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Arithmetic Progressions

Definition:

By an arithmetic progression of $m$ terms, we mean a finite sequence of the form

The real number $a$ is called the first term of the arithmetic progression, and the real number $d$ is called the difference of the arithmetic progression.

Consider the sequence of numbers

The property of this sequence is that the difference between successive terms is constant and equal to 2.

Here we have: $a = 1$; $d = 2$.

Consider the sequence of numbers:

The property of this sequence is that the difference between successive terms is constant and equal to 3.

Here we have: $a = 2$; $d = 3$.

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Arithmetic Sequences Problems with Solutions

Arithmetic sequences are used throughout mathematics and applied to engineering, sciences, computer sciences, biology and finance problems. A set of problems and exercises involving arithmetic sequences, along with detailed solutions are presented.

Review of Arithmetic Sequences

The formula for the n th term a n of an arithmetic sequence with a common difference d and a first term a 1 is given by \[ a_n = a_1 + (n - 1) d \] The sum s n of the first n terms of an arithmetic sequence is defined by \[ s_n = a_1 + a_2 + a_3 + ... + a_n \] and is is given by \[ s_n = \dfrac{n (a_1 + a_n)}{2} \] Arithmetic Series Online Calculator . An online calculator to calculate the sum of the terms in an arithmetic sequence.

Problems with Solutions

The first term of an arithmetic sequence is equal to 6 and the common difference is equal to 3. Find a formula for the n th term and the value of the 50 th term Solution to Problem 1: Use the value of the common difference d = 3 and the first term a 1 = 6 in the formula for the n th term given above \( a_n = a_1 + (n - 1) d \\ = 6 + 3 (n - 1) \\ = 3 n + 3 \) The 50 th term is found by setting n = 50 in the above formula. \[ a_{50} = 3 (50) + 3 = 153 \]

The first term of an arithmetic sequence is equal to 200 and the common difference is equal to -10. Find the value of the 20 th term Solution to Problem 2: Use the value of the common difference d = -10 and the first term a 1 = 200 in the formula for the n th term given above and then apply it to the 20 th term a 20 = 200 + (-10) (20 - 1 ) = 10

An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term. Solution to Problem 3: We use the n th term formula for the 6 th term, which is known, to write a 6 = 52 = a 1 + 10 (6 - 1 ) The above equation allows us to calculate a 1 . a 1 = 2 Now that we know the first term and the common difference, we use the n th term formula to find the 15 th term as follows. a 15 = 2 + 10 (15 - 1) = 142

An arithmetic sequence has a its 5 th term equal to 22 and its 15 th term equal to 62. Find its 100 th term. Solution to Problem 4: We use the n th term formula for the 5 th and 15 th terms to write a 5 = a 1 + (5 - 1 ) d = 22 a 15 = a 1 + (15 - 1 ) d = 62 We obtain a system of 2 linear equations where the unknown are a 1 and d. Subtract the right and left term of the two equations to obtain 62 - 22 = 14 d - 4 d Solve for d. d = 4 Now use the value of d in one of the equations to find a 1 . a 1 + (5 - 1 ) 4 = 22 Solve for a 1 to obtain. a 1 = 6 Now that we have calculated a 1 and d we use them in the n th term formula to find the 100 th formula. a 100 = 6 + 4 (100 - 1 )= 402

Find the sum of all the integers from 1 to 1000. Solution to Problem 5: The sequence of integers starting from 1 to 1000 is given by 1 , 2 , 3 , 4 , ... , 1000 The above sequence has 1000 terms. The first term is 1 and the last term is 1000 and the common difference is equal to 1. We have the formula that gives the sum of the first n terms of an arithmetic sequence knowing the first and last term of the sequence and the number of terms (see formula above). s 1000 = 1000 (1 + 1000) / 2 = 500500

Find the sum of the first 50 even positive integers. Solution to Problem 6: The sequence of the first 50 even positive integers is given by 2 , 4 , 6 , ... The above sequence has a first term equal to 2 and a common difference d = 2. We use the n th term formula to find the 50 th term a 50 = 2 + 2 (50 - 1) = 100 We now the first term and last term and the number of terms in the sequence, we now find the sum of the first 50 terms s 50 = 50 (2 + 100) / 2 = 2550

Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5. Solution to Problem 7: The first few terms of a sequence of positive integers divisible by 5 is given by 5 , 10 , 15 , ... The above sequence has a first term equal to 5 and a common difference d = 5. We need to know the rank of the term 1555. We use the formula for the n th term as follows 1555 = a 1 + (n - 1 )d Substitute a 1 and d by their values 1555 = 5 + 5(n - 1 ) Solve for n to obtain n = 311 We now know that 1555 is the 311 th term, we can use the formula for the sum as follows s 311 = 311 (5 + 1555) / 2 = 242580

Find the sum S defined by \[ S = \sum_{n=1}^{10} (2n + 1 / 2) \] Solution to Problem 8: Let us first decompose this sum as follows \( S = \sum_{n=1}^{10} (2n + 1 / 2) \) \( = 2 \sum_{n=1}^{10} n + \sum_{n=1}^{10} (1/2) \) The term ∑ n is the sum of the first 10 positive integers. The 10 first positive integers make an arirhmetic sequence with first term equal to 1, it has n = 10 terms and its 10 th term is equal to 10. This sum is obtained using the formula s n = n (a 1 + a n ) / 2 as follows 10(1+10)/2 = 55 The term ∑ (1 / 2) is the addition of a constant term 10 times and is given by 10(1/2) = 5 The sum S is given by S = 2(55) + 5 = 115

Answer the following questions related to arithmetic sequences: a) Find a 20 given that a 3 = 9 and a 8 = 24 b) Find a 30 given that the first few terms of an arithmetic sequence are given by 6,12,18,... c) Find d given that a 1 = 10 and a 20 = 466 d) Find s 30 given that a 10 = 28 and a 20 = 58 e) Find the sum S defined by \[ S = \sum_{n=1}^{20}(3n - 1 / 2) \] f) Find the sum S defined by \[ S = \sum_{n=1}^{20}0.2 n + \sum_{j=21}^{40} 0.4 j \]

Solutions to Above Exercises

a) a 20 = 60 b) a 30 = 180 c) d = 24 d) s 30 = 1335 e) 1380 f) 286

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Arithmetic and Geometric Progressions

In this mini-lesson, we will explore the world of arithmetic progression and geometric progression in math. You will get to learn about the arithmetic progression formula, geometric progression formula, sum of arithmetic progression, geometric progression sum, and other interesting facts around the topic. You can also check out the playful calculators to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page.

Let's start by answering a simple question. Have you ever observed patterns in your day-to-day life?

For example, Olivia and Betty are walking along a street to visit their friend, Hailey.

As they walk, they observe a pattern of house numbers on the street.

These numbers are in ascending order.

Here, these house numbers form a sequence.

Let's go ahead and learn about two types of sequences: arithmetic progression and geometric progression. 

Lesson Plan

What Is Definition And General Notations of Arithmetic Progression and Geometric Progression?

