Arithmetic Series. Definition of an arithmetic series. The sum of the terms in an arithmetic...

Post on 03-Jan-2016

248 views 2 download

Tags:

Transcript of Arithmetic Series. Definition of an arithmetic series. The sum of the terms in an arithmetic...

Arithmetic Series

Definition of an arithmetic series.

The sum of the terms in an arithmetic sequence.

ArithmeticSequence

ArithmeticSeries

4, 7, 10, 13 4 + 7 + 10 + 13

-10, -4, 2 -10 + -4 + 2

27

67

10 7

14 7

, , ,27

67

10 7

14 7

+ + +

Example 1. During Kwanzaa one candle is lit the first night, two on the second night, and so forth for seven nights.How many candles are lit in all.1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

The symbol Sn is used to representthe sum of the first n terms of a series. The above represents S7.

Suppose we write S7 in two different orders and find the sum

S7 = 1 + 2 + 3 + 4 + 5 + 6 + 7S7 = 7 + 6 + 5 + 4 + 3 + 2 + 1

2S7 = 8 + 8 + 8 + 8 + 8 + 8 + 8

7 sums of 8

S7 = 1 + 2 + 3 + 4 + 5 + 6 + 7S7 = 7 + 6 + 5 + 4 + 3 + 2 + 1

2S7 = 8 + 8 + 8 + 8 + 8 + 8 + 8

7 sums of 82S7 = 7(8) S7 =

7(8) 2

Now analyze this expression in terms of Sn.

S7 =7(8) 2

Now analyze this expression in terms of Sn.

7 represents n and 8 represents the sum of the first and last terms, a1+an.

Thus we can replace the expressionwith

Sn = n(a1 + an) 2

This formula can be used to findthe sum of any arithmetic series.

Sum of an arithmetic series.

The sum of the first n terms of an arithmetic series is given by

Sn = n(a1 + an) 2

where n is a positive integer.

Example 1. Find the sum of the first 50 positive even integers.

= 25(102)

Sn = n(a1 + an) 2

S50 = 50(2 + 100) 2

= 2550

We have discovered in a previous lesson that an = a1 + (n-1)d

Using this formula and substitutiongives us another version of the formula for the sum of an arithmeticsequence.

Sn = n(a1 + an) 2

replace an

Sn = n(a1 + {a1 +(n-1)d}) 2

Sn = n(2a1 +(n-1)d) 2

This formula is useful when we do not know the value of the lastterm.

Example 2.Find the sum of the first 40 termsof an arithmetic series in which a1 = 70 and d = -21

The series is 70 + 49 + 27 + ...

Sn = n(2a1 +(n-1)d) 2

Example 2. Sum of first 40 termsa1 = 70 and d = -21The series is 70 + 49 + 27 + ...

Sn = n(2a1 +(n-1)d) 2

Sn = 40(2(70) +(40-1)(-21)) 2

= -13580

Example 3. PhysicsWhen an object is dropped it falls16 feet in the first second, 48 feetin the second second, and 80 feet in the third second.

How many feet would if fall inthe 20th second.

Example 3. Object falling 16, 48, and 80 feet in 1, 2, and 3 seconds.How many feet would if fall inthe 20th second.

16 48 80

32 32

Common difference is 32

Example 3. Object falling 16, 48, and 80 feet in 1, 2, and 3 seconds.How many feet would if fall inthe 20th second. d = 32

an = a1 + (n-1)d

a20 = 16 + (20-1)32

a20 = 624

Example 4. Refer to example 3How many feet would a free fallingHow many feet would a free fallingobject fall in 20 seconds?object fall in 20 seconds?

Sn = n(2a1 +(n-1)d) 2

S20 = 20(2(16) + (20-1)(32)) 2

S20 = 6400

Example 5.Find the first three terms of an Find the first three terms of an arithmetic series in which aarithmetic series in which a11 = 13, = 13,

aann = 157, and S = 157, and Snn = 1445 = 1445

We are given aWe are given a11, a, ann, and S, and Snn..

Therefore we use the formulaTherefore we use the formula

Sn = n(a1 + an) 2

and solve for n.and solve for n.

Example 5. Find the first 3 termsif aif a11 = 13, a = 13, ann = 157, and S = 157, and Snn = 1445 = 1445

Sn = n(a1 + an) 2

solve for n.solve for n.

1445 = n(13 + 157) 2

1445 = n(170) 2

1445 = 85n

n = 17

Example 5. Find the first 3 termsif aif a11 = 13, a = 13, ann = 157, and S = 157, and Snn = 1445 = 1445

n = 17, now find d. n = 17, now find d.

an = a1 + (n-1)d

157 = 13 + (17-1)d

157 = 13 + 16d144 = 16d

d = 9

Example 5. Find the first 3 termsif aif a11 = 13, a = 13, ann = 157, and S = 157, and Snn = 1445 = 1445

n = 17, d = 9. n = 17, d = 9.

a2 = a1 + d

a2 = 13 + 9

a2 = 22

a3 = a2 + da3 = 22 + 9a3 = 31

The first 3 terms are 13, 22, and 31.

To simplify writing out series weuse sigma or summation notation.2 + 4 + 6 + 8 + ... + 20 is written

∑ 2n.10

n=1This expression is read

the sum of 2n as n increases from the sum of 2n as n increases from one to ten.one to ten.

Last value of n

First value of n

∑ 2n.10

n=1

Formula for the seq

The variable below the ∑ sigma iscalled the index of summation. Theupper limit is the upper limit of theindex.

∑ 2n.10

n=1

The variable below the ∑ sigma is called the index of summation. The upper limit is the upper limit of the index.

To generate the terms of the series, successively replace the index of summation with consecutive integers as values of n. In this seriesn = 1,2,3, and so on, through 10.

Example 6.Write the terms of

and find the sum.

∑ (2k + 5) 7

k=3

∑ (2k + 5) 7

k=3

= (2•3+5) + (2•4+5) +(2•5+5) + (2•6+5) +(2•7+5)

= 11+13+15+17+19= 75