential growth, e.g. bacteria in a Petri dish (or in your leftovers if you find Petri dishes not "every day life" enough) ;

the intensity of radioactivity after $n$ years of a given radioactive material (with application to determining the age of mommies!).

• 1 $\begingroup$ As a mod whose 'home' is Money.SE a +1 for the first 2 examples being money related. $\endgroup$ –  JTP - Apologise to Monica Commented Jun 4, 2016 at 18:24

I tutored a student who came with a kind of problem I had never seen before and found quite refreshing. It was something like:

A child is being pushed on a swing by their father, reaching a maximum height of 4 feet. The father stops pushing, and the maximum height of the swing decreases by 15% on each successive swing.

I don't remember the question itself, but the main idea was that the sequence of maximum heights of the child was a geometric sequence! (Perhaps the goal was to find the total vertical distance traveled by the child.)

You get a related example considering a bouncing ball:

It turns out that a bouncing object loses an approximately constant fraction of its remaining energy with each bounce, and in turn the sequence of maximum heights is (approximately) geometric!

More info on the latter here .

I like to explain why arithmetic and geometric progressions are so ubiquitous. Using the examples other people have given.

Geometric progressions happen whenever each agent of a system acts independently. For example population growth each couple do not decide to have another kid based on current population. So population growth each year is geometric. Each radioactive atom independently disintegrates, which means it will have fixed decay rate. In other words that is why there is "half-life" of a radioactive element, in a fixed amount of time it becomes half. Email chains, Interest rate, etc are more examples of the same kind.

On the other end global/singular decisions give arithmetic progressions. If you add a fixed amount to your piggy bank each week that is arithmetic progression. The child who swings extra each time is likely to give only a constant extra force each time, so it is not likely for that to be geometric, it will be an arithmetic progression. There are exceptions of course like the ball bouncing is geometric even though it is singular because of coefficient of restitution. In general singular decisions can be anything - but typically arithmetic.

In reality, these are ideal cases, most of the natural phenomenon will have both global and local influencers. Making it somewhere in between arithmetic and geometric progressions.

If the population is already huge having another kid might not be so conducive. So the population growth will stop when overall resources get limited. Thomas Malthus wrote that all life forms, including humans, have a propensity to exponential population growth when resources are abundant but that actual growth is limited by available resources.

Tumour growth, the growth rate is exponential unless it becomes so large that it cannot get food to grow effectively. So it starts of exponentially and stops completely. A more precise statement is known as Gompertz Law of Mortality - "rate of decay falls exponentially with current size".

Even radioactive decay is not really immune, there is something called Quantum anti-Zeno effect if you wanna go wiki hopping.

https://en.wikipedia.org/wiki/Malthusian_growth_model

https://en.wikipedia.org/wiki/Gompertz_function

They might be interested to know about both Moore's Law and "Nielsen's Law" . You've probably heard about Moore's Law , where computer complexity doubles about every two and a half years.

Internet bandwidth seems to also have a doubling time, in this case that doubling time is 21 months.

https://www.nngroup.com/articles/law-of-bandwidth/

When I think of a geometric sequence, I think of something where the initial input value = 1, not 0. Most interest problems would start at time = 0, so I would exclude these unless you said something like "let x = $ in bank at beginning of each year". Then your first input value would be 1.

Also, geometric sequences have a domain of only natural numbers (1,2,3,...), and a graph of them would be only points and not a continuous curved line. So again, a problem about earned interest might not be a perfect example, since you can withdraw your money at any instant and not only at whole number year values.

The best one I have come up with is tile values in the game 2048.

• tile 4 = 16
• tile 5 = 32
• tile 6 = 64

and so on...

Arithmetics Examples

Time on clock, each minute hand that the second hand covers is 5 seconds.

The point at which a runner passes the finish line in a 3000 metre race.

Sound waves or waves in the sea are sinusoids, so they can repeat their pattern for the range of the sinusoid.

Geometric examples

Half life of carbon or any element. You can then show how all the carbon 14 is depleted over thousands of years.

A graph where logs is used is easy to read and can be almost linear, whereas if there is a geometric increase you can't even plot it on paper. Kids love this one, and understand it very quickly.

Puzzle: If a frog is 1 metre from a door and jumps halfway, and then jumps halfway again continuing to half its jump each time, will it ever reach the door? This displays limits and geometric decreasing functions and the idea that decreasing jumps results in infinitesimally small results over time.

Showing kids how much they pay on a mortgage at 5% interest rate for 10 years, 20 years and 30 years is very insightful and clearly displays geometric increase.

Not the answer you're looking for? Browse other questions tagged examples mathematics-in-daily-life series or ask your own question .

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9.3 Geometric Sequences

Learning objectives.

In this section, you will:

• Find the common ratio for a geometric sequence.
• List the terms of a geometric sequence.
• Use a recursive formula for a geometric sequence.
• Use an explicit formula for a geometric sequence.

Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be $26,520 after one year; $27,050.40 after two years; $27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way.

Finding Common Ratios

The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio . The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.

Definition of a Geometric Sequence

A geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If a 1 a 1 is the initial term of a geometric sequence and r r is the common ratio, the sequence will be

Given a set of numbers, determine if they represent a geometric sequence.

• Divide each term by the previous term.
• Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.

Is the sequence geometric? If so, find the common ratio.

• ⓐ 1 , 2 , 4 , 8 , 16 , ... 1 , 2 , 4 , 8 , 16 , ...
• ⓑ 48 , 12 , 4 ,  2 , ... 48 , 12 , 4 ,  2 , ...

