**Series**

A **series** is something we obtain from a sequence by adding all the terms together.

For example, suppose we have the sequence

The series we obtain from this is

*u*

_{1},

*u*

_{2},

*u*

_{3}, …,

*u*

_{n}and we write

*S*for the sum of these

_{n}*n*terms. So although the ideas of a ‘sequence’ and a ‘series’ are related, there is an important distinction between them.

For example, let us consider the sequence of numbers

*n*.

Then

*S*

_{1}=1, as it is the sum ofjust the first term on its own. The sum of the first two terms is

*S*

_{2}=1+2=3. Continuing, we get

*S*

_{3}=1+2+3=6,

*S*

_{4}=1+2+3+4=10,

and so on.

Key Point

A series is a sum of the terms in a sequence. If there arenterms in the sequence and we evaluate the sum then we often writeSfor the result, so that_{n}S=_{n}u_{1}+u_{2}+u_{3}+⋯+u_{n}

**Arithmetic Series**

A **series** is a sum of the terms in a sequence.

An arithmetic series is the sum of the terms in an arithmetic sequence.

For example, an arithmetic sequence is: 5, 8, 11, 14, …

The related arithmetic series is: 5+8+11+14+…

The term, *S _{n}*, is used to represent the sum of the first

*n*terms of a series. The

*n*th term of an arithmetic series is the

*n*th term of the related arithmetic sequence.

For the arithmetic series above:

*S*

_{1}=

*t*

_{1};

*S*

_{2}=

*t*

_{1}+

*t*

_{2};

*S*

_{3}=

*t*

_{1}+

*t*

_{2}+

*t*

_{3};

*S*

_{4}=

*t*

_{1}+

*t*

_{2}+

*t*

_{3}+

*t*

_{4}

*S*

_{1}=5;

*S*

_{2}=5+8;

*S*

_{3}=5+8+11;

*S*

_{4}=5+8+11+14

*S*

_{2}=13;

*S*

_{3}=24;

*S*

_{4}=38

These are called

*partial sums*.

If there are only a few terms,

*S*can be determined using mental math.

_{n}To develop a rule to determine

*S*, use algebra.

_{n}Write the sum on one line, reverse the order of the terms on the next line, then add vertically. Write the sum as a product.

Solve for

*S*.

_{n}*S*=½

_{n}*n*(

*t*

_{1}+

*t*)

_{n}The rule above is used when

*t*

_{1}and

*t*are known. Substitute for

_{n}*t*to write

_{n}*S*in a different way, so it can be used when

_{n}*t*and the common difference,

_{n}*n*, are known.

*S*=½

_{n}*n*(

*t*

_{1}+

*t*)

_{n}Substitute:

*t*=

_{n}*t*

_{1}+

*d*(

*n*-1)

*S*=½

_{n}*n*(

*t*

_{1}+

*t*

_{1}+

*d*(

*n*-1))

Combine like terms.

*S*=½

_{n}*n*(2

*t*

_{1}+

*d*(

*n*-1))

The Sum ofnTerms of an Arithmetic Series

For an arithmetic series with 1st term,t, common difference,_{n}d, andnth term,t, the sum of the first_{n}nterms,S, is:_{n}S=½_{n}n(t_{1}+t) or_{n}S=½_{n}n(2t_{1}+d(n-1))

In the last section, we wrote the sequence of minutes that Ken walked each day for two weeks:

Since the difference between each pair of consecutive times is a constant, 5, the sequence is an arithmetic sequence. The total length of time that Ken walked in two weeks is the sum of the terms of this sequence:

In general, if *a*_{1}, *a*_{1}+*d*, *a*_{1}+2*d*, …, *a*_{1}+(*n*-1)*d* is an arithmetic sequence with *n* terms, then:

_{i=1}

^{n}[

*a*

_{1}+(

*i*-1)

*d*] =

*a*

_{1}+(

*a*

_{1}+

*d*)+(

*a*

_{1}+2

*d*)+⋯+[

*a*

_{1}+(

*n*-1)

*d*]

This sum is called an *arithmetic series*.

definition

Anarithmetic seriesis the indicated sum of the terms of an arithmetic sequence.

Once a given series is defined, we can refer to it simply as S (for sigma). *S _{n}* is called the

**and represents the sum of the first**

*n*th partial sum*n*terms of the sequence.

We can find the number of minutes that Ken walked in 14 days by adding the 14 numbers or by observing the pattern of this series. Begin by writing the sum first in the order given and then in reverse order.