Definition of arithmetic progression .

An arithmetic progression is a sequence of numbers in which each successive term is a sum of its preceding term and a fixed number.

This fixed number is called a common difference.

Definition of Geometric Progression

A geometric progression is a sequence where every term bears a constant ratio  to its preceding term.

General Notations

We generally use the notation \(a_{n}\) to represent the \(n^{\text{th}}\) term in an arithmetic progression or a geometric progression.

The common difference is denoted by \(d\) and the common ratio is denoted by \(r\).

General Form and Nth Terms

Let's learn the general forms for arithmetic progression and geometric progression.

Arithmetic Progression: Formula

Let the first term of the sequence be \(a\) and the common difference be \(d\).

Then, \(n^{\text{th}}\) term of the sequence is given by: 

The nth term is the arithmetic sequence explicit formula and can be useful to find any term of the progression.

Experiment with the simulation below to determine the \(n^{\text{th}}\) term of different arithmetic progressions.

Change the sliders for \(a\) and \(d\) to get different sequences and change the slider for \(n\) to get the number of terms of that sequence.

Geometric Progression: Formula

Let the first term of the sequence be \(a\) and the common ratio be \(r\).

Then, \(n^{\text{th}}\) term of the sequence is given by:

Sum Of N Terms

Sum Of Arithmetic Progression

Consider an arithmetic progression  whose first term is \(a\) and the common difference is \(d\). 

The sum of the first \(n\) terms of an arithmetic progression is:

Now, we already know that \(n^{\text{th}}\) term of an arithmetic progression is \(\begin{align}a_{n}=a+(n-1)d\end{align}\).

So, the sum formula of an arithmetic progression can be written as 

\[\begin{align}S_n&=\frac{n}{2}(2 a+(n-1) d)\\&=\frac{n}{2}(a+a+(n-1) d)\\&=\frac{n}{2}(a+a_n)\end{align}\]

So, another formula to find the sum of the first \(n\) terms of an arithmetic progression is:

Geometric Progression: Sum

Consider a geometric progression whose first term is \(a\) and the common ratio is \(r\). 

The sum of the first \(n\) terms of a geometric progression is:

In an arithmetic progression, each successive term is obtained by adding the common difference to its preceding term.

In a geometric progression, each successive term is obtained by multiplying the common ratio to its preceding term.

• The sum of arithmetic progression whose first term is \(a\) and common difference is \(d\) can be calculated using one of the following formulas: \[\begin{align}S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm]S_n &= \frac{n}{2}(a_1+a_n) \end{align} \]
• The sum of geometric progression whose first term is \(a\) and common ratio is \(r\) can be calculated using the formula: \[\begin{align}S_n=\dfrac{a(1-r^n)}{1-r}\end{align} \]

Solved Examples

A taxi charges $2 for the first mile and $1.5 for each subsequent mile.

How much Katie needs to pay to the taxi driver if she travels 5 miles?

The taxi fare for the first few miles are $2, $3.5, $5, ...

It forms an arithmetic progression with a common difference, \(d=$1.5\) and first term, \(a=$2\)

The 5th term of the sequence will give the taxi fare for the first 5 miles.

The 5th term of the sequence will be given by \(a+4d\).

\[\begin{align}a+4d&=$2+4($1.5)\\&=$2+$6\\&=$8\end{align}\]

Mushi put $30 in her piggy bank when she was 7 years old.

She increased the amount on her each successive birthday by $3.

How much money will she have on her 12th birthday?

The amount in Mushi's piggy bank follows the pattern of $30, $33, $36, and so on.

The succeeding terms are obtained by adding a fixed number, that is, $3

This fixed number is a common difference.

The total money collected on her 12th birthday will be the sum of the first 6 terms on this arithmetic progression.

\[\begin{align}S_n&=\frac{n}{2}(2 a+(n-1) d)\\S_{6}&=\frac{6}{2}(60+(6-1)3)\\&=3\times 75\\&=225\end{align}\]

Jack is a fitness freak.

He started practicing by doing 10 pushups on his first day.

Every day, he did 5 pushups extra than the previous day.

How many pushups did he do on the last day of the first week?

The number of pushups represents the arithmetic progression given by 10, 15, 20, and so on.

The last day of the week is the 7th term of this sequence.

The 7th term of the sequence will be given by \(a+6d\).

\[\begin{align}a+6d&=10+6(5)\\&=40\end{align}\]

Look at the pattern shown below.

Observe that each square is half of the size of the square next to it.

Which sequence does this pattern represent?

Let's write the sequence represented in the figure.

\[1, \dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8}, \dfrac{1}{16},...\]

Every successive term is obtained by dividing its preceding term by 2

The sequence exhibits a common ratio of \(\dfrac{1}{2}\)

Hailey's teacher asks her to find the \( 10^{\text{th}}\) term of the sequence: 1, 4, 16, 64, ...

Can you help her?

Observe that \(\dfrac{4}{1}=\dfrac{16}{4}=\dfrac{64}{16}=4\)

Here, \(r=4\) is a common ratio.

So, the given sequence represents the geometric progression.

The 10th term of the sequence will be given by \(ar^{9}\).

\[\begin{align}ar^{9}&=1\times 4^9\\&=4^9\end{align}\]

A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on.

Can you determine the number of ancestors during the 8 generations preceding his own?

The number of ancestors in each generation represents the term of a geometric progression.

So, \(a=2\), \(r=2\), and \(n=10\).

We will use the geometric progression sum formula.

\[\begin{align}S_n&=\dfrac{a(1-r^n)}{1-r}\\S_{10}&=\dfrac{2\left(1-2^{10}\right)}{1-2}\\&=2(2^{10}-1)\\&=2046\end{align}\]

Interactive Questions

Here are a few activities for you to practice.

Select/type your answer and click the "Check Answer" button to see the result.

Let's Summarize

We hope you enjoyed learning about sequences with the examples and practice questions. Now, you will be able to easily remember the formulas of sequence and solve problems on sequences in math, which include arithmetic sequence, geometric sequence, harmonic sequence, and other types of sequences.

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Frequently Asked Questions (FAQs)

1. what is a geometric progression in mathematics.

A geometric progression is a sequence where every term bears a constant ratio to its preceding term.

2. How do you know if a sequence is arithmetic or geometric?

If each successive term of a sequence is less than the preceding term by a fixed number, then the sequence is arithmetic.

If each successive term of a sequence is a product of the preceding term and a fixed number, then the sequence is geometric.

3. What is the nth term of a geometric sequence?

The nth term of a geometric sequence with first term \(a\) and the common ratio \(r\) is given by \(a_{n}=ar^{n-1}\).

4. What is the nth term of an arithmetic sequence?

The nth term of an arithmetic sequence with first term \(a\) and the common difference \(d\) is given by \(a_{n}=a+(n-1)d\).

5. What is an arithmetic progression in mathematics?

An arithmetic progression is a sequence of numbers in which each successive term is a sum of its preceding term and a fixed number.