Divide each term by the previous term to determine whether a common ratio exists.

The sequence is geometric because there is a common ratio. The common ratio is 2.

The sequence is not geometric because there is not a common ratio.

The graph of each sequence is shown in Figure 1 . It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.

If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio?

No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio.

Writing Terms of Geometric Sequences

Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is a 1 = − 2 a 1 = − 2 and the common ratio is r = 4, r = 4, we can find subsequent terms by multiplying − 2 ⋅ 4 − 2 ⋅ 4 to get − 8 − 8 then multiplying the result − 8 ⋅ 4 − 8 ⋅ 4 to get − 32 − 32 and so on.

The first four terms are { –2 ,  –8 ,  –32 ,  –128 } . { –2 ,  –8 ,  –32 ,  –128 } .

Given the first term and the common factor, find the first four terms of a geometric sequence.

• Multiply the initial term, a 1 , a 1 , by the common ratio to find the next term, a 2 . a 2 .
• Repeat the process, using a n = a 2 a n = a 2 to find a 3 a 3 and then a 3 a 3 to find a 4, a 4, until all four terms have been identified.
• Write the terms separated by commons within brackets.

Writing the Terms of a Geometric Sequence

List the first four terms of the geometric sequence with a 1 = 5 a 1 = 5 and r = –2. r = –2.

Multiply a 1 a 1 by − 2 − 2 to find a 2 . a 2 . Repeat the process, using a 2 a 2 to find a 3 , a 3 , and so on.

The first four terms are { 5 , –10 , 20 , –40 } . { 5 , –10 , 20 , –40 } .

List the first five terms of the geometric sequence with a 1 = 18 a 1 = 18 and r = 1 3 . r = 1 3 .

Using Recursive Formulas for Geometric Sequences

A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.

Recursive Formula for a Geometric Sequence

The recursive formula for a geometric sequence with common ratio r r and first term a 1 a 1 is

Given the first several terms of a geometric sequence, write its recursive formula.

• State the initial term.
• Find the common ratio by dividing any term by the preceding term.
• Substitute the common ratio into the recursive formula for a geometric sequence.

Write a recursive formula for the following geometric sequence.

The first term is given as 6. The common ratio can be found by dividing the second term by the first term.

Substitute the common ratio into the recursive formula for geometric sequences and define a 1 . a 1 .

The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown in Figure 2

Do we have to divide the second term by the first term to find the common ratio?

No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio.

Using Explicit Formulas for Geometric Sequences

Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.

Let’s take a look at the sequence { 18 ,  36 ,  72 ,  144 ,  288 ,  ... } . { 18 ,  36 ,  72 ,  144 ,  288 ,  ... } . This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is

The graph of the sequence is shown in Figure 3 .

Explicit Formula for a Geometric Sequence

The n n th term of a geometric sequence is given by the explicit formula :

Writing Terms of Geometric Sequences Using the Explicit Formula

Given a geometric sequence with a 1 = 3 a 1 = 3 and a 4 = 24 , a 4 = 24 , find a 2 . a 2 .

The sequence can be written in terms of the initial term and the common ratio r . r .

Find the common ratio using the given fourth term.

Find the second term by multiplying the first term by the common ratio.

The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power.

Given a geometric sequence with a 2 = 4 a 2 = 4 and a 3 = 32 a 3 = 32 , find a 6 . a 6 .

Writing an Explicit Formula for the n n th Term of a Geometric Sequence

Write an explicit formula for the n th n th term of the following geometric sequence.

The first term is 2. The common ratio can be found by dividing the second term by the first term.

The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.

The graph of this sequence in Figure 4 shows an exponential pattern.

Write an explicit formula for the following geometric sequence.

Solving Application Problems with Geometric Sequences

In real-world scenarios involving geometric sequences, we may need to use an initial term of a 0 a 0 instead of a 1 . a 1 . In these problems, we can alter the explicit formula slightly by using the following formula:

In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.

• ⓐ Write a formula for the student population.
• ⓑ Estimate the student population in 2020.

The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.

Let P P be the student population and n n be the number of years after 2013. Using the explicit formula for a geometric sequence we get

We can find the number of years since 2013 by subtracting.

We are looking for the population after 7 years. We can substitute 7 for n n to estimate the population in 2020.

The student population will be about 374 in 2020.

A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.

• ⓐ Write a formula for the number of hits.
• ⓑ Estimate the number of hits in 5 weeks.

Access these online resources for additional instruction and practice with geometric sequences.

• Geometric Sequences
• Determine the Type of Sequence
• Find the Formula for a Sequence

9.3 Section Exercises

What is a geometric sequence?

How is the common ratio of a geometric sequence found?

What is the procedure for determining whether a sequence is geometric?

What is the difference between an arithmetic sequence and a geometric sequence?

Describe how exponential functions and geometric sequences are similar. How are they different?

For the following exercises, find the common ratio for the geometric sequence.

1 , 3 , 9 , 27 , 81 , ... 1 , 3 , 9 , 27 , 81 , ...

− 0.125 , 0.25 , − 0.5 , 1 , − 2 , ... − 0.125 , 0.25 , − 0.5 , 1 , − 2 , ...

− 2 , − 1 2 , − 1 8 , − 1 32 , − 1 128 , ... − 2 , − 1 2 , − 1 8 , − 1 32 , − 1 128 , ...

For the following exercises, determine whether the sequence is geometric. If so, find the common ratio.

− 6 , − 12 , − 24 , − 48 , − 96 , ... − 6 , − 12 , − 24 , − 48 , − 96 , ...