*S*

_{14}=20+25+30+35+40+45+50+55+60+65+70+75+80+85

*S*

_{14}=85+80+75+70+65+60+55+50+45+40+35+30+25+20

Note that for this arithmetic series:

*a*

_{1}+

*a*

_{14}=

*a*

_{2}+

*a*

_{13}=

*a*

_{3}+

*a*

_{12}=

*a*

_{4}+

*a*

_{11}=

*a*

_{5}+

*a*

_{10}=

*a*

_{6}+

*a*

_{9}=

*a*

_{7}+

*a*

_{8}

Add the sums together, combining corresponding terms. The sum of each pair is 105 and there are 14/2 or 7 pairs.

Therefore, the total number of minutes that Ken walked in 14 days is (7×105) or 735 minutes.

Does a similar pattern exist for every arithmetic series? Consider the general arithmetic series With *n* terms, *a _{n}*=

*a*

_{1}+(

*n*-1)

*d*. List the terms of the series in ascending order from

*a*

_{1}to

*a*and in descending order from

_{n}*a*to

_{n}*a*

_{1}.

In general, for any arithmetic series with

*n*terms there are ½

*n*pairs of terms whose sum is

*a*

_{1}+

*a*:

_{n}*S*=

_{n}*n*(

*a*

_{1}+

*a*)=

_{n}*n*(2

*a*

_{1}+(

*n*-1)

*d*)

*S*=½

_{n}*n*(

*a*

_{1}+

*a*)=½

_{n}*n*(2

*a*

_{1}+(

*n*-1)

*d*)

Often we have a sequence of numbers and want to know their sum. For an example, we return to the oil drops on the racetrack from the start of the previous section on arithmetic sequences. The distance covered by the car each second illustrated the concept of an arithmetic sequence.

The total distance covered by the car is the sum of the individual distances covered each second. So after one second the car has travelled 10 m, after two seconds the car has travelled 10+18 m=28 m, after three seconds the car has travelled a total distance of 10+18+26 m=54 m, and so on.

A series,

S, is the sum of a sequence of_{n}ntermst_{1}+t_{2}+t_{3}+…+t._{n}

Thus:

S_{1}=t_{1} |

S_{2}=t_{1}+t_{2} |

S_{3}=t_{1}+t_{2}+t_{3} |

S=_{n}t_{1}+t_{2}+t_{3}+⋯+t_{(n-2)}+t_{(n-1)}+t_{n} |

For an arithmetic sequence, the sum of the first *n* terms, *S _{n}*, can be written in two ways.

1. The first term in the arithmetic sequence is a, the common difference is

*d*, and the last term — that is, the

*n*th term — in the sequence is

*ℓ*.

*S*=

_{n}*a*+(

*a*+

*d*)+(

*a*+2

*d*)+…+(

*a*+(

*n*-2)

*d*)+(

*a*+(

*n*-1)

*d*)

=

*a*+(

*a*+

*d*)+(

*a*+2

*d*)+…+(

*ℓ*-2

*d*)+(

*ℓ*–

*d*)+

*ℓ*…[1]

2. We can write the sum

*S*in reverse order starting with the

_{n}*n*th term and summing back to the first term a:

*S*=

_{n}*ℓ*+(

*ℓ*–

*d*)+(

*ℓ*-2

*d*)+⋯+(

*a*+2

*d*)+(

*a*+

*d*)+

*a*⋯ [2]

If we add equation [1] and equation [2] together and recognise that there are

*n*terms, each of which equals (

*a*+

*ℓ*), we get:

*S*=(

_{n}*a*+

*ℓ*)+(

*a*+

*ℓ*)+⋯

*n*times

2

*S*=

_{n}*n*(

*a*+

*ℓ*)

and so:

*S*=½

_{n}*n*(

*a*+

*ℓ*)

or since

*ℓ*is the

*n*th term,

*ℓ*=

*a*+(

*n*-1)

*d*so

*S*=½

_{n}*n*(2

*a*+(

*n*-1)

*d*)

The sum of the first

nterms in an arithmetic sequence is given byS=½_{n}n(a+ℓ)

whereais the first term andℓis the last term; or alternately, sinceℓ=a+(n-1)d, byS=½_{n}n(2a+(n-1)d)

whereais the first term anddis the common difference.

If we know the first term, a, the common difference, *d*, and the number of terms, *n*, that we wish to add together, we can calculate the sum directly without having to add up all the individual terms.

It is worth while also to note that *S*_{(n+1)}=*S _{n}*+

*t*

_{(n+1)}. This tells us that the next term in the series

*S*

_{(n+1)}is the present sum,

*S*, plus the next tenn in the sequence,

_{n}*t*

_{(n+1)}. This

result is useful in spreadsheets where one column gives the sequence and an adjacent

column is used to give the series.