6. How to find the sum of arithmetic progression?

The sum of arithmetic progression with first term \(a\) and the common difference \(d\) is given by following formulas:

• \(S_n=\frac{n}{2}(2 a+(n-1) d)\)
• \(S_n=\frac{n}{2}(a+a_{n})\)

7. How to find the sum of geometric progression?

The sum of geometric progression with first term \(a\) and the common ratio \(r\) is given by \(S_n=\dfrac{a(1-r^n)}{1-r}\).

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Arithmetic Sequence Problems with Solutions – Mastering Series Challenges

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Arithmetic Sequences Practice Problems and Solutions

Calculating terms in an arithmetic sequence, solving problems involving arithmetic sequences.

An arithmetic sequence is a series where each term increases by a constant amount, known as the common difference . I’ve always been fascinated by how this simple pattern appears in many mathematical problems and real-world situations alike.

Understanding this concept is fundamental for students as it not only enhances their problem-solving skills but also introduces them to the systematic approach of sequences in math .

The first term of an arithmetic sequence sets the stage, while the common difference dictates the incremental steps that each subsequent term will follow. This can be mathematically expressed as $a_n = a_1 + (n – 1)d$.

Whether I’m calculating the nth term or the sum of terms within a sequence , these formulas are the tools that uncover solutions to countless arithmetic sequence problems. Join me in unraveling the beauty and simplicity of arithmetic sequences ; together, we might just discover why they’re considered the building blocks in the world of mathematics .

When I work with arithmetic sequences , I always keep in mind that they have a unique feature: each term is derived by adding a constant value, known as the common difference , to the previous term. Let’s explore this concept through a few examples and problems.

Example 1: Finding a Term in the Sequence

Given the first term, $a_1$ of an arithmetic sequence is 5 and the common difference ( d ) is 3, what is the 10th term $a_{10}$?

Here’s how I determine it: $a_{10} = a_1 + (10 – 1)d ] [ a_{10} = 5 + 9 \times 3 ] [ a_{10} = 5 + 27 ] [ a_{10} = 32$

So, the 10th term is 32.

Sequence A: If $a_1 = 2 $and ( d = 4 ), find $a_5$.

Sequence B: For $a_3 = 7 $ and $a_7 = 19$, calculate the common difference ( d ).

I calculate $a_5$ by using the formula: $a_n = a_1 + (n – 1)d $ $ a_5 = 2 + (5 – 1) \times 4 $ $a_5 = 2 + 16 $ $a_5 = 18$

To find ( d ), I use the formula: $a_n = a_1 + (n – 1)d$ Solving for ( d ), I rearrange the terms from $a_3$ and $a_7$: $d = \frac{a_7 – a_3}{7 – 3}$ $d = \frac{19 – 7}{4}$ $d = \frac{12}{4}$ [ d = 3 ]

Here’s a quick reference table summarizing the properties of arithmetic sequences :

Remember these properties to solve any arithmetic sequence problem effectively!

In an arithmetic sequence , each term after the first is found by adding a constant, known as the common difference ( d ), to the previous term. I find that a clear understanding of the formula helps immensely:

$a_n = a_1 + (n – 1)d$

Here, $a_n$ represents the $n^{th}$term, $a_1$ is the first term, and ( n ) is the term number.

Let’s say we need to calculate the fourth and fifth terms of a sequence where the first term $a_1 $ is 8 and the common difference ( d ) is 2. The explicit formula for this sequence would be $ a_n = 8 + (n – 1)(2) $.

To calculate the fourth term $a_4 $: $a_4 = 8 + (4 – 1)(2) = 8 + 6 = 14$

For the fifth term ( a_5 ), just add the common difference to the fourth term: $a_5 = a_4 + d = 14 + 2 = 16$

Here’s a table to illustrate these calculations:

Remember, the formula provides a direct way to calculate any term in the sequence, known as the explicit or general term formula. Just insert the term number ( n ) and you’ll get the value for $a_n$. I find this methodical approach simplifies the process and avoids confusion.

When I approach arithmetic sequences , I find it helpful to remember that they’re essentially lists of numbers where each term is found by adding a constant to the previous term. This constant is called the common difference, denoted as ( d ). For example, in the sequence 3, 7, 11, 15, …, the common difference is ( d = 4 ).

To articulate the ( n )th term of an arithmetic sequence, $a_n $, I use the fundamental formula:

$a_n = a_1 + (n – 1)d $

In this expression, $a_1$ represents the first term of the sequence.

If I’m solving a specific problem—let’s call it Example 1—I might be given $a_1 = 5 $and ( d = 3 ), and asked to find $a_4 $. I’d calculate it as follows:

$a_4 = 5 + (4 – 1) \times 3 = 5 + 9 = 14$

In applications involving arithmetic series, such as financial planning or scheduling tasks over weeks, the sum of the first ( n ) terms often comes into play. To calculate this sum, ( S_n ), I rely on the formula:

$S_n = \frac{n}{2}(a_1 + a_n)$

Now, if I’m asked to work through Example 3, where I need the sum of the first 10 terms of the sequence starting with 2 and having a common difference of 5, the process looks like this:

$a_{10} = 2 + (10 – 1) \times 5 = 47$ $S_{10} = \frac{10}{2}(2 + 47) = 5 \times 49 = 245$

Linear functions and systems of equations sometimes bear a resemblance to arithmetic sequences, such as when I need to find the intersection of sequence A and sequence B. This would involve setting the nth terms equal to each other and solving the resulting linear equation.

Occasionally, arithmetic sequences can be mistaken for geometric sequences , where each term is found by multiplying by a constant. It’s important to differentiate between them based on their definitions.

For exercises, it’s beneficial to practice finding nth terms, and sums , and even constructing sequences from given scenarios. This ensures a robust understanding when faced with a variety of problems involving arithmetic sequences .

In exploring the realm of arithmetic sequences , I’ve delved into numerous problems and their corresponding solutions. The patterns in these sequences—where the difference between consecutive terms remains constant—allow for straightforward and satisfying problem-solving experiences.

For a sequence with an initial term of $a_1 $ and a common difference of ( d ), the $n^{th}$term is given by $a_n = a_1 + (n – 1)d $.

I’ve found that this formula not only assists in identifying individual terms but also in predicting future ones. Whether calculating the $50^{th}$term or determining the sum of the first several terms, the process remains consistent and is rooted in this foundational equation.

In educational settings, arithmetic sequences serve as an excellent tool for reinforcing the core concepts of algebra and functions. Complexity varies from basic to advanced problems, catering to a range of skill levels. These sequences also reflect practical real-world applications, such as financial modeling and computer algorithms, highlighting the relevance beyond classroom walls.

Through practicing these problems, the elegance and power of arithmetic sequences in mathematical analysis become increasingly apparent. They exemplify the harmony of structure and progression in mathematics —a reminder of how simple rules can generate infinitely complex and fascinating patterns.