5 , 5.2 , 5.4 , 5.6 , 5.8 , ... 5 , 5.2 , 5.4 , 5.6 , 5.8 , ...

− 1 , 1 2 , − 1 4 , 1 8 , − 1 16 , ... − 1 , 1 2 , − 1 4 , 1 8 , − 1 16 , ...

6 , 8 , 11 , 15 , 20 , ... 6 , 8 , 11 , 15 , 20 , ...

0.8 , 4 , 20 , 100 , 500 , ... 0.8 , 4 , 20 , 100 , 500 , ...

For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio.

a 1 = 8 , r = 0.3 a 1 = 8 , r = 0.3

a 1 = 5 , r = 1 5 a 1 = 5 , r = 1 5

For the following exercises, write the first five terms of the geometric sequence, given any two terms.

a 7 = 64 , a 10 = 512 a 7 = 64 , a 10 = 512

a 6 = 25 , a 8 = 6.25 a 6 = 25 , a 8 = 6.25

For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio.

The first term is 2, 2, and the common ratio is 3. 3. Find the 5 th term.

The first term is 16 and the common ratio is − 1 3 . − 1 3 . Find the 4 th term.

For the following exercises, find the specified term for the geometric sequence, given the first four terms.

a n = { − 1 , 2 , − 4 , 8 , ... } . a n = { − 1 , 2 , − 4 , 8 , ... } . Find a 12 . a 12 .

a n = { − 2 , 2 3 , − 2 9 , 2 27 , ... } . a n = { − 2 , 2 3 , − 2 9 , 2 27 , ... } . Find a 7 . a 7 .

For the following exercises, write the first five terms of the geometric sequence.

a 1 = − 486 , a n = − 1 3 a n − 1 a 1 = − 486 , a n = − 1 3 a n − 1

a 1 = 7 , a n = 0.2 a n − 1 a 1 = 7 , a n = 0.2 a n − 1

For the following exercises, write a recursive formula for each geometric sequence.

a n = { − 1 , 5 , − 25 , 125 , ... } a n = { − 1 , 5 , − 25 , 125 , ... }

a n = { − 32 , − 16 , − 8 , − 4 , ... } a n = { − 32 , − 16 , − 8 , − 4 , ... }

a n = { 14 , 56 , 224 , 896 , ... } a n = { 14 , 56 , 224 , 896 , ... }

a n = { 10 , − 3 , 0.9 , − 0.27 , ... } a n = { 10 , − 3 , 0.9 , − 0.27 , ... }

a n = { 0.61 , 1.83 , 5.49 , 16.47 , ... } a n = { 0.61 , 1.83 , 5.49 , 16.47 , ... }

a n = { 3 5 , 1 10 , 1 60 , 1 360 , ... } a n = { 3 5 , 1 10 , 1 60 , 1 360 , ... }

a n = { − 2 , 4 3 , − 8 9 , 16 27 , ... } a n = { − 2 , 4 3 , − 8 9 , 16 27 , ... }

a n = { 1 512 , − 1 128 , 1 32 , − 1 8 , ... } a n = { 1 512 , − 1 128 , 1 32 , − 1 8 , ... }

a n = − 4 ⋅ 5 n − 1 a n = − 4 ⋅ 5 n − 1

a n = 12 ⋅ ( − 1 2 ) n − 1 a n = 12 ⋅ ( − 1 2 ) n − 1

For the following exercises, write an explicit formula for each geometric sequence.

a n = { − 2 , − 4 , − 8 , − 16 , ... } a n = { − 2 , − 4 , − 8 , − 16 , ... }

a n = { 1 , 3 , 9 , 27 , ... } a n = { 1 , 3 , 9 , 27 , ... }

a n = { − 4 , − 12 , − 36 , − 108 , ... } a n = { − 4 , − 12 , − 36 , − 108 , ... }

a n = { 0.8 , − 4 , 20 , − 100 , ... } a n = { 0.8 , − 4 , 20 , − 100 , ... }

a n = { − 1.25 , − 5 , − 20 , − 80 , ... } a n = { − 1.25 , − 5 , − 20 , − 80 , ... }

a n = { − 1 , − 4 5 , − 16 25 , − 64 125 , ... } a n = { − 1 , − 4 5 , − 16 25 , − 64 125 , ... }

a n = { 2 , 1 3 , 1 18 , 1 108 , ... } a n = { 2 , 1 3 , 1 18 , 1 108 , ... }

a n = { 3 , − 1 , 1 3 , − 1 9 , ... } a n = { 3 , − 1 , 1 3 , − 1 9 , ... }

For the following exercises, find the specified term for the geometric sequence given.

Let a 1 = 4 , a 1 = 4 , a n = − 3 a n − 1 . a n = − 3 a n − 1 . Find a 8 . a 8 .

Let a n = − ( − 1 3 ) n − 1 . a n = − ( − 1 3 ) n − 1 . Find a 12 . a 12 .

For the following exercises, find the number of terms in the given finite geometric sequence.

a n = { − 1 , 3 , − 9 , ... , 2187 } a n = { − 1 , 3 , − 9 , ... , 2187 }

a n = { 2 , 1 , 1 2 , ... , 1 1024 } a n = { 2 , 1 , 1 2 , ... , 1 1024 }

For the following exercises, determine whether the graph shown represents a geometric sequence.

For the following exercises, use the information provided to graph the first five terms of the geometric sequence.

a 1 = 1 , r = 1 2 a 1 = 1 , r = 1 2

a 1 = 3 , a n = 2 a n − 1 a 1 = 3 , a n = 2 a n − 1

a n = 27 ⋅ 0.3 n − 1 a n = 27 ⋅ 0.3 n − 1

Use recursive formulas to give two examples of geometric sequences whose 3 rd terms are 200. 200.