An arithmetic series is the sum of the terms of an arithmetic sequence.

For example:

21, 23, 25, 27, …, 49 is a finite arithmetic sequence.

21+23+25+27+…+49 is the corresponding arithmetic series.

Sum of a Finite Arithmetic Series

If the first term is *u*_{1} and the common difference is *d*, the terms are *u*_{1}, *u*_{1}+*d*, *u*_{1}+2*d*, *u*_{1}+3*d*, and so on.

Suppose that *u _{n}* is the final term of an arithmetic series.

So,

*S*=

_{n}*u*

_{1}+(

*u*

_{1}+

*d*)+(

*u*

_{1}+2

*d*)+⋯+(

*u*-2

_{n}*d*)+(

*u*–

_{n}*d*)+

*u*

_{n}But

*S*=

_{n}*u*+(

_{n}*u*–

_{n}*d*)+(

*u*-2

_{n}*d*)+⋯+(

*u*

_{1}+2

*d*)+(

*u*

_{1}+

*d*)+

*u*

_{1}{reversing them}

Adding these two equations vertically we get

The sum of a finite arithmetic series with first term

u_{1}, common differenced, and last termuis_{n}S=½_{n}n(u_{1}+u) or_{n}S=½_{n}n(2u_{1}+(n-1)d).

(How). Do arithmetic series apply to amphitheaters?

The first amphitheaters were built for contests between gladiators. Modern amphitheaters are usually used for the performing arts. Amphitheaters generally get wider as the distance from the stage increases.

Suppose a small amphitheater can seat 18 people in the first row and each row can seat 4 more people than the previous row.

Study Tip

Indicated Sum

The sum of a series is the result when the terms of the series are added. An indicated sum is the expression that illustrates the series, which includes the terms + or – .

ARITHMETIC SERIES

The numbers of seats in the rows of the amphitheater form an arithmetic sequence. To find the number of people who could sit in the first four rows, add the first four terms of the sequence. That sum is 18+22+26+30 or 96. A series is an indicated sum of the terms of a sequence. Since 18, 22, 26, 30 is an arithmetic sequence, 18+22+26+30 is an arithmetic series. Below are some more arithmetic sequences and the corresponding arithmetic series.

Arithmetic Sequence | Arithmetic Series |
---|---|

-9, -3, 3 | -9+(-3)+3 |

⅜, 8/8, 13/8, 18/8 | ⅜+8/8+13/8+18/8 |

*S _{n}* represents the sum of the first

*n*terms of a series. For example,

*S*

_{4}is the sum of the first four terms.

To develop a formula for the sum of any arithmetic series, consider the series below.

*S*_{9}=4+11+18+25+32+39+46+53+60

Write *S*_{9} in two different orders and add the two equations.

An arithmetic series

*S*has

_{n}*n*terms, and the sum of the first and last terms is

*a*

_{1}+

*a*. Thus, the formula

_{n}*S*=½

_{n}*n*(

*a*

_{1}+

*a*) represents the sum of any arithmetic series.

_{n}Key Concept: Sum of an Arithmetic Series

The sum

Sof the first_{n}nterms of an arithmetic series is given byS=½_{n}n[2a_{1}+(n-1)d] orS=½_{n}n(a_{1}+a)_{n}

Challenge:

Let the sum of *n*, 2*n*, 3*n* terms of an A.P. be *S*_{1}, *S*_{2} and *S*_{3}, respectively, show that *S*_{3}=3(*S*_{2}–*S*_{1}).

Answer:

Let *a* and *b* be the first term and the common difference of the A.P. respectively.

Therefore,

*S*

_{1}=½

*n*[2

*a*+

*d*(

*n*-1)]…(1)

*S*

_{2}=½⋅2

*n*[2

*a*+

*d*(2

*n*-1)]…(2)

*S*

_{3}=½⋅3

*n*[2

*a*+

*d*(3

*n*-1)]…(3)

From (1) and (2), we obtain

*S*

_{2}–

*S*

_{1}=(

*an*+½⋅3

*n*

_{2}

*d*-½

*n*

*d*)

=½

*n*(2

*a*+3

*n*

*d*–

*d*)

=½

*n*(2

*a*+

*d*(3

*n*-1))

3(

*S*

_{2}–

*S*

_{1})=½⋅3

*n*[2

*a*+

*d*(3

*n*-1)]

∴3(

*S*

_{2}–

*S*

_{1})=

*S*

_{3}[From (3)]

Hence, the given result is proved.

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