# Arithmetic Series

The sum of an indicated number of terms in a sequence is called a **series**. The series obtained by adding the terms of an arithmetic progression is called an **a****rithmetic series**.

For example, the sum of the first seven terms of the sequence $$\left\{ {{n^2}} \right\}$$ is the series,

\[1 + 4 + 9 + 16 + 25 + 36 + 49\]

The above series is also called the $$7th$$ partial sum of the sequence $$\left\{ {{n^2}} \right\}$$.

If the number of terms in a series is finite, then the series is called a finite series, while a series consisting of an unlimited number of terms is known as an infinite series.

__The sum of the first n terms of an arithmetic series__

For any sequence $$\left\{ {{a_n}} \right\}$$, we have

$${S_n} = {a_1} + {a_2} + {a_3} + \cdots + {a_n}$$

If $$\left\{ {{a_n}} \right\}$$ is an A.P., then $${S_n}$$ can be written with usual notation as:

$${S_n} = {a_1} + \left( {{a_1} + d} \right) + \left( {{a_1} + 2d} \right) + \cdots + \left( {{a_n} – 2d} \right) + \left( {{a_n} – d} \right) + {a_n}$$ — (1)

If we write the terms of the series in the reverse order, the sum of $$n$$ terms remains the same; that is

$${S_n} = {a_n} + \left( {{a_n} – d} \right) + \left( {{a_n} – 2d} \right) + \cdots + \left( {{a_1} + 2d} \right) + \left( {{a_1} + d} \right) + {a_n}$$ — (2)

Adding (1) and (2), we get

$$2{S_n} = \left( {{a_1} + {a_n}} \right) + \left( {{a_1} + {a_n}} \right) + \cdots + \left( {{a_1} + {a_n}} \right) + \left( {{a_1} + {a_n}} \right)$$

$$ = \left( {{a_1} + {a_n}} \right) + \left( {{a_1} + {a_n}} \right) + \left( {{a_1} + {a_n}} \right) + \cdots {\text{ to}}\;n{\text{ terms}}$$

$$ = n\left( {{a_1} + {a_n}} \right)$$

Thus, $${S_n} = \frac{n}{2}\left( {{a_1} + {a_n}} \right)$$ — (3)

If in (3) we replace $${a_n}$$ by its value as $${a_n} = {a_1} + \left( {n – 1} \right)d$$, then we obtain another useful rule for $${S_n}$$ as

$${S_n} = \frac{n}{2}\left[ {{a_1} + {a_1} + \left( {n – 1} \right)d} \right]$$

$${S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n – 1} \right)d} \right]$$ — (4)

__Example__:

Sum up the following series: $$51 + 50 + 49 + \cdots + 21$$

__Solution__:

Here $${a_1} = 51$$, $$d = – 1$$, $${a_n} = 21$$

To find $$n$$ first, we have $${a_n} = {a_1} + \left( {n – 1} \right)d$$

$$\therefore $$ $$21 = 51 + \left( {n – 1} \right)\left( { – 1} \right)$$

$$\therefore $$ $$21 = 52 – n$$

$$n = 52 – 21 = 31$$

Now using, we have

$${S_n} = \frac{n}{2}\left( {{a_1} + {a_{n}}} \right) = \frac{{31}}{2}\left( {51 + 21} \right) = \frac{{31}}{2} \times 72 = 1116$$

__Example__:

Find the sum of the first $$n$$ terms of the arithmetic series $$4 + 9 + 14 + \cdots $$. Also find the sum of the first $$17$$ terms.

__Solution__:

Here $${a_1} = 4$$ and $$d = 9 – 4 = 5$$

If $${S_n}$$ is the sum of the first $$n$$ terms, then

$${S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n – 1} \right)d} \right]$$

$$ = \frac{n}{2}\left[ {2\left( 4 \right) + \left( {n – 1} \right)\left( 5 \right)} \right] = \frac{n}{2}\left[ {8 + 5n – 5} \right] = \frac{n}{2}\left( {5n + 3} \right)$$

$${S_n} = \frac{{n\left( {5n + 3} \right)}}{2}$$ —— (1)

For the sum of the first $$17$$ terms, we put $$n = 17$$ in (1), i.e.

$${S_{17}} = \frac{{17\left[ {5\left( {17} \right) + 3} \right]}}{2} = \frac{{17\left( {85 + 3} \right)}}{2} = \frac{{17\left( {88} \right)}}{2} = 748$$