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Arithmetic Progression Questions

The Arithmetic Progression questions and answers are given here for students to help them better grasp the concept. Arithmetic Progressions (AP) is one of the most important concepts in Maths and it is included in higher education. NCERT standards will be followed for preparing these questions. The problems given here will cover both the basics and more complex topics, for students of all levels. Students can practise the arithmetic progression questions, and can then cross verify their answers with the provided solutions. To learn more about Arithmetic progression, click here .

Here, we have provided the Arithmetic Progression questions and answers with complete explanations.

Arithmetic Progression Questions with Solutions

1. Find the common difference for the following AP: 10, 20, 30, 40, 50.

Given AP: 10, 20, 30, 40, 50

Common difference:

d = 20 – 10 = 10

d = 30 – 20 = 10

d = 40 – 30 = 10

d = 50 – 40 = 10.

Hence, the common difference for the sequence, 10, 20, 30, 40, 50 is 10.

2. Is a, 2a, 3a, 4a, … an arithmetic progression?

Given sequence: a, 2a, 3a, 4a, …

To check whether the given sequence is AP or not, we have to find the common difference.

Hence, d = 2a – a = a

d = 3a – 2a = a

d = 4a – 3a = a

Hence, the common difference is “a”.

Therefore, the sequence a, 2a, 3a, 4a,… is an arithmetic progression.

3. Prove that 7, 11, 15, 19, 23 is an AP.

Given sequence: 7, 11, 15, 19, 23.

To prove that the sequence is AP, find the common difference between two consecutive terms.

d = 11 – 7 = 4

d = 15 – 11 = 4

d = 19 – 15 = 4

d = 23 – 19 = 4

Hence, 7, 11, 15, 19, 23 is an AP with a common difference of 4.

4. The sequence 28, 22, x, y, 4 is an AP. Find the values of x and y.

Given AP: 28, 22, x, y, 4

Here, first term, a = 28

Common difference, d = 22 – 28 = -6

Hence, x = 22 – 6 = 16

y = 16 – 6 = 10.

Hence, the values of x and y are 16 and 10, respectively.

5. What is the nth term of an AP 9, 13, 17, 21, 25, …?

Given AP: 9, 13, 17, 21, 25, …

Here, a = 9

d = 13 – 9 = 4

The formula to find the nth term of an AP is a + (n-1)d.

The nth term of an AP = 9 + (n-1)4

= 9 + 4n -4

Hence, the nth term of AP 9, 13, 17, 21, 25, … is 4n+5.

6. Find the 5th term of the arithmetic progression 1, 4, 7, ….

Given AP: 1, 4, 7, …

d = 4 – 1 = 3

As we know,

The nth term of AP = a + (n-1)d

Hence, 5th term of AP = 1 + (5-1)3

Hence, the 5th term of AP is 13.

Also, check: Arithmetic Progression Class 10 Notes .

7. Find the 17th term of AP 4, 9, 14, …

Given AP: 4, 9, 14, …

Here, a = 4

d = 9 – 4 = 5

Now, substitute the values in the formula a+(n-1)d,

17th term of AP = 4+(17-1)5

= 4+80 = 84

Hence, the 17th term of AP is 84.

8. If the first, second and last terms of the AP are 5, 9, 101, respectively, find the total number of terms in the AP.

Given: First term, a = 5

Common difference, d = 9 – 5 = 4

Last term, a n = 101

Now, we have to find the value of “n”.

Hence, a n = a+(n-1)d

Substituting the values, we get

5+(n-1)4 = 101

5+4n-4 = 101

n=100/4 = 25

Hence, the number of terms in the AP is 25.

9. What is the general term of the series, 4, 7, 10, 13, …?

Given sequence is 4, 7, 10, 13, …

d = 7-4 = 3

The general term of an AP is:

a n = a+(n-1)d

a n = 4 +(n-1)3

a n = 4 + 3n-3

Therefore, the general term of the series 4, 7, 10, 13 is 3n+1.

10. Which term of AP 27, 24, 21, … is 0?

Given AP: 27, 24, 21, …

Here, a = 27

d = 24 – 27 = -3.

Also given that a n = 0

Now, we have to find the value of n.

0 = 27 +(n-1)(-3)

0 = 27 -3n +3

0 = 30 – 3n

Hence, n = 10

Therefore, the 10th term of AP is 0.

Practice Questions

• The sequence 12b, 8b, 4b is in AP. Find the sum of the first 18 terms.
• The first three terms of a sequence are 8, y, 18. Find the value of y so that the sequence becomes an Arithmetic progression.
• In an Arithmetic progression, the ratio of the 7th term to the 10th term is -1. If the 16th term is -15, find the 3rd term.

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Arithmetic-Geometric Progression

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• Rishi Gupta
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An arithmetic-geometric progression (AGP) is a progression in which each term can be represented as the product of the terms of an arithmetic progressions (AP) and a geometric progressions (GP).

In the following series, the numerators are in AP and the denominators are in GP:

\[\large \dfrac{\color{blue}{1}}{\color{red}{2}}+\dfrac{\color{blue}{2}}{\color{red}{4}}+\dfrac{\color{blue}{3}}{\color{red}{8}}+\dfrac{\color{blue}{4}}{\color{red}{16}}+\dfrac{\color{blue}{5}}{\color{red}{32}}+\cdots= \, ?\]

Arithmetic-geometric progressions are nice to work with because their sums can be evaluated easily, and this tool is used in a variety of contest problems.

Sum of AGP up to Infinite Terms

Applications.

Let's start with a few simple definitions of the concepts that we will repeatedly use.

Arithmetic-Geometric Progression (AGP) : This is a sequence in which each term consists of the product of an arithmetic progression and a geometric progression. In variables, it looks like

\[ a , (a+d) r , (a+2d) r^2 , (a+3d)r^3, \ldots , \left[ a + (n-1) d \right] r^{n-1},\]

where \(a\) is the initial term, \(d\) is the common difference, and \(r\) is the common ratio.

General term of AGP: The \(n^{\text{th}}\) term of the AGP is obtained by multiplying the corresponding terms of the arithmetic progression (AP) and the geometric progression (GP). So, in the above sequence the \(n^{\text{th}}\) term is given by

\[t_n=\left[ a + (n-1) d \right] r ^ { n -1}.\]

Sum of terms of AGP: The sum of the first \(n\) terms of the AGP is \( S_n=\displaystyle \sum_{k= 1 } ^ {n} \left[ a + (k-1) d \right] r ^ { k-1 }\), which can be solved further to obtain

\[S_n=\dfrac{a -\left[ a+(n - 1)d\right] r^{n}}{1 - r}+\dfrac{dr(1 - r^{n-1 })}{(1 - r)^2}.\]

Sum to infinity of AGP: If \( |r| < 1 \), then the sum to infinity is given by

\[ S_\infty = \dfrac{ a } { 1 - r } + \dfrac{ dr } { (1-r)^2 }. \]

If we wanted to find the sum of an AGP, we could manually calculate the total. However, this could be a tedious process, and the terms will quickly get too large or too small for us to easily sum them. Let's find a more general approach, and we start by looking at an example.

