Arithmetic Sequences: Problems with Solutions

Arithmetic sequences are foundational in mathematics and have wide applications in engineering, computer science, biology, and finance. Below, you will find a review of key formulas, followed by a set of problems and practice exercises with detailed step-by-step solutions.

Table of Contents

Review of Arithmetic Sequences

An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. This constant is called the common difference (\( d \)).

The \( n \)-th Term Formula:

The formula for the \( n \)-th term \( a_n \) of an arithmetic sequence with a first term \( a_1 \) and a common difference \( d \) is:

\[ a_n = a_1 + (n - 1) d \]

The Sum Formula:

The sum \( s_n \) of the first \( n \) terms of an arithmetic sequence (\( s_n = a_1 + a_2 + a_3 + \dots + a_n \)) is calculated by:

\[ s_n = \dfrac{n (a_1 + a_n)}{2} \]

An equivalent form using the common difference is:

\[ s_n = \dfrac{n[2a_1 + (n-1)d]}{2} \]

Key Facts and Terminology

Tip: You can use our Arithmetic Series Online Calculator to quickly calculate the sum of terms in any sequence.

Step-by-Step Solved Problems

Problem 1: Finding a Specific Term

Show Solution

Question: 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 50th term.

Solution

Use the common difference \( d = 3 \) and the first term \( a_1 = 6 \) in the \( n \)-th term formula:

\[ a_n = a_1 + (n - 1) d \]

\[ a_n = 6 + 3 (n - 1) \]

\[ a_n = 3n + 3 \]

To find the 50th term, substitute \( n = 50 \) into the formula:

\[ a_{50} = 3(50) + 3 = 153 \]

Verification: The sequence begins 6, 9, 12, 15, …, and the 50th term is indeed \( 6 + 49(3) = 153 \). ✓

Problem 2: Negative Common Difference

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Question: The first term of an arithmetic sequence is equal to 200 and the common difference is equal to -10. Find the value of the 20th term.

Solution

Use \( d = -10 \) and \( a_1 = 200 \) in the \( n \)-th term formula:

\[ a_{20} = a_1 + (20 - 1) d \]

\[ a_{20} = 200 + 19(-10) \]

\[ a_{20} = 200 - 190 = 10 \]

Problem 3: Finding the First Term

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Question: An arithmetic sequence has a common difference equal to 10 and its 6th term is equal to 52. Find its 15th term.

Solution

Use the formula for the 6th term to find \( a_1 \):

\[ a_6 = a_1 + 10(6 - 1) \]

\[ 52 = a_1 + 50 \]

\[ a_1 = 2 \]

Now that we have the first term and the common difference, we calculate the 15th term:

\[ a_{15} = 2 + 10(15 - 1) \]

\[ a_{15} = 2 + 140 = 142 \]

Problem 4: Solving a System of Linear Equations

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Question: An arithmetic sequence has its 5th term equal to 22 and its 15th term equal to 62. Find its 100th term.

Solution

Write out the equations for the 5th and 15th terms to create a system of two linear equations:

\[ a_5 = a_1 + 4d = 22 \]

\[ a_{15} = a_1 + 14d = 62 \]

Subtract the first equation from the second equation to eliminate \( a_1 \):

\[ (a_1 + 14d) - (a_1 + 4d) = 62 - 22 \]

\[ 10d = 40 \implies d = 4 \]

Substitute \( d = 4 \) back into the first equation to find \( a_1 \):

\[ a_1 + 4(4) = 22 \]

\[ a_1 + 16 = 22 \implies a_1 = 6 \]

Finally, find the 100th term:

\[ a_{100} = 6 + 4(100 - 1) \]

\[ a_{100} = 6 + 4(99) = 402 \]

Problem 5: Summing Consecutive Integers

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Question: Find the sum of all integers from 1 to 1000.

Solution

The sequence is: \( 1, 2, 3, 4, \dots, 1000 \).

This is an arithmetic sequence with \( n = 1000 \) terms, \( a_1 = 1 \), and \( a_{1000} = 1000 \). Using the sum formula:

\[ s_n = \dfrac{n(a_1 + a_n)}{2} \]

\[ s_{1000} = \dfrac{1000(1 + 1000)}{2} \]

\[ s_{1000} = 500 \times 1001 = 500500 \]

Historical note: This is the same technique attributed to Carl Friedrich Gauss, who famously summed the integers from 1 to 100 in seconds as a child.

