Arithmetic Sequences and Series
This tutorial explains arithmetic sequences and their sums. It covers the nth-term and sum formulas, the recursive form, the arithmetic mean, and six worked examples with step-by-step solutions.
What Is an Arithmetic Sequence?
An arithmetic sequence is a sequence in which each term is obtained by adding the same constant to the previous term. This constant is called the common difference and is denoted by \( d \).
Arithmetic sequences are linear: plotting \(a_n\) against \(n\) produces points that lie on a straight line.
Essential Formulas for Arithmetic Sequences
Recursive form:
\[ a_1 = \text{given}, \qquad a_n = a_{n-1} + d \quad (n \ge 2) \]
Explicit (nth-term) formula:
\[ a_n = a_1 + (n - 1)d \]
Arithmetic mean: for three consecutive terms \(a\), \(b\), \(c\), the middle term is the average:
\[ b = \frac{a + c}{2} \]
Sum of the first \(n\) terms (arithmetic series):
\[ S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}\left[\,2a_1 + (n-1)d\,\right] \]
Worked Examples with Detailed Solutions
Example 1: Identifying a Sequence
Question: Consider the sequence \(0, 6, 12, 18, 24, \ldots\). Is it arithmetic? What is its common difference?
View solution
Subtract consecutive terms:
\[ 6 - 0 = 6, \qquad 12 - 6 = 6, \qquad 18 - 12 = 6 \]
The difference is constant, so the sequence is arithmetic with common difference \( d = 6 \).
Example 2: Finding the Term Number
Question: Which term of the arithmetic sequence \(2, 5, 8, \ldots\) equals \(227\)?
View solution
Here \(a_1 = 2\) and \(d = 3\). Using \(a_n = a_1 + (n-1)d\):
\[ 227 = 2 + (n-1)\cdot 3 \]
\[ 225 = 3(n-1) \]
\[ n - 1 = 75 \quad\Longrightarrow\quad n = 76 \]
The 76th term is 227.
Example 3: Sum of Consecutive Odd Integers
Question: How many consecutive odd integers, starting from \(9\), must be added to obtain a sum of \(15{,}860\)?
View solution
First term \(a_1 = 9\); consecutive odd integers differ by \(d = 2\). Apply the sum formula:
\[ S_n = \frac{n}{2}\left[2(9) + (n-1)\cdot 2\right] = 15860 \]
\[ \frac{n}{2}\,(18 + 2n - 2) = 15860 \]
\[ \frac{n}{2}\,(2n + 16) = 15860 \]
\[ n(n + 8) = 15860 \]
\[ n^2 + 8n - 15860 = 0 \]
Factor the quadratic: \(15860 = 122 \cdot 130\), so
\[ (n - 122)(n + 130) = 0 \quad\Longrightarrow\quad n = 122 \ \text{ or }\ n = -130 \]
Since \(n\) must be positive, \(n = 122\). 122 consecutive odd integers must be added.
Example 4: Finding Terms from the Sum Formula
Question: The sum of the first \(n\) terms of an arithmetic sequence is \(S_n = 2n^2 + 5n\). Find the first three terms.
View solution
For \(n \ge 2\), the nth term satisfies \(a_n = S_n - S_{n-1}\). (Also \(a_1 = S_1\).)
\[ a_n = (2n^2 + 5n) - \left[2(n-1)^2 + 5(n-1)\right] \]
Expand the bracket:
\[ 2(n-1)^2 + 5(n-1) = 2(n^2 - 2n + 1) + 5n - 5 = 2n^2 + n - 3 \]
Subtract:
\[ a_n = (2n^2 + 5n) - (2n^2 + n - 3) = 4n + 3 \]
Now substitute \(n = 1, 2, 3\):
- \( a_1 = 4(1) + 3 = 7 \)
- \( a_2 = 4(2) + 3 = 11 \)
- \( a_3 = 4(3) + 3 = 15 \)
The first three terms are 7, 11, 15.
Example 5: Three Consecutive Terms Given Their Sum and Product
Question: The sum of three consecutive terms of an arithmetic sequence is 27, and their product is 585. Find the three numbers.
View solution
Let the three terms be \(x\), \(x + d\), and \(x + 2d\).
Step 1 — Sum equation:
\[ x + (x+d) + (x+2d) = 27 \]
\[ 3x + 3d = 27 \quad\Longrightarrow\quad x + d = 9 \tag{1} \]
Step 2 — Product equation:
\[ x(x+d)(x+2d) = 585 \tag{2} \]
From (1), \(x + d = 9\) and \(x = 9 - d\), so \(x + 2d = 9 + d\). Substitute into (2):
\[ (9-d)(9)(9+d) = 585 \]
\[ 9(81 - d^2) = 585 \]
\[ 81 - d^2 = 65 \quad\Longrightarrow\quad d^2 = 16 \quad\Longrightarrow\quad d = \pm 4 \]
If \(d = 4\): the terms are \(5, 9, 13\). If \(d = -4\): the terms are \(13, 9, 5\).
The three numbers are 5, 9, and 13 (in either ascending or descending order).
Example 6: Finding \(x\) and \(d\) from Three Algebraic Terms
Question: The first three terms of an arithmetic sequence are \(x\), \(\frac{5x}{4}\), and \(\frac{9}{2}\). Find \(x\) and the common difference \(d\).
View solution
Since the terms are consecutive, the differences between consecutive terms must be equal:
\[ \frac{5x}{4} - x = \frac{9}{2} - \frac{5x}{4} \]
Simplify the left side (\( \frac{5x}{4} - \frac{4x}{4} = \frac{x}{4} \)):
\[ \frac{x}{4} = \frac{9}{2} - \frac{5x}{4} \]
Use a common denominator on the right:
\[ \frac{x}{4} = \frac{18 - 5x}{4} \]
Equal denominators ⇒ equal numerators:
\[ x = 18 - 5x \quad\Longrightarrow\quad 6x = 18 \quad\Longrightarrow\quad x = 3 \]
The common difference is
\[ d = \frac{5x}{4} - x = \frac{15}{4} - 3 = \frac{15 - 12}{4} = \frac{3}{4} \]
\(x = 3\) and \(d = \dfrac{3}{4}\), so the sequence begins \(3, \tfrac{15}{4}, \tfrac{9}{2}\).
Practice Problems
Try these on your own, then reveal the solution.
Problem 1
Find the 20th term of the arithmetic sequence \(4, 11, 18, \ldots\).
View solution
\(a_1 = 4\), \(d = 7\), so \(a_{20} = 4 + 19 \cdot 7 = 137\).
Problem 2
Find the sum of the first 50 positive even integers.
View solution
\(a_1 = 2\), \(d = 2\), \(n = 50\): \(S_{50} = \dfrac{50}{2}\,(2 + 100) = 25 \cdot 102 = 2550\).
Problem 3
In an arithmetic sequence, \(a_4 = 14\) and \(a_{10} = 32\). Find \(a_1\) and \(d\).
View solution
Subtract: \(a_{10} - a_4 = 6d = 18\), so \(d = 3\). Then \(a_1 = a_4 - 3d = 14 - 9 = 5\). The sequence is \(5, 8, 11, \ldots\).
Related Practice
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