Find the sum of the series \(1 \cdot 2 +2 \cdot 2^2 +3 \cdot 2^3+\cdots+100 \cdot 2^{100}\). The terms of this sequence are too large for us to want to attempt to sum them manually. Let the sum of the series be \(S\), then \[S= 1 \cdot 2+2 \cdot 2^2+3 \cdot 2^3+\cdots+100 \cdot 2^{100}.\] On multiplying \(S\) by 2, we get \[2S= 1\cdot 2^2+2 \cdot 2^3+\cdots+99 \cdot 2^{100}+100 \cdot 2^{101}.\] Now subtracting \(2S\) from \(S\), we get \[ \begin{array} { rlllllllll} S&= 1 \cdot 2&+2 \cdot 2^2&+3 \cdot 2^3&+\cdots+100 \cdot 2^{100} \\ 2S&= 0 &+1\cdot 2^2&+2 \cdot 2^3&+\cdots+99 \cdot 2^{100}&+&100 \cdot 2^{101}\\ \hline S(1-2)& = 1 \cdot 2&+1 \cdot 2^2&+1 \cdot 2^3&+\cdots+1 \cdot 2^{100}&-&100 \cdot 2^{101}.\\ \end{array}\] Then we have \[\begin{align} -S&=2\dfrac{(2^{100}-1)}{2-1}-100 \cdot 2^{101} \qquad (\text{since the first 100 terms are in GP})\\ S&=100 \cdot 2^{101}-2 \cdot 2^{100}+2\\ S&=200 \cdot 2^{100} - 2\cdot 2^{100}+2\\ S&=198 \cdot 2^{100}+2. \ _\square \end{align}\]

In this problem, the crucial step was to multiply by the common ratio and subtract the sequences, which allowed us to reduce it to a GP which we are familiar with. Now let's derive a general formula for the sum of terms of an AGP with initial term \(a\), common difference \(d\) and common ratio \(r\):

The sum of the first \(n\) terms of an AGP is given by \[\begin{align} S_n &= \displaystyle \sum_{k = 1}^n \left[a + (k - 1) d\right] r^{k-1} \\\\ &= \dfrac{a -\left [a+(n - 1)d\right] r^{n}}{1 - r}+\dfrac{dr(1 - r^{n-1 })}{(1 - r)^2}. \end{align}\]

We'll use the same method of subtraction to find the sum of AGP that we used in the above example to prove this theorem:

To get the sum of the first \(n\) terms of an AGP, we need to find the value of \[S= a+(a+d)r+(a+2d)r^2+\cdots+[a+(n-1)d]r^{n-1}.\] Now let's multiply \(S\) by \(r\), then we get \[Sr= 0 +ar+(a+d)r^2+\cdots+[a+(n-2)d]r^{n-1}+[a+(n-1)d]r^{n}.\] Subtracting \(Sr\) from \(S\), we get \[\begin{array} { rlllllllll} S&= a+(a+d)r&+(a+2d)r^2&+\cdots+[a+(n-1)d]r^{n-1}\\ Sr&= 0 +ar&+(a+d)r^2&+\cdots+[a+(n-2)d]r^{n-1}&+[a+(n-1)d]r^{n}\\ \hline S(1-r)& = a+dr&+dr^2&+\cdots+dr^{n-1}&-[a+(n-1)d]r^{n}.\\ \end{array}\] If we were to exclude the first term and the last term, then the rest is a sum of geometric progression with first term \(dr\) and common ratio \(r\). Thus \[\begin{align} S(1-r)&=a-[a+(n-1)d]r^{n}+\dfrac{dr(1-r^{n-1})}{1-r} \\ S&=\dfrac{a-[a+(n-1)d]r^{n}}{1-r}+\dfrac{dr(1-r^{n-1})}{(1-r)^2}. \ _\square \end{align}\]

Let's try one problem to practice the above method:

\[1+2 \cdot 2+ 3 \cdot 2^2 + 4 \cdot 2^3 + \cdots+ 100 \cdot 2^{99}= \, ?\]

Now that we've found the sum of finitely many terms, let's consider the case of infinitely many terms. We certainly cannot manually sum up infinite terms, so we will have to find a general approach. We start by discussing the problem you encountered at the top of this page:

\[\large \dfrac{\color{blue}{1}}{\color{red}{2}}+\dfrac{\color{blue}{2}}{\color{red}{4}}+\dfrac{\color{blue}{3}}{\color{red}{8}}+\dfrac{\color{blue}{4}}{\color{red}{16}}+\dfrac{\color{blue}{5}}{\color{red}{32}}+\cdots=\, ?\] Let us assume the given series to be \(S\), then \[S=\dfrac 12 +\dfrac 24 +\dfrac 38+\dfrac{4}{16}+\dfrac{5}{32}+\cdots.\] On multiplying \(S\) by \(\frac 12\), we get \[ \dfrac S2=\dfrac 14 +\dfrac 28 +\dfrac{3}{16}+\dfrac{4}{32}+\dfrac{5}{64}+\cdots.\] Now subtracting \(\frac S2\) from \(S\), we get \[ \begin{array} { rlllllllll} S&=\dfrac 12 &+\dfrac 24 &+\dfrac 38 &+\dfrac{4}{16} &+\dfrac{5}{32}+ \cdots \\ \dfrac S2&=0&+\dfrac 14 & +\dfrac 28 & +\dfrac{3}{16}&+\dfrac{4}{32}+\dfrac{5}{64}+\cdots \\ \hline S \left(1- \dfrac 12 \right)& =\dfrac 12& +\dfrac 14 & + \dfrac 18 & +\dfrac{1}{16} & +\dfrac{1}{32} +\cdots \\ \Rightarrow \dfrac S2&=\dfrac 12 &+\dfrac 14 &+ \dfrac 18 &+\dfrac {1}{16} &+\dfrac {1}{32} +\cdots, \end{array}\] which is a GP. Therefore, using the formula for the sum of infinite terms of GP, we get \[S=2 \times \left( \dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}} \right)=2. \ _\square \]

We are now ready to state the sum of an infinite AGP, and will present the proof below:

The sum of infinite terms of an AGP is given by \(S_{\infty}=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}\) , where \(|r|<1\).