Problem 6: Summing Even Integers

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Question: Find the sum of the first 50 even positive integers.

Solution

The sequence is: \( 2, 4, 6, \dots \).

Here, \( a_1 = 2 \) and \( d = 2 \). First, find the 50th term (\( a_{50} \)):

\[ a_{50} = 2 + 2(50 - 1) = 100 \]

Now, calculate the sum of the first 50 terms:

\[ s_{50} = \dfrac{50(2 + 100)}{2} \]

\[ s_{50} = 25 \times 102 = 2550 \]

Problem 7: Summing Multiples of 5

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Question: Find the sum of all positive integers from 5 to 1555 inclusive, that are divisible by 5.

Solution

The sequence is: \( 5, 10, 15, \dots, 1555 \).

We know \( a_1 = 5 \) and \( d = 5 \). First, determine how many terms (\( n \)) are in the sequence:

\[ 1555 = a_1 + (n - 1)d \]

\[ 1555 = 5 + 5(n - 1) \]

\[ 1550 = 5(n - 1) \implies 310 = n - 1 \implies n = 311 \]

Now, compute the sum of these 311 terms:

\[ s_{311} = \dfrac{311(5 + 1555)}{2} \]

\[ s_{311} = \dfrac{311(1560)}{2} = 311 \times 780 = 242580 \]

Problem 8: Evaluating a Summation

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Question: Find the sum \( S \) defined by:

\[ S = \sum_{n=1}^{10} \left(2n + \frac{1}{2}\right) \]

Solution

Decompose the summation into two parts:

\[ S = 2 \sum_{n=1}^{10} n + \sum_{n=1}^{10} \frac{1}{2} \]

The first term \( \sum_{n=1}^{10} n \) is the sum of the first 10 positive integers (\( a_1 = 1, a_{10} = 10 \)):

\[ s_{10} = \dfrac{10(1 + 10)}{2} = 55 \]

The second term adds the constant \( 1/2 \) exactly 10 times:

\[ 10 \times \dfrac{1}{2} = 5 \]

Finally, substitute these values back into \( S \):

\[ S = 2(55) + 5 = 115 \]

Verification: Expanding the first few terms: \( 2.5 + 4.5 + 6.5 + \dots + 20.5 \). There are 10 terms with average \( (2.5 + 20.5)/2 = 11.5 \), so \( S = 10 \times 11.5 = 115 \). ✓

Problem 9: An Arithmetic Sequence Application

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Question: A theater has 25 rows of seats. The first row has 12 seats, and each subsequent row has 3 more seats than the row in front of it. How many total seats are in the theater?

Solution

This is an arithmetic sequence with \( a_1 = 12 \), \( d = 3 \), and \( n = 25 \). First, find the number of seats in the last row:

\[ a_{25} = 12 + 3(25 - 1) = 12 + 72 = 84 \]

Now compute the total number of seats using the sum formula:

\[ s_{25} = \dfrac{25(12 + 84)}{2} = \dfrac{25 \times 96}{2} = 25 \times 48 = 1200 \]

Answer: The theater has 1,200 seats total.

Practice Exercises

Try solving the following questions related to arithmetic sequences on your own:

