What Are Arithmetic Series?

An arithmetic series is a series in which you have to add a specific difference $d$ to get the next term of the series. Arithmetic series are pretty straightforward, and all the exercises can be solved by using three formulas. Check them out here:

Formula

Arithmetic series

Finding the difference in the series:

d = a_{n + 1} - a_{n}

Finding the $n$ th term of the series:

a_{n} = a_{1} + (n - 1) d

Finding the sum:

S_{n} = \frac{a_{1} + a_{n}}{2} \cdot n

Example 1

You have this arithmetic series

2 + 4 + 6 + 8 + \dots

Find the difference, a formula for the $n^{t h}$ term, and the sum of the first 50 terms.

In order to solve this exercise you have use the formulas for arithmetic series. You can find the difference by subtracting two consecutive numbers from the series. You then get:

d = a_{n + 1} - a_{n} = 4 - 2 = 2 .

The formula for finding the $n$ th term is given by

\begin{array}{l} a_{n} & = a_{1} + (n - 1) d \\ = 2 + (n - 1) \cdot 2 \\ = 2 + 2 n - 2 \\ = 2 n . \end{array}

Finally, you can find the sum of the first 50 terms by finding $a_{50}$ . This can be done by using the formula for the $n$ th term where $n = 50$ . You then get:

\begin{array}{l} a_{50} & = 2 + (50 - 1) \cdot 2 \\ = 2 + 49 \cdot 2 \\ = 100 . \end{array}

a_{50} = 2 + (50 - 1) \cdot 2 = 2 + 49 \cdot 2 = 100 .

You can now find the sum of the first 50 terms:

S_{50} = \frac{2 + 100}{2} \cdot 50 = 51 \cdot 50 = 2550 .

Example 2

You know that the third number in an arithmetic series is 7 and the sixth number is 12. Find the difference $d$ and the first term $a_{1}$ .

This is a classical case where you have to solve the exercise with the help of an equation system. You know that the formula for the $n$ th term in the arithmetic series is $a_{n} = a_{1} + (n - 1) d$ . If you change the formula by using the information given from the exercise, you will find that $a_{3} = 7$ and $a_{6} = 12$ . Thus, you can use this as a starting point for your two equations.

Firstly, $a_{3} = 7$ gives you

\begin{align} a_{1} + (3 - 1) d & = 7, \\ a_{1} + 2 d & = 7, \\ a_{1} & = 7 - 2 d . & (1) \end{align}

Secondly, $a_{6} = 12$ gives you

\begin{array}{l} a_{1} + (6 - 1) d & = 12, \\ a_{1} + 5 d & = 12 . \end{array}

By replacing $a_{1}$ from Equation (1) you get

\begin{array}{l} (7 - 2 d) + 5 d & = 12, \\ 3 d & = 5 | : 3, \\ d & = \frac{5}{3} . \end{array}

Finally, you can replace d in Equation (1) and get:

a_{1} = 7 - 2 \cdot \frac{5}{3} = \frac{21}{3} - \frac{10}{3} = \frac{11}{3} .

You have now $a_{1} = \frac{11}{3}$ and $d = \frac{5}{3}$ .

Example 3

A phone subscription has a $$ 3$ starting price, and the price per minute is $$ 2$ . $a_{n}$ is the price of a conversation lasting $n$ minutes. Find an arithmetic series for the price of the subscription. What is the difference $d$ ? Find a formula for the sum of the sequence.

If you have a one-minute conversation, the price will be $3 + 2 \cdot 1 = 5 $$ , if you have a tow-minutes conversation, the price will be $3 + 2 \cdot 2 = 7 $$ , if you have a three-minutes conversation, the price will be $3 + 2 \cdot 3 = 9 $$ . Generally, if you have a conversation lasting $n$ minutes, the price will cost

3 + 2 \cdot $ n .

You can count this as an arithmetic series where the price for a minute is the term $a_{1}$ , the price for two minutes is the term $a_{2}$ , the price for three minutes is the term $a_{3}$ and the price for $n$ minutes is the term $a_{n}$ . The arithmetic series is therefore:

5 + 7 + 9 + \dots + (3 + 2 \cdot n)

You can now see that each term is increasing by 2, in such a manner that the difference is $d = 2$ (price per minute).

In order to find a formula for the sum of the sequence you have to change the terms in the arithmetic series formula. This gives you:

\begin{array}{l} S_{n} & = \frac{a_{1} + a_{n}}{2} \cdot n \\ = \frac{5 + (3 + 2 \cdot n)}{2} \cdot n \\ = \frac{8 + 2 n}{2} \cdot n \\ = \frac{2 (4 n + n^{2})}{2} \\ = 4 n + n^{2} \end{array}

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