Use explicit formulas to give two examples of geometric sequences whose 7 th terms are 1024. 1024.

Find the 5 th term of the geometric sequence { b , 4 b , 16 b , ... } . { b , 4 b , 16 b , ... } .

Find the 7 th term of the geometric sequence { 64 a ( − b ) , 32 a ( − 3 b ) , 16 a ( − 9 b ) , ... } . { 64 a ( − b ) , 32 a ( − 3 b ) , 16 a ( − 9 b ) , ... } .

At which term does the sequence { 10 , 12 , 14.4 , 17.28 , ... } { 10 , 12 , 14.4 , 17.28 , ... } exceed 100 ? 100 ?

At which term does the sequence { 1 2187 , 1 729 , 1 243 , 1 81 ... } { 1 2187 , 1 729 , 1 243 , 1 81 ... } begin to have integer values?

For which term does the geometric sequence a n = − 36 ( 2 3 ) n − 1 a n = − 36 ( 2 3 ) n − 1 first have a non-integer value?

Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10 th term.

Use the explicit formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8 th term.

Is it possible for a sequence to be both arithmetic and geometric? If so, give an example.

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• Book title: College Algebra 2e
• Publication date: Dec 21, 2021
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra-2e/pages/9-3-geometric-sequences

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Lesson: Applications of Geometric Sequences and Series Mathematics • Second Year of Secondary School

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In this lesson, we will learn how to solve real-world applications of geometric sequences and series, where we will find the common ratio, the nth term explicit formula, the order and value of a specific sequence term, and the sum of a given number of terms.

Lesson Plan

Students will be able to

• model a real-world problem using a geometric sequence,
• apply methods from geometric sequences to solve real-world problems, for example, modeling a bouncing ball.

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Geometric Sequences – Examples and Practice Problems

Geometric sequences have the main characteristic of having a common ratio, which is multiplied by the last term to find the next term. Any term in a geometric sequence can be found using a formula.

Here, we will look at a summary of geometric sequences and we will explore its formula. In addition, we will see several examples with answers and exercises to solve to practice these concepts.

Relevant for …

Exploring examples with answers of geometric sequences.

See examples

Summary of geometric sequences

Geometric sequences – examples with answers, geometric sequences – practice problems.

Geometric sequences are sequences in which the next number in the sequence is found by multiplying the previous term by a number called the  common ratio . The common ratio is denoted by the letter  r .

Depending on the common ratio, the geometric sequence can be increasing or decreasing. If the common ratio is greater than 1, the sequence is increasing and if the common ratio is between 0 and 1, the sequence is decreasing:

We can find any number in the geometric sequence using the geometric sequence formula:

We can find the common ratio by dividing any term by the previous term:

$latex r=\frac{a_{n}}{a_{n-1}}$

Find the next term in the geometric sequence: 4, 8, 16, 32,  ? .

First, we have to find the common ratio of the geometric progression. To do this, we divide a term by the previous term:

• $latex \frac{32}{16}=2$
• $latex \frac{16}{8}=2$
• $latex \frac{8}{4}=2$

Therefore, the common ratio is 2. To find the next term, we multiply the last term by the common ratio: $latex 32\times 2=64$.

What is the next term in the geometric sequence? 3, 15, 75, 375,  ? .

We start by finding the common ratio for the geometric progression. Then, we divide each term by its previous term:

• $latex \frac{375}{75}=5$
• $latex \frac{75}{15}=5$
• $latex \frac{15}{3}=5$

We see that the common ratio is 5. We find the next term by multiplying the last term by the common ratio : $latex 375 \times 5=1875$.

Determine the next term in the geometric sequence: 48, 24, 12, 6,  ? .

Again, we start by finding the common ratio in the progression:

• $latex \frac{6}{12}=0.5$
• $latex \frac{12}{24}=0.5$
• $latex \frac{24}{48}=0.5$

In this case, we see that the common ratio is between 0 and 1, so the progression is slowing down. The next term in the geometric progression is $latex 6\times 0.5=3$.

What is the value of the 6th term of a geometric sequence where the first term is 3 and the common ratio is 2?

We have the following values:

• First term: $latex a_{1}=3$
• Common ratio: $latex r=2$
• Position of term: $latex n=6$

Then, we can use the formula for geometric sequences with the given values:

$latex a_{n}=a_{1}(r^{n-1})$

$latex a_{6}=3(2^{6-1})$

$latex a_{6}=3(2^{5})$

$latex a_{6}=5(32)$

$latex a_{6}=160$

Find the 12th term in the geometric sequence: 5, 15, 45, 135, …

In this case, we have to use the formula of geometric progressions $latex a_{n}=a_{1}({{r}^{n-1}})$. Therefore, we have to identify the first term, the common reason and the position of the term:

• First term: $latex a_{1}=5$
• Common ratio: $latex r=3$
• Position of term: $latex n=12$

Now, we substitute this data into the formula:

$latex a_{n}=a_{1}({{r}^{n-1}})$

$latex a_{12}=5({{3}^{12-1}})$

$latex a_{12}=5({{3}^{11}})$

$latex a_{12}=5(177147)$

$latex a_{12}=885 735$

We see that we have a very large number. Geometric progressions tend to grow rapidly depending on the common proportion.