  1. Find \( a_{20} \) given that \( a_3 = 9 \) and \( a_8 = 24 \).
  2. Find \( a_{30} \) given that the first few terms of an arithmetic sequence are: \( 6, 12, 18, \dots \)
  3. Find the common difference \( d \) given that \( a_1 = 10 \) and \( a_{20} = 466 \).
  4. Find the sum \( s_{30} \) given that \( a_{10} = 28 \) and \( a_{20} = 58 \).
  5. Find the sum \( S \) defined by: \( \displaystyle S = \sum_{n=1}^{30} \left(3n - \frac{1}{2}\right) \)
  6. Find the sum \( S \) defined by: \( \displaystyle S = \sum_{n=1}^{20} 0.2n + \sum_{j=21}^{40} 0.4j \)
View Solutions to Exercises
  1. \( a_{20} = 60 \)
    From \( a_3 = a_1 + 2d = 9 \) and \( a_8 = a_1 + 7d = 24 \), subtracting gives \( 5d = 15 \implies d = 3 \), so \( a_1 = 3 \). Then \( a_{20} = 3 + 19(3) = 60 \).
  2. \( a_{30} = 180 \)
    \( a_1 = 6 \), \( d = 6 \), so \( a_{30} = 6 + 29(6) = 180 \).
  3. \( d = 24 \)
    \( 466 = 10 + 19d \implies 456 = 19d \implies d = 24 \).
  4. \( s_{30} = 1335 \)
    From \( a_{10} = a_1 + 9d = 28 \) and \( a_{20} = a_1 + 19d = 58 \), subtracting gives \( 10d = 30 \implies d = 3 \), so \( a_1 = 1 \). Then \( a_{30} = 1 + 29(3) = 88 \), and \( s_{30} = 30(1+88)/2 = 15 \times 89 = 1335 \).
  5. \( S = 1380 \)
    \( S = 3\sum_{n=1}^{30} n - 30 \cdot \frac{1}{2} = 3(465) - 15 = 1395 - 15 = 1380 \).
  6. \( S = 286 \)
    First sum: \( 0.2\sum_{n=1}^{20} n = 0.2(210) = 42 \). Second sum: \( 0.4\sum_{j=21}^{40} j = 0.4[(20 \cdot 41/2) - (20 \cdot 21/2)] = 0.4(410 - 210) = 0.4(200) = 80 \). Wait — let me redo: \( \sum_{j=21}^{40} j = \sum_{j=1}^{40} j - \sum_{j=1}^{20} j = 820 - 210 = 610 \). So \( 0.4(610) = 244 \). Total \( S = 42 + 244 = 286 \). ✓

Common Mistakes to Avoid

  1. Using \( n \) instead of \( n-1 \) in the \( n \)-th term formula. The correct formula is \( a_n = a_1 + (n-1)d \), not \( a_1 + nd \).
  2. Confusing the common difference with the first term. In \( a_n = a_1 + (n-1)d \), \( a_1 \) is the first term and \( d \) is the common difference.
  3. Forgetting the sign of \( d \). A negative common difference produces a decreasing sequence.
  4. Miscounting the number of terms. For a sequence from \( a_1 \) to \( a_n \), there are exactly \( n \) terms, where \( n = \dfrac{a_n - a_1}{d} + 1 \).
  5. Using the sum formula with an incorrect \( a_n \). If you need the sum of the first \( n \) terms but don't know \( a_n \), compute \( a_n \) first using \( a_n = a_1 + (n-1)d \), then apply the sum formula.

Frequently Asked Questions

What is an arithmetic sequence?

An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. This constant is called the common difference, usually denoted by \( d \). For example, \( 3, 7, 11, 15, \ldots \) is arithmetic with \( d = 4 \).

What is the formula for the \( n \)-th term of an arithmetic sequence?

The \( n \)-th term of an arithmetic sequence with first term \( a_1 \) and common difference \( d \) is given by \( a_n = a_1 + (n - 1)d \).

How do you find the sum of the first \( n \) terms of an arithmetic sequence?

The sum of the first \( n \) terms is given by \( s_n = \dfrac{n(a_1 + a_n)}{2} \), where \( a_1 \) is the first term and \( a_n \) is the \( n \)-th term. This can also be written as \( s_n = \dfrac{n[2a_1 + (n-1)d]}{2} \) using the common difference.

How do you find the common difference of an arithmetic sequence?

The common difference \( d \) is found by subtracting any term from the following term: \( d = a_{n+1} - a_n \). For instance, in the sequence \( 5, 12, 19, 26 \), the common difference is \( 12 - 5 = 7 \).

What is the difference between an arithmetic sequence and an arithmetic series?

An arithmetic sequence is an ordered list of terms with a common difference. An arithmetic series is the sum of the terms of an arithmetic sequence. For example, \( 3, 7, 11, 15 \) is a sequence, while \( 3 + 7 + 11 + 15 = 36 \) is the corresponding series.

Can the common difference be negative or zero?

Yes. If \( d \gt 0 \), the sequence is increasing. If \( d \lt 0 \), the sequence is decreasing. If \( d = 0 \), all terms are equal. Any real number is a valid common difference.

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