It is clear that if \( |r | \geq 1 \), then the term \( [a+(n-1)d]r^{n-1}\) gets arbitrarily large, and hence the sum does not converge . So we just have to consider the case of \( |r| < 1 \). We will be using the fact that in this case, \[ \lim_{n\rightarrow \infty} r^n = 0 . \] As we have discussed earlier, the sum of the first \(n\) terms in an AGP is given by \[S_n = \sum_{k = 1}^n \left[a + (k - 1) d\right] r^{k - 1} = \dfrac{a -\left [a+(n - 1)d\right] r^n}{1 - r}+\dfrac{dr(1 - r^{n - 1})}{(1 - r)^2}. \] Taking the limit as \(n\) goes to infinity, we get \[\begin{align} \lim_{n \to \infty}S_{n} &= \dfrac{a-0}{1-r}+\dfrac{dr(1-0)}{(1-r)^2}\\ &=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}, \text{ where } |r|<1. \ _\square \end{align}\]

With this formula, we can quickly find the sum of infinitely many terms of a suitable AGP. Let's practice with more examples:

Calculate the value of the sum \(\displaystyle\sum_{i=1}^{\infty} \dfrac{i}{7^i}\). The sum can be expanded as \[S=\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\dfrac{4}{7^4}+\cdots .\] To get the above series into the AGP form, we have to factor out the term \(\dfrac 17\) and then the series can be written as \[S=\left( \dfrac 17 \right) \left( 1+\dfrac{2}{7}+\dfrac{3}{7^2}+\dfrac{4}{7^3}+\cdots \right) .\] Clearly, this is an infinite AGP with \(a=1, r=\frac{1}{7}, d=1\). Using the above formula, the sum \(S\) is \[\begin{align} \left( \dfrac 17 \right) \left( \dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2} \right) &=\left( \dfrac 17 \right) \left( \dfrac{1}{1-\dfrac{1}{7}}+\dfrac{ 1 \times\frac{1}{7}}{\left(1-\dfrac{1}{7}\right)^2} \right) \\ &=\dfrac 17 \left( \dfrac 76+\dfrac{7}{36} \right)\\ &=\dfrac{7}{36}. \ _\square \end{align} \]

Calculate the value of the summation \(\displaystyle\sum_{i=1}^{\infty} \dfrac{2i-1}{2^{i+1}}\). Solution 1: Writing down the given summation as the difference of two summations, we get \[\sum_{i=1}^{\infty} \dfrac{2i-1}{2^{i+1}}=\displaystyle\sum_{i=1}^{\infty} \left(\dfrac{2i}{2^{i+1}}-\dfrac{1}{2^{i+1}}\right)=\displaystyle\sum_{i=1}^{\infty}\dfrac{i}{2^i}-\displaystyle\sum_{i=1}^{\infty}\dfrac{1}{2^{i+1}}.\] The first summation is an AGP with \( a =\frac{1}{2} \), \( d = 1 \), and \( r = \frac{1}{2} \). So the sum to infinity is \( \frac{ \frac{1}{2} } { 1 - \frac{1}{2} } + \frac{ 1 \times \frac{1}{2} } { ( 1- \frac{1}{2} ) ^ 2 } = 2 \). The second summation is a geometric progression with the sum to infinity of \( \frac { \frac{1}{4} } { 1 - \frac{1}{2} } = \frac{1}{2} \). Hence, the total is \( 2 - \frac{1}{2} = 1.5 \ _\square \). Solution 2: The given series can be written as \[\dfrac 14+\dfrac 38 +\dfrac{5}{16}+\dfrac{7}{32}+\cdots .\] Multiplying and dividing the series by \(4\), we get \[ \dfrac{1}{4} \left( 1+\dfrac{3}{2}+ \dfrac{5}{4}+\dfrac{7}{8}+ \cdots \right) .\] Now clearly, the series is an AGP with \(a=1,d=2,r=\frac 12\). Hence, using the formula for the sum of infinite terms of AGP, we get \[S=\dfrac 14 \left( \dfrac{1}{1-\dfrac 12}+\dfrac{2 \cdot \dfrac 12}{ \left( 1-\dfrac 12 \right)^2} \right) =\dfrac 14 \left( 2+4 \right)=1.5. \ _\square \]

Solving the problems below will check if you have a grip over the concepts and problem solving:

Find the value of \(p\) given

\[ 3+\dfrac{1}{4}(3+p)+\dfrac{1}{4^{2}}(3+2p)+\dfrac{1}{4^3}(3+3p)+\cdots =8.\]

The value of \(\displaystyle \sum_{n=1}^ \infty \frac{ 2n}{ 3^n } \) can be expressed in the form \( \frac{a}{b} \), where \(a\) and \(b\) are coprime positive integers. Find \( a - b \).

In this section we will work out some examples and problems based on applications of AGP:

1. Recognizing an AGP

Prove the following identity: \[\left( x^{n-1}+\frac{1}{x^{n-1}} \right) + 2 \left( x^{n-2} +\frac{1}{x^{n-2}} \right) +\cdots+(n-1) \left( x +\frac 1x \right) +n=\frac{1}{x^{n-1}} \left( \dfrac{x^n-1}{x-1} \right)^2.\] We will start proving by rewriting the LHS of the identity as \[\begin{align} &\left( x^{n-1}+\frac{1}{x^{n-1}} \right) + 2 \left( x^{n-2} +\frac{1}{x^{n-2}} \right) +\cdots+(n-1) \left( x +\frac 1x \right) +n\\ &=\left( \frac{1}{x^{n-1}}+\frac{2}{x^{n-2}}+\cdots+\frac{n-1}{x} \right) + \left[ x^{n-1}+2x^{n-2}+\cdots+(n-1)x \right] +n. \end{align}\] Multiplying and dividing the expression in the parenthesis by \(x^{n}\), we get \[\dfrac{1}{x^{n}} \left( x+2x^2+\cdots+(n-1)x^{n-1} \right) +\left[ x^{n-1}+2x^{n-2}+\cdots+(n-1)x \right] +n.\] Using the formula for the sum of terms of an AGP for the expression in the parenthesis, we get \[\dfrac{x \left( (n-1)x^n-nx^{n-1}+1\right)}{x^n(1-x)^2}+\left[ x^{n-1}+2x^{n-2}+\cdots+(n-1)x \right] +n.\] Now, since the expression in the bracket can be obtained from the first term in the above expression by replacing \(x\) with \(\frac 1x\), we have \[\dfrac{x [ (n-1)x^n-nx^{n-1}+1]}{x^n(1-x)^2} +\dfrac{\left( \frac 1x \right) \left[ (n-1)\left( \frac 1x \right)^n-n\left( \frac 1x \right)^{n-1}+1 \right]}{\left( \frac 1x \right)^n \left[ 1-\left( \frac 1x \right) \right]^2} +n.\] Simplifying this further gives \[\dfrac{1}{x^{n-1}} \left( \dfrac{x^n-1}{x-1} \right)^2=\text{RHS}.\] Hence proved. \(_\square\)

What is the expected number of coin flips before we get the first head? Let \(P(n)\) be the probability of gettinng first head in \(n\) flips, then \(P(n)=\left(\dfrac{1}{2}\right)^n\). Hence, the expected number of coin flips is \(\displaystyle\sum_{n=1}^{\infty} nP(n)=\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{2^n}=2. \ _\square\)

Let's see if you can solve the following problem.

If we roll one die, what is the expected number of dice throws before we get the first 6?

Note: the roll of the 6 is included in the "number of throws before we get the first 6."