Find the 8th term in the geometric sequence 8, 32, 128, 512, …

Again, we start by identifying the first term, the common ratio, and the position of the term to be used with the formula:

• First term: $latex a_{1}=8$
• Common ratio: $latex r=4$
• Position of term: $latex n=8$

Now, we use the formula with these values:

$latex a_{8}=8({{4}^{8-1}})$

$latex a_{8}=8({{4}^{7}})$

$latex a_{8}=8(16384)$

$latex a_{8}=131072$

Find the 10th term in the geometric sequence: 168, 84, 42, 21, …

In this case, we have a decreasing geometric progression, so we expect the common ratio to be between 0 and 1:

• First term: $latex a_{1}=168$
• Common ratio: $latex r=0.5$
• Position of term: $latex n=10$

We use the formula to find the term 10:

$latex a_{10}=168({{0.5}^{10-1}})$

$latex a_{10}=168({{0.5}^{9}})$

$latex a_{10}=168(0.001953)$

$latex a_{10}=0.328$

Find the 7th term in the geometric sequence: 540, 180, 60, 20, …

Similar to the previous example, here we have a decreasing geometric progression, so the common ratio must be between 0 and 1:

• First term: $latex a_{1}=540$
• Common ratio: $latex r=\frac{1}{3}$
• Possiion of term: $latex n=7$

We use these values to substitute in the formula:

$latex a_{7}=540({{\left( \frac{1}{3}\right)}^{7-1}})$

$latex a_{7}=540({{\left( \frac{1}{3}\right)}^{6}})$

$latex a_{7}=540(0.0013717)$

$latex a_{7}=0.7407$

If the 4th term of a geometric sequence is 16 and the 7th term is 128, what is the 11th term?

In this case, we know neither the value of the first term nor the common ratio. However, we can start by forming the following equations:

$latex a_{4}=a_{1}(r^{4-1})$

$latex 16=a_{1}(r^{3})~~~[1]$

$latex a_{7}=a_{1}(r^{7-1})$

$latex 128=a_{1}(r^{6})~~~[2]$

If we divide equation 2 by equation 1, we have:

$$\frac{128}{16}=\frac{a_{1}(r^{6})}{a_{1}(r^{3})$$

$latex 8=r^{3}$

$latex r=2$

If we consider the 7th term as the 1st term, the 11th term is now the 5th term:

• First term: $latex a_{1}=128$
• Position of term: $latex n=5$

Using these values in the formula, we have:

$latex a_{5}=128(2^{5-1})$

$latex a_{5}=128(2^4)$

$latex a_{5}=128(16)$

$latex a_{5}=2048$

Then, the 11th term of the given sequence is 2048.

A geometric sequence has a 3rd term equal to 256 and an 8th term equal to -8. What is the value of the 14th term?

Similar to the previous example, we can find the common ratio by forming the following equations:

$latex a_{3}=a_{1}(r^{3-1})$

$latex 256=a_{1}(r^{2})~~~[1]$

$latex a_{8}=a_{1}(r^{8-1})$

$latex -8=a_{1}(r^{7})~~~[2]$

Now we divide them to obtain:

$$\frac{-8}{256}=\frac{a_{1}(r^{7})}{a_{1}(r^{2})$$

$$-\frac{1}{32}=r^{5}$$

$$r=-\frac{1}{2}$$

Considering the 8th term as the first term, the 14th term corresponds to the 7th term. Then:

• First term: $latex a_{1}=-8$
• Common ratio: $latex r=-\frac{1}{2}$
• Position of term: $latex n=7$

Using the formula, we have:

$$a_{7}=-8(-\frac{1}{2}^{7-1})$$

$$a_{7}=-8(-\frac{1}{2}^6)$$

$$a_{7}=-8(\frac{1}{64})$$

$$a_{7}=-\frac{1}{8}$$

Then, the 14th term of the given sequence is $latex -\frac{1}{8}$.

In a geometric sequence, the 4th term is 135 and 7th term is 3645. What is the value of the 15th term?

Write the answer in the input box.

Interested in learning more about sequences? Take a look at these pages:

• Arithmetic and Geometric Sequences
• Examples of Arithmetic Sequences

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Geometric sequence word problems

This lesson will show you how to solve a variety of geometric sequence word problems.

Example #1:

The stock's price of a company is not doing well lately. Suppose the stock's price is 92% of its previous price each day. What is the stock's price after 10 days if the stock was worth $2500 right before it started to go down?  

To solve this problem, we need the geometric sequence formula shown below.

a n  = a 1  × r (n - 1)

a 1  = original value of the stock  = 2500

a 2  = value of the stock after 1 day

a 11  = value of the stock after 10 days

a 11 = 2500 × (0.92) (11 - 1)

a 11  = 2500 × (0.92) 10

a 11  = 2500 × 0.434

a 11  = $1085

The stock's price is about 1085 dollars.

Example #2:

The third term of a geometric sequence is 45 and the fifth term of the geometric sequence is 405. If all the terms of the sequence are positive numbers, find the 15th term of the geometric sequence.

Solution To solve this problem, we need the geometric sequence formula shown below.

a n  = a 1  × r (n - 1)

Find the third term

a 3  = a 1  × r (3 - 1)

a 3  = a 1  × r 2

Since the third term is 45,  45 = a 1  × r 2 ( equation 1 )

Find the fifth term

a 5 = a 1  × r (5 - 1)

a 5  = a 1  × r 4

Since the fifth term is 405,  405 = a 1  × r 4 ( equation 2 )

Divide equation 2 by equation 1 .

