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