Arithmetic Series

The sum of an indicated numbers of terms in a sequence is called a Series. The series obtained by adding the terms of an arithmetic progression is called 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 named as the 7th partial sum of the sequence \left\{ {{n^2}} \right\}.

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

Sum of 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 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( {82}  \right)}}{2} = 697

Comments

comments