problem solving in math arithmetic sequence

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 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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How to Solve Arithmetic Sequences? (+FREE Worksheet!)

Do you want to know how to solve Arithmetic Sequences problems? you can do it in few simple and easy steps.

Related Topics

• How to Solve Finite Geometric Series
• How to Solve Infinite Geometric Series
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Step by step guide to solve Arithmetic Sequences problems

• A sequence of numbers such that the difference between the consecutive terms is constant is called arithmetic sequence. For example, the sequence \(6, 8, 10, 12, 14\), … is an arithmetic sequence with common difference of \(2\).
• To find any term in an arithmetic sequence use this formula: \(\color{blue}{x_{n}=a+d(n-1)}\)
• \(a =\) the first term ,\(d =\) the common difference between terms , \(n =\) number of items

Arithmetic Sequences – Example 1:

Find the first three terms of the sequence. \(a_{17}=38,d=3\)

First, we need to find \(a_{1}\) or a. Use arithmetic sequence formula: \(\color{blue}{x_{n}=a+d(n-1)}\) If \(a_{8}=38\), then \(n=8\). Rewrite the formula and put the values provided: \(x_{n}=a+d(n-1)→38=a+3(3-1)=a+6\), now solve for \(a\). \(38=a+6→a=38-6=32\), First three terms: \(32,35,38\)

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Arithmetic sequences – example 2:.

Given the first term and the common difference of an arithmetic sequence find the first five terms. \(a_{1}=18,d=2\)

Use arithmetic sequence formula: \(\color{blue}{x_{n}=a+d(n-1)}\) If \(n=1\) then: \(x_{1}=18+2(1)→x_{1}=18\) First five terms: \(18,20,22,24,26\)

Arithmetic Sequences – Example 3:

Given the first term and the common difference of an arithmetic sequence find the first five terms. \(a_{1}=24,d=2\)

Use arithmetic sequence formula: \(\color{blue}{x_{n}=a+d(n-1)}\) If \(n=1\) then: \(x_{1}=22+2(1)→x_{1}=24\) First five terms: \(24,26,28,30,32\)

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Arithmetic sequences – example 4:.

Find the first five terms of the sequence. \(a_{17}=152,d=4\)

First, we need to find \(a_{1}\) or \(a\). Use arithmetic sequence formula: \(\color{blue}{x_{n}=a+d(n-1)}\) If \(a_{17}=152\), then \(n=17\). Rewrite the formula and put the values provided: \(x_{n}=a+d(n-1)→152=a+4(17-1)=a+64\), now solve for \(a\). \(152=a+64→a=152-64=88\), First five terms: \(88,92,96,100,104\)

Given the first term and the common difference of an arithmetic sequence find the first five terms and the explicit formula.

• \(\color{blue}{a_{1} = 24, d = 2}\)
• \(\color{blue}{a_{1} = –15, d = – 5}\)
• \(\color{blue}{a_{1} = 18, d = 10}\)
• \(\color{blue}{a_{1 }= –38, d = –100}\)

Download Arithmetic Sequences Worksheet

• First Five Terms \(\color{blue}{: 24, 26, 28, 30, 32, Explicit: a_{n} = 22 + 2n}\)
• First Five Terms \(\color{blue}{: –15, –20, –25, –30, –35, Explicit: a_{n} = –10 – 5n}\)
• First Five Terms \(\color{blue}{: 18, 28, 38, 48, 58, Explicit: a_{n} = 8 + 10n}\)
• First Five Terms \(\color{blue}{: –38, –138, –238, –338, –438, Explicit: a_{n} = 62 – 100n}\)

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Arithmetic Sequence Problems

There are many problems we can solve if we keep in mind that the  n th term of an arithmetic sequence can be written in the following way: a n  = a 1 +(n - 1)d Where a 1  is the first term, and  d  is the common difference. For example, if we are told that the first two terms add up to the fifth term, and that the common difference is 8 less than the first term we can take this equation: a 1  + a 2  = a 5 and rewrite it as follows: a 1  + [a 1  + d] = [a 1  + 4d] This leads to a 1  = 3d. Combine this with d = a 1  - 8, and we have: a 1  = 3(a 1  - 8) or a 1  = 12. This leads to d = 4, and from this information, we can find any other term of the sequence.

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