\(\text{}\)

2. Extension of the summation method

Evaluate \(\displaystyle\sum_{i=1}^\infty \frac{ i^2} { 2^i } \). If we had to describe this summation, we would call it a "quadratic-geometric progression", because the numerator is a quadratic \( i^2 \). We will use a different approach to reduce this to a "linear-geometric progression", which is an AGP. Let the sum be \(S,\) then since the common ratio is \( \frac{1}{2} \), we will multiply \(S\) by \( \frac{1}{2} .\) Subtracting \(\frac{1}{2}S\) from \(S\) gives \[ \begin{array} { r lllllll} S& =\frac 12 & +\frac {4}{4} & +\frac{9}{8} &+\frac{16}{16} &+\frac{25}{32} &+\cdots \\ \frac{1}{2} S& = & + \frac 14 & +\frac {4}{8} & +\frac{9}{16} &+\frac{16}{32} &+\cdots \\ \hline \frac{1}{2} S & =\frac 12 & +\frac {3}{4} & +\frac{5}{8} &+\frac{7}{16} &+\frac{9}{32} &+\cdots .\\ \end{array} \] We may recognize the AGP here, but let \(T=\frac{1}{2} S\) and continue with this procedure of taking the difference: \[ \begin{array} { r lllllll} T & =\frac 12 & +\frac {3}{4} & +\frac{5}{8} &+\frac{7}{16} &+\frac{9}{32} &+\cdots \\ \frac{1}{2} T & = & + \frac 14 & +\frac {3}{8} & +\frac{5}{16} &+\frac{7}{32} &+\cdots \\ \hline \frac{1}{2} T & = \frac{1}{2} & + \frac{2}{4} & + \frac{2}{8} & + \frac{2}{16} & + \frac{2}{32} & + \cdots .\\ \end{array} \] Now, observe that with the exception of the first term, we get a GP with initial term \( \frac{2}{4} \) and common ratio \( \frac{1}{2} \). As it turns out, the first term would often not fit into the pattern of the sequence, and we just got lucky previously. We thus get \[ \frac{1}{2} T = \frac{1}{2} + \frac{ \frac{2}{4} } { 1 - \frac{1}{2} } = \frac{1}{2} + 1 = \frac{3}{2} . \] Therefore, \( T = 3 ,\) which implies \( S = 2T = 6. \ _\square \)

Observe that when we take the difference of terms in a quadratic sequence, we will end up with a linear sequence. This holds more generally: When we take the difference of terms in a degree \(n\) sequence, we will get a degree \(n-1 \) sequence. This is explored in detail in Method of Differences . We will use this idea repeatedly, to work with such "polynomial-geometric progression."

Evaluate \(\displaystyle\sum_{i=1}^\infty \frac{ i^3} { 3^i } \). Observe that we have a "cubic-geometric progression" with common ratio \( \frac{1}{3} \). Let's multiply by \( \frac{1}{3} \) and take the difference: \[ \begin{array} { r lllll} S & = \frac{1}{3} & + \frac{8}{9} & + \frac{ 27}{27} & + \frac{ 64}{ 81} & + \frac{ 125}{243} & + \cdots \\ \frac{1}{3} S & = & + \frac{1}{9} & + \frac{8}{27} & + \frac{ 27}{81} & + \frac{ 64}{ 243} & + \cdots \\ \hline \frac{2}{3} S & = \frac{1}{3} & + \frac{7}{9} & + \frac{ 19}{27} & + \frac{ 37}{ 81} & + \frac{ 61}{243} & + \cdots. \\ \end{array} \] We thus get \( \frac{2}{3} S - \frac{1}{3} =\displaystyle \sum_{i=1}^\infty \frac{ 3i^2+3i+1}{3^{i+1} } \), which is a "quadratic-geometric progression". Set this to be \(T\). Then multiplying by \( \frac{1}{3} \) and taking the difference gives \[ \begin{array} { r lllll} T & = \frac{7}{9} & + \frac{ 19}{27} & + \frac{ 37}{ 81} & + \frac{ 61}{243} & + \cdots \\ \frac{1}{3} T & = & + \frac{7}{27} & + \frac{ 19}{81} & + \frac{ 37}{ 243} & + \cdots \\ \hline \frac{2}{3} T & = \frac{7}{9} & + \frac{12}{27} & + \frac{ 18}{81} & + \frac{24}{243} & + \cdots. \\ \end{array} \] We thus get \( \frac{2}{3} T - \frac{7}{9} =\displaystyle \sum_{i=1}^\infty \frac{ 6(i+1)} { 3^{i+2}} \), which is a "linear-geometric progression. Set this to be \( U \). Then multiplying by \( \frac{1}{3} \) and taking the difference gives \[ \begin{array} { r lllll} U & = \frac{12}{27} & + \frac{ 18}{81} & + \frac{24}{243} & + \cdots \\ \frac{1}{3} U & = & + \frac{12}{81} & + \frac{ 18}{342} & + + \cdots \\ \hline \frac{2}{3} U & = \frac{12}{27} & + \frac{6}{81} & + \frac{6}{ 243} & + \cdots. \\ \end{array} \] We then get \( \frac{2}{3} U - \frac{12}{27} = \displaystyle \sum_{i=1}^\infty \frac{6}{3^{i+3} } ,\) which is a geometric progression with an infinite sum of \( \frac{ \frac{6}{81} } { 1 - \frac{1}{3} } = \frac{1}{9} \). This gives us \[\begin{align} \frac{2}{3} U &= \frac{12}{27} + \frac{1}{9} &&\Rightarrow U = \frac{5}{6} \\ \frac{2}{3} T &= \frac{7}{9} + \frac{5}{6} &&\Rightarrow T = \frac{29}{12} \\ \frac{2}{3} S &= \frac{1}{3} + \frac{29}{12} &&\Rightarrow S = \frac{33}{8}. \ _\square \end{align}\]

Now, you're ready to solve the following problems on your own. Good luck!

\[W=2+5x+9x^2+14x^3+20x^4+\cdots \]

Given \(|x|<1\), find the value of the series \(W\).

Clarification : \(2,5,9,14,20,\ldots\) follows a \(2^{\text{nd}}\)-degree polynomial function, i.e. their second finite difference is a constant.

\[ \dfrac{\color{magenta}{1^2}}{\color{indigo}{1}}+\dfrac{\color{magenta}{5^2}}{\color{indigo}{11}}+\dfrac{\color{magenta}{9^2}}{\color{indigo}{(11)^2}}+\dfrac{\color{magenta}{13^2}}{\color{indigo}{(11)^3}}+\dfrac{\color{magenta}{17^2}}{\color{indigo}{(11)^4}}+\cdots= \, ?\]

\[ S = 0.1 + 0.02 + 0.003 + 0.0004 + \cdots + \dfrac n{10^n} + \cdots \]

Given that the infinite sum \( S\) can be expressed as \( \frac ab\), where \(a\) and \(b\) are coprime positive integers, find \(a+b\).

To enhance your problem solving skills on arithmetic progressions , geometric progressions , and arithmetic-geometric progression , you can check out the following wikis:

• Arithmetic and Geometric Progressions: Problem Solving

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Arithmetic Progression

A progression is another term for sequence . Therefore, Arithmetic Progressions (also known as Arithmetic Sequences) are special sequences defined by the property that the difference between any two consecutive terms of the sequence are constant. Whereas the rule for regular sequences is that the difference between consecutive terms has to have some kind of relationship, it may or may not be constant. The relationship may not even be of a ‘difference’ nature but may be of a ‘multiplicative’ nature. This is not so for Arithmetic sequences, the relationship has to be of a difference nature only. This common difference is denoted by the letter d .