(a 1  × r 4 ) / (a 1  × r 2 ) = 405 / 45

Cancel a 1 since it is both on top and at the bottom of the fraction.

r 4 / r 2 = 9

r = ±√9

r = ±3

Use r  = 3, and equation 1 to find a 1

45 = a 1  × (3) 2

45 = a 1  × 9

a 1 = 45 / 9 = 5

Since all the terms of the sequence are positive numbers, we must use r = 3 if we want all the terms to be positive numbers.

Let us now find a 15

a 15 = 5 × (3) (15 - 1)

a 15 = 5 × (3) 14  

a 15  = 5 × 4782969 

a 15  =  23914845

Challenging geometric sequence word problems

Example #3:

Suppose that the magnification of a PDF file on a desktop computer is increased by 15% for each level of zoom. Suppose also that the original length of the word " January " is 1.2 cm. Find the length of the word " January " after 6 magnifications.

a 1  = original length of the word  = 1.2 cm

a 2 = length of the word after 1 magnification

a 7 = length of the word after 6 magnifications

r = 1 + 0.15 = 1.15

a 7  = 1.2 × (1.15) (7 - 1)

a 7 = 1.2 × (1.15) 6

a 7 = 1.2 × 2.313

a 7  = 2.7756

After 6 magnifications, the length of the word "January" is 2.7756 cm.

Notice that we added 1 to 0.15. Why did we do that? Let us not use the formula directly so you can see the reason behind it. Study the following carefully !

Day 1 : a 1 = 1.2

Day 2 : a 2 = 1.2 + 1.2 (0.15) = 1.2 (1 + 0.15)

Day 3 : a 3 =  1.2(1 + 0.15) + [ 1.2(1 + 0.15) ]0.15 =  1.2(1 + 0.15) (1 + 0.15) = 1.2(1 + 0.15) 2

Day 7 : a 7 = 1.2(1 + 0.15) 6

Suppose that you want a reduced copy of a photograph. The actual length of the photograph is 10 inches. If each reduction is 64% of the original, how many reductions, will shrink the photograph to 1.07 inches.

a 1  = original length of the photograph  = 10 inches

a 2  = length of the photograph after 1 reduction

n = number of reductions = ?

1.07 = 10 × (0.64) (n - 1)

Divide both sides by 10

1.07 / 10 = [10 × (0.64) (n - 1) ] / 10

0.107 = (0.64) (n - 1)

Notice that you have an exponential equation to solve. The biggest challenge then is knowing how to solve exponential equations !

Take the natural log of both sides of the equation.

ln(0.107) = ln[(0.64) (n - 1) ]

Use the power property of logarithms .

ln(0.107) = (n - 1)ln(0.64)

Divide both sides of the equation by ln(0.64)

ln(0.107) / ln(0.64) = (n - 1)ln(0.64) / ln(0.64)

n - 1 = ln(0.107) / ln(0.64)

Use a calculator to find ln(0.107) and ln(0.64)

n - 1 = -2.23492644452 \ -0.44628710262

n - 1 = 5.0078

n = 1 + 5.0078

Therefore, you will need 6 reductions.

Geometric sequence

Arithmetic sequence word problems

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Geometric Series & Applications

These lessons, with videos, examples and step-by-step solutions, help High School students learn to derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems. For example, calculate mortgage payments.

Related Pages Geometric Sequence Common Core (Algebra) Common Core for Mathematics

Suggested Learning Targets

Develop the formula for the sum of a finite geometric series when the ratio is not 1.

Use the formula to solve real world problems such as calculate mortgage payments.

Common Core: HSA-SSE.B.4

The following diagrams show to derive the formula for the sum of a finite geometric series. Scroll down the page for more examples and solutions of geometric series.

Derive Formula for finite geometric series

Geometric Series (How to Find the Sum of a Sequence) How can you find the sum of a geometric series when you’re given only the first few terms and the last one? There are two formulas, and I show you how to do it. You need to find “n” for the last term.

Geometric series sum to figure out mortgage payments Figuring out the formula for fixed mortgage payments using the sum of a geometric series.

Geometric Series Application - Total Income with Doubling Pay This video provides an application problem that can be modeled by the sum of an a geometric sequence dealing with total income with pay doubling everyday.

Examples: A company offers to pay you $0.10 for the first day, $0.20 for the second day, $0.40 for the third day, $0.80 for the fourth day, and so on. What would your total income be after 1 month (30 days)?

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WORD PROBLEMS IN GEOMETRIC SEQUENCE

Problem 1 :

A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?

Starting salary   =  60,000 

Every year 5% of annual salary is increasing.

Second year salary  =  60000 + 5% of 60000

  =  60000(1 + 5%)

Third year salary  =  60000 + 5% of 60000(1 + 5%)

  =  60000(1 + 5%)(1 + 5%)

  =  60000(1 + 5%) 2

By continuing in this way,    salary after 5 years is

  =   60000(1 + 5%)5

  =   60000(105/100) 5

  =   60000(1.05) 5 

  =  76577

Problem 2 :

Sivamani is attending an interview for a job and the company gave two offers to him.

Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.

Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

What is his salary in the 4th year with respect to the offers A and B?

a = 20000, r = 0.06 and n = 5

Starting salary   =  20,000 

Every year 6% of annual salary is increasing.

Starting salary  =  20000

Second year salary  =  20000 + 6% of 20000

  =  20000(1 + 6%)

By continuing in this way, we get

4th year salary  =  20000(1 + 6%) 3

=  20000(1.06) 3

=  22820

a = 22000, r = 0.03 and n = 5

Starting salary   =  22,000 

Every year 3% of annual salary is increasing.

