problem solving arithmetic series

Arithmetic Sequence Practice Problems

More practice problems with the arithmetic sequence formula.

Direction: Read each arithmetic sequence question carefully, then answer with supporting details.

Arithmetic Sequence Practice Problems with Answers

1) Tell whether the sequence is arithmetic or not. Explain why or why not.

Sequence A : [latex] – 1,{\rm{ }} – 3,{\rm{ }} – 5,{\rm{ }} – 7,{\rm{ }} \ldots [/latex]

Sequence B : [latex] – 3,{\rm{ }}0,{\rm{ }}4,{\rm{ }}7,{\rm{ }} \ldots [/latex]

Sequence A is an arithmetic sequence since every pair of consecutive terms has a common difference of [latex]-2[/latex],  that is, [latex]d=-2[/latex].

On the other hand, sequence B is not an arithmetic sequence. There’s no common difference among the pairs of consecutive terms in the sequence.

2) Find the next term in the sequence below.

The sequence has a common difference of [latex]5[/latex]. To get to the next term, add the previous term by [latex]5[/latex]. For example, from [latex]4[/latex] to [latex]9[/latex], you add [latex]5[/latex] to [latex]4[/latex] to get to [latex]9[/latex]. That is, [latex]4 + 5 = 9[/latex].

3) Find the next two terms in the sequence below.

The common difference among adjacent terms is [latex]\large- {1 \over 3}[/latex] . Use this value to get the two succeeding terms. Make sure that you correctly add or subtract fractions with different denominators. If you need a refresher lesson on that, check it out .

4) If a sequence has a first term of [latex]{a_1} = 12[/latex] and a common difference [latex]d=-7[/latex]. Write the formula that describes this sequence. Use the formula of the arithmetic sequence.

Since we know the values of the first term, [latex]{a_1} = 12[/latex], and the common difference, [latex]d=-7[/latex], the only thing we need to do is substitute these values in the formula.

Or, you may further simplify your answer by getting rid of the parenthesis and combining like terms. Both solutions should be acceptable. If you’re not sure, ask your teacher.

5) Write the formula describing the sequence [latex]6,{\rm{ }}14,{\rm{ }}22,{\rm{ }}30,{\rm{ }}…[/latex]

The first term is [latex]6[/latex], which means [latex]{a_1} = 6[/latex]. The common difference is [latex]d=8[/latex].

Substitute these values to arrive at the required formula:

or in a more simplified form;

6) Find the 45 th term of the arithmetic sequence [latex] – \,9,{\rm{ }} – 2,{\rm{ }}5,{\rm{ }}12,{\rm{ }}…[/latex] 

Find the rule that defines the sequence using the arithmetic sequence formula. The first term is [latex]{a_1} = -9[/latex] while the common difference is [latex]d=7[/latex].

Plug these values in the formula, we get

Now we can find the 45 th term,

7) Write the formula of a sequence with two given terms, [latex]{a_5} = -32[/latex], and [latex]{a_{18}} = 85[/latex].

Use the information of each term to construct an equation with two unknowns using the arithmetic sequence formula.

• For [latex]{a_5} = -32[/latex]
• For [latex]{a_{18}} = 85[/latex]

Solve the system of equations using the  Elimination Method . Multiply Equation # 1 by [latex]−1[/latex] and add it to Equation #2 to eliminate [latex]{a_1}[/latex].

After finding the value of the common difference, it is now easy to find the value of the first term. Back substitute [latex]d=9[/latex] to any of the two equations to find [latex]{a_1}[/latex].

We’ll use Equation #1 for this,

Since [latex]{a_1} = -68[/latex] and [latex]d=9[/latex], the formula we’re looking for is,

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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 Progressions: Very Difficult Problems with Solutions