As an example, take a look at the sequence below;

for the sequence below to be an arithmetic progression, the difference between consecutive terms has to be some constant d such that:

since that relationship must hold true, we can move further and say that

the same applies to a 4

the same also applies to the last term a n where by a n-1 is the second last term

So now let us take a replace the terms in the sequence with their corresponding terms that we have obtained

From the above new sequence, we should observe that to get the next term, you always add the first term to the product of the common difference and one less than the position of the next term.

i.e. if the next term is denoted by the letter n, then

The above relationship should always hold true as the relationship between consecutive terms in an arithmetic progression. You can check that it is true by substituting for the values of the sequence we’ve been working with.

This is what makes arithmetic progressions such special sequences, all you need to know is the first term and the common difference d and then you’ll be able to find any term in the sequence without having to find its preceding term.

For example, given that the common difference in an arithmetic progression is 5 and the first term is 3, find the 10th and 25th terms of the sequence.

The common relationship between all terms in an arithmetic progression is given by

Arithmetic Series

Since there exist Arithmetic Sequences, Arithmetic Series also exist and are the sums of the terms in arithmetic sequences.

We have already seen that a series is given in the following form

where S n is the notation for the sum of a series.

We have also established that the above terms can also be written in terms of the first term and the common difference

but the above can also be written in terms of the last term as follows

If we were to add the two sums we would obtain the following

The above is obtained after some tedious addition and collecting like terms but it turns out as a nice simple expression

So we now have an expression for the sum in terms of only the first and last terms as

but we can simplify this further since we already established a way of expressing the any term in terms of the first term and the common difference d

so now we have two options of formula to use to find the sum of an arithmetic progression.

Remembering from the section on Series, the notation for summation:

Examples of Arithmetic Progression

Find the sum of the Arithmetic Progression below given that the total number of terms is 15

Since we haven’t been told what the common difference textbf{d} is, we need to find that first

Next to find the sum, all we have to do is substitute in the formula since we already have the first and last terms

Given that the following is an Arithmetic Progression with 20 terms, find its sum

Like in the previous example, we’re not given the common difference, and in this case we don’t even have the last term. We could choose to find both or we could just find the difference and use only this and the first term to find the sum.

The formula below only utilizes the first term and the common difference

Substituting:

Quiz on Arithmetic Progression

The first step is to list out the sequence for the 12 days:

• On the first day, you receive 1 gift
• On the second day you receive 1 + 2 = 3 gifts
• On the third day you receive 1 + 2 + 3 = 6
• On the fourth day you receive 1 + 2 + 3 + 4 = 6
• And so on, until you reach day 12

The sequence for the gifts received on each day becomes:

To find the total number of gifts, just change the sequence into a series and find its sum

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2013 AMC 12A Problems/Problem 14

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The sequence

Since the sequence is arithmetic,

Video Solution by OmegaLearn

https://youtu.be/RdIIEhsbZKw?t=944

~ pi_is_3.14

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Mathematicians Are Edging Close to Solving One of the World's 7 Hardest Math Problems

And there’s $1 million at stake.

• In new research, mathematicians have narrowed down one of the biggest outstanding problems in math.
• Huge breakthroughs in math and science are usually the work of many people over many years.
• Seven math problems were given a $1 million bounty each in 2000, and just one has been solved so far.

The “Millennium Problems” are seven infamously intractable math problems laid out in the year 2000 by the prestigious Clay Institute, each with $1 million attached as payment for a solution. They span all areas of math , as the Clay Institute was founded in 1998 to push the entire field forward with financial support for researchers and important breakthroughs.

But the only solved Millennium Problem so far, the Poincare conjecture, illustrates one of the funny pitfalls inherent to offering a large cash prize for math. The winner, Grigori Perelman, refused the Clay prize as well as the prestigious Fields Medal. He withdrew from mathematics and public life in 2006, and even in 2010, he still insisted his contribution was the same as the mathematician whose work laid the foundation on which he built his proof, Richard Hamilton.

Math, all sciences, and arguably all human inquiries are filled with pairs or groups that circle the same finding at the same time until one officially makes the breakthrough. Think about Sir Isaac Newton and Gottfried Leibniz, whose back-and-forth about calculus led to the combined version of the field we still study today. Rosalind Franklin is now mentioned in the same breath as her fellow discoverers of DNA, James Watson and Francis Crick. Even the Bechdel Test for women in media is sometimes called the Bechdel-Wallace Test, because humans are almost always in collaboration.

That’s what makes this new paper so important. Two mathematicians—Larry Guth of the Massachusetts Institute of Technology (MIT) and James Maynard of the University of Oxford—collaborated on the new finding about how certain polynomials are formed and how they reach out into the number line. Maynard is just 37, and won the Fields Medal himself in 2022. Guth, a decade older, has won a number of important prizes with a little less name recognition.

The Riemann hypothesis is not directly related to prime numbers , but it has implications that ripple through number theory in different ways (including with prime numbers). Basically, it deals with where and how the graph of a certain function of complex numbers crosses back and forth across axes. The points where the function crosses an axis is called a “zero,” and the frequency with which those zeroes appear is called the zero density.

In the far reaches of the number line, prime numbers become less and less predictable (in the proverbial sense). They are not, so far, predictable in the literal sense—a fact that is an underpinning of modern encryption , where data is protected by enormous strings of integers made by multiplying enormous prime numbers together. The idea of a periodic table of primes, of any kind of template that could help mathematicians better understand where and how large primes cluster together or not, is a holy grail.

In the new paper, Maynard and Guth focus on a new limitation of Dirichlet polynomials. These are special series of complex numbers that many believe are of the same type as the function involved in the Riemann hypothesis involves. In the paper, they claim they’ve proven that these polynomials have a certain number of large values, or solutions , within a tighter range than before.

In other words, if we knew there might be an estimated three Dirichlet values between 50 and 100 before, now we may know that range to be between 60 and 90 instead. The eye exam just switched a blurry plate for a slightly less blurry one, but we still haven’t found the perfect prescription. “If one knows some more structure about the set of large values of a Dirichlet polynomial, then one can hope to have improved bound,” Maynard and Guth conclude.

No, this is not a final proof of the Riemann hypothesis. But no one is suggesting it is. In advanced math, narrowing things down is also vital. Indeed, even finding out that a promising idea turns out to be wrong can have a lot of value—as it has a number of times in the related Twin Primes Conjecture that still eludes mathematicians.

In a collaboration that has lasted 160 years and counting, mathematicians continue to take each step together and then, hopefully, compare notes.

Caroline Delbert is a writer, avid reader, and contributing editor at Pop Mech. She's also an enthusiast of just about everything. Her favorite topics include nuclear energy, cosmology, math of everyday things, and the philosophy of it all. 

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AI Math Solver Homework Helper 17+

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