Starting salary  =  22000

Second year salary  =  22000 + 3% of 22000

  =  22000(1 + 3%)

4th year salary  =  22000(1 + 3%) 3

=  22000(1.03) 3

=  24040

Problem 3 :

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that x b−c × y c−a × z a−b = 1 .

Since a, b and c are in A.P,

b - a  = c - b  = d (common difference)

We need to prove,

x b−c  × y c−a  × z a−b  = 1 

Let us try to convert the powers in terms of one variable.

2b  =  c + a - a + a

2b  =  c - a + 2a

2(b - a)  =  c - a

2d  =  c - a 

If c - b  =  d, then b - c = -d

If b - a  =  d, then a - b = -d

x b−c  × y c−a  × z a−b  =   x −d  × y 2d  × z − d    ---(1)

y =  √xz

By applying the value of y in (1)

 =   x −d  × ( √xz) 2d  × z − d  

 =  x −d  × ( xz) d  × z − d  

  =  x −d + d   z -d + d

  =  1

Hence proved.

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Real-Life Applications of Geometric and Arithmetic Sequences

• First Online: 10 December 2016

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• Ellina Grigorieva 2  

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Over the millenia, legends have developed around mathematical problems involving series and sequences. One of the most famous legends about series concerns the invention of chess. According to the legend, an Indian king summoned the inventor and suggested that he choose the award for the creation of an interesting and wise game. The king was amazed by the “modest” request from the inventor who asked to give him for the first cell of the chessboard 1 grain of wheat, for the second—2 grains, for the third—4 grains, for the fourth—twice as much as in the previous cell, etc. As a result, the total number of grains per 64 cells of the chessboard would be so huge that the king would have to plant it everywhere on the entire surface of the Earth including the space of the oceans, mountains, and deserts and even then would not have enough!

Have you ever thought of how archeologists in the movies, such as Indiana Jones, can predict the age of different artifacts? Do not you know that the age of artifacts in real life can be established by the amount of the radioactive isotope of Carbon 14 in the artifact? Carbon 14 has a very long half-lifetime which means that each half-lifetime of 5730 years or so, the amount of the isotope is reduced by half. Hence, these consecutive amounts of Carbon 14 are the terms of a decreasing geometric progression with common ratio of ½.

This chapter is for those who want to see applications of arithmetic and geometric progressions to real life. There are many applications for sciences, business, personal finance, and even for health, but most people are unaware of these. We will familiarize you with these by giving you five mini-projects and some related problems associated with the concepts afterwards.

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About this chapter

Grigorieva, E. (2016). Real-Life Applications of Geometric and Arithmetic Sequences. In: Methods of Solving Sequence and Series Problems. Birkhäuser, Cham. https://doi.org/10.1007/978-3-319-45686-7_4

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8.2: Problem Solving with Arithmetic Sequences

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Arithmetic sequences, introduced in Section 8.1, have many applications in mathematics and everyday life. This section explores those applications.

Example 8.2.1

A water tank develops a leak. Each week, the tank loses \(5\) gallons of water due to the leak. Initially, the tank is full and contains \(1500\) gallons.

• How many gallons are in the tank \(20\) weeks later?
• How many weeks until the tank is half-full?
• How many weeks until the tank is empty?

This problem can be viewed as either a linear function or as an arithmetic sequence. The table of values give us a few clues towards a formula.

The problem allows us to begin the sequence at whatever \(n\)−value we wish. It’s most convenient to begin at \(n = 0\) and set \(a_0 = 1500\).

Therefore, \(a_n = −5n + 1500\)

Since the leak is first noticed in week one, \(20\) weeks after the initial week corresponds with \(n = 20\). Use the formula where \(\textcolor{red}{n = 20}\):

\(a_{20} = −5(\textcolor{red}{20}) + 1500 = −100 + 1500 = 1400\)

Therefore, \(20\) weeks later, the tank contains \(1400\) gallons of water.

• How many weeks until the tank is half-full? A half-full tank would be \(750\) gallons. We need to find \(n\) when \(\textcolor{red}{a_n = 750}\).

\(\begin{array} &750 &= −5n + 1500 &\text{Substitute \(a_n = 750\) into the general term.} \\ 750 − 1500 &= −5n + 1505 − 1500 &\text{Subtract \(1500\) from each side of the equation.} \\ −750 &= −5n &\text{Simplify each side of the equation.} \\ \dfrac{−750}{−5} &= \dfrac{−5n}{−5} &\text{Divide both sides by \(−5\).} \\ 150 &= n & \end{array}\)

Since \(n\) is the week-number, this answer tells us that on week \(150\), the tank is half full. However, most people would better understand the answer if stated in the following way, “The tank is half full after 150 weeks.” This answer sounds more natural and is preferred.

• How many weeks until the tank is empty? The tank is empty when \(a_n = 0\) gallons. Find \(n\) such that \(\textcolor{red}{a_n = 0}\).

\(\begin{array}& 0 &= −5n + 1500 &\text{Substitute \(a_n=0\) into the general term.} \\ 0 − 1500 &= −5n + 1500 − 1500 &\text{Subtract \(1500\) from each side of the equation.} \\ −1500 &= −5n &\text{Simplify.} \\ \dfrac{−1500}{−5} &= \dfrac{−5n}{−5} &\text{Divide both sides by \(−5\).} \\ 300 &= n & \end{array}\)

Since \(n\) is the week-number, this answer tells us that on week \(300\), the tank is empty. However, most people would better understand the answer if stated in the following way, “ The tank is empty after 300 weeks. ” This answer sounds more natural and is preferred.

Example 8.2.2

Three stages of a pattern are shown below, using matchsticks. Each stage requires a certain number of matchsticks. If we keep up the pattern…

• How many matchsticks are required to make the figure in stage \(34\)?
• What stage would require \(220\) matchsticks?

Let’s create a table of values. Let \(n =\) stage number, and let \(a_n =\) the number of matchsticks used in that stage. Then note the common difference.

Find the value \(a_0\):

\(\begin{array} &a_0 + 3 &= 4 \\ a_0 + 3 − 3 &= 4 − 3 \\ a_0 &= 1 \end{array}\)

The general term of the sequence is:

\(a_n = 3n + 1\)

• Compute \(a_{34}\) to find the number of matchsticks in stage \(34\):

\(a_{34} = 3(\textcolor{red}{34}) + 1 = 103\).

There are \(103\) matchsticks in stage \(34\).

• What stage would require \(220\) matchsticks? We are looking for the stage-number, given the number of matchsticks. Find \(n\) if \(a_n = 220\).

\(\begin{array} &220 &= 3n + 1 \\ 219 &= 3n \\ 73 &= n \end{array}\)

Answer Stage \(73\) would require \(220\) matchsticks.

Example 8.2.3

Cory buys \(5\) items at the grocery store with prices \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\) which is an arithmetic sequence. The least expensive item is \($1.89\), while the total cost of the \(5\) items is \($12.95\). What is the cost of each item?

Put the \(5\) items in order of expense: least to most and left to right. Because it is an arithmetic sequence, each item is \(d\) more dollars than the previous item. Each item’s price can be written in terms of the price of the least expensive item, \(a_1\), and \(a_1 = $1.89\).

The diagram above gives \(5\) expressions for the costs of the \(5\) items in terms of \(a_1\) and the common difference is \(d\).

\(\begin{array} &a_1 + a_2 + a_3 + a_4 + a_5 &= 12.95 &\text{Total cost of \(5\) items is \($12.95\).} \\ a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + (a_1 + 4d) &= 12.95 &\text{See diagram for substitutions.} \\ 5s_1 + 10d &= 12.95 &\text{Gather like terms.} \\ 5(1.89) + 10d &= 12.95 &a_1 = 1.89. \\ 9.45 + 10d &= 12.95 &\text{Simplify.} \\ 9.45 + 10d − 9.45 &= 12.95 − 9.45 &\text{Subtract \(9.45\) from each side of equation.} \\ 10d &= 3.50 &\text{Simplify. Then divide both sides by \(10\).} \\ d &= 0.35 &\text{The common difference is \($0.35\).} \end{array}\)

Now that we know the common difference, \(d = $0.35\), we can answer the question.

The price of each item is as follows: \($1.89, $2.24, $2.59, $2.94, $3.29\).

Try It! (Exercises)

1. ZKonnect cable company requires customers sign a \(2\)-year contract to use their services. The following describes the penalty for breaking contract: Your services are subject to a minimum term agreement of \(24\) months. If the contract is terminated before the end of the \(24\)-month contract, an early termination fee is assessed in the following manner: \($230\) termination fee is assessed if contract is terminated in the first \(30\) days of service. Thereafter, the termination fee decreases by \($10\) per month of contract.

• If Jack enters contract with ZKonnect on April 1 st of \(2021\), but terminates the service on January 10 th of \(2022\), what are Jack’s early termination fees?
• The general term \(a_n\) describes the termination fees for the stated contract. Describe the meaning of the variable \(n\) in the context of this problem. Find the general term \(a_n\).
• Is the early termination fee a finite sequence or an infinite sequence? Explain.
• Find the value of \(a_{13}\) and interpret its meaning in words.

2. A drug company has manufactured \(4\) million doses of a vaccine to date. They promise additional production at a rate of \(1.2\) million doses/month over the next year.

• How many doses of the vaccine, in total, will have been produced after a year?
• The general term \(a_n\) describes the total number of doses of the vaccine produced. Describe the meaning of the variable \(n\) in the context of this problem. Find the general term\(a_n\).
• Find the value of \(a_8\) and interpret its meaning in words.

3. The theater shown at right has \(22\) seats in the first row of the “A Center” section. Each row behind the first row gains two additional seats.

• Let \(a_n = 22 + 2n\), starting with \(n = 0\). Give the first \(10\) values of this sequence.
• Using \(a_n = 22 + 2n\), Find the value of \(a_{10}\) and interpret its meaning in words in the context of this problem. Careful! Does \(n=\) row number?
• How many seats, in total, are in “A Center” section if there are \(12\) rows in the section?

4) Logs are stacked in a pile with \(48\) logs on the bottom row and \(24\) on the top row. Each row decreases by three logs.

• The stack, as described, has how many rows of logs?
• Write the general term \(a_n\) to describe the number of logs in a row in two different ways. Each general term should produce the same sequence, regardless of its starting \(n\)-value.

i. Start with \(n = 0\).

ii. Start with \(n = 1\).

5) The radii of the target circle are an arithmetic sequence. If the area of the innermost circle is \(\pi \text{un}^2\) and the area of the entire target is \(49 \pi \text{un}^2\), what is the area of the blue ring? [The formula for area of a circle is \(A = \pi r^2\)].

6) Three stages of a pattern are shown below, using matchsticks. Each stage adds another triangle and requires a certain number of matchsticks. If we keep up the pattern…

• What stage would require \(325\) matchsticks?

7) Three stages of a pattern are shown below, using matchsticks. Each stage requires a certain number of matchsticks. If we keep up the pattern…

• How many matchsticks are required to make the figure in stage \(22\)?
• What stage would require \(424\) matchsticks?