5.1.5 arithmetic sequences and sums

Arithmetic Sequences

A sequence a1, a2 , a3 , … is an arithmetic sequence

if an = d*n + c, i.e. it is defined by a linear formula.




Example A. The sequence of odd numbers

a1= 1, a2= 3, a3= 5, a4= 7, …

is an arithmetic sequence because an = 2n – 1.





a1= 1, a2= 3, a3= 5, a4= 7, …


Fact: If a1, a2 , a3 , …is an arithmetic sequence and that

an = d*n + c then the difference between any two

neighboring terms is d, i.e. ak+1 – ak = d.





a1= 1, a2= 3, a3= 5, a4= 7, …






In example A, 3 – 1 = 5 – 3 = 7 – 5 = 2 = d.




a1= 1, a2= 3, a3= 5, a4= 7, …






The following theorem gives the converse of the above fact

and the main formula for arithmetic sequences.

In example A, 3 – 1 = 5 – 3 = 7 – 5 = 2 = d.




a1= 1, a2= 3, a3= 5, a4= 7, …






Theorem: If a1, a2 , a3 , …an is a sequence such that

an+1 – an = d for all n, then a1, a2, a3,… is an arithmetic

sequence



In example A, 3 – 1 = 5 – 3 = 7 – 5 = 2 = d.




a1= 1, a2= 3, a3= 5, a4= 7, …








sequence and the formula for the sequence is

an = d(n – 1) + a1.



In example A, 3 – 1 = 5 – 3 = 7 – 5 = 2 = d.




a1= 1, a2= 3, a3= 5, a4= 7, …








sequence and the formula for the sequence is

an = d(n – 1) + a1.

This is the general formula of arithmetic sequences.



In example A, 3 – 1 = 5 – 3 = 7 – 5 = 2 = d.

Arithmetic Sequences Given the description of an arithmetic sequence, we use the

general formula to find the specific formula for that sequence.

Example B. Given the sequence 2, 5, 8, 11, …

a. Verify it is an arithmetic sequence.





It's arithmetic because 5 – 2 = 8 – 5 = 11 – 8 = … = 3 = d.






b. Find the (specific) formula that represents this sequence.







Plug a1 = 2 and d = 3, into the general formula

an = d(n – 1) + a1








an = d(n – 1) + a1

we get

an = 3(n – 1) + 2








an = d(n – 1) + a1

we get

an = 3(n – 1) + 2

an = 3n – 3 + 2








an = d(n – 1) + a1

we get

an = 3(n – 1) + 2

an = 3n – 3 + 2

an = 3n – 1 the specific formula.








an = d(n – 1) + a1

we get

an = 3(n – 1) + 2

an = 3n – 3 + 2


c. Find a1000.








an = d(n – 1) + a1

we get

an = 3(n – 1) + 2

an = 3n – 3 + 2


c. Find a1000.

Set n = 1000 in the specific formula,








an = d(n – 1) + a1

we get

an = 3(n – 1) + 2

an = 3n – 3 + 2


c. Find a1000.

Set n = 1000 in the specific formula, we get

a1000 = 3(1000) – 1 = 2999.



Arithmetic Sequences To use the arithmetic general formula to find the specific

formula, we need the first term a1 and the difference d.



Example C. Given a1, a2 , a3 , …an arithmetic sequence with

d = -4 and a6 = 5, find a1, the specific formula and a1000.



Set d = –4 in the general formula an = d(n – 1) + a1,





Set d = –4 in the general formula an = d(n – 1) + a1, we get

an = –4(n – 1) + a1.






an = –4(n – 1) + a1.

Set n = 6 in this formula,






an = –4(n – 1) + a1.

Set n = 6 in this formula, we get

a6 = -4(6 – 1) + a1 = 5






an = –4(n – 1) + a1.


a6 = -4(6 – 1) + a1 = 5

-20 + a1 = 5






an = –4(n – 1) + a1.


a6 = -4(6 – 1) + a1 = 5

-20 + a1 = 5

a1 = 25






an = –4(n – 1) + a1.


a6 = -4(6 – 1) + a1 = 5

-20 + a1 = 5

a1 = 25

To find the specific formula , set 25 for a1 in an = -4(n – 1) + a1






an = –4(n – 1) + a1.


a6 = -4(6 – 1) + a1 = 5

-20 + a1 = 5

a1 = 25


an = -4(n – 1) + 25






an = –4(n – 1) + a1.


a6 = -4(6 – 1) + a1 = 5

-20 + a1 = 5

a1 = 25


an = -4(n – 1) + 25

an = -4n + 4 + 25






an = –4(n – 1) + a1.


a6 = -4(6 – 1) + a1 = 5

-20 + a1 = 5

a1 = 25


an = -4(n – 1) + 25

an = -4n + 4 + 25

an = -4n + 29






an = –4(n – 1) + a1.


a6 = -4(6 – 1) + a1 = 5

-20 + a1 = 5

a1 = 25


an = -4(n – 1) + 25

an = -4n + 4 + 25

an = -4n + 29



To find a1000, set n = 1000 in the specific formula




an = –4(n – 1) + a1.


a6 = -4(6 – 1) + a1 = 5

-20 + a1 = 5

a1 = 25


an = -4(n – 1) + 25

an = -4n + 4 + 25

an = -4n + 29



To find a1000, set n = 1000 in the specific formula

a1000 = –4(1000) + 29 = –3971

Example D. Given that a1, a2 , a3 , …is an arithmetic sequence

with a3 = -3 and a9 = 39, find d, a1 and the specific formula.




Set n = 3 and n = 9 in the general arithmetic formula

an = d(n – 1) + a1,





an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3





an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3





an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


a9 = d(9 – 1) + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3

Subtract these equations:

8d + a1 = 39

) 2d + a1 = -3


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


8d + a1 = 39

) 2d + a1 = -3

6d = 42


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


8d + a1 = 39

) 2d + a1 = -3

6d = 42

d = 7


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


8d + a1 = 39

) 2d + a1 = -3

6d = 42

d = 7

Put d = 7 into 2d + a1 = -3,


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


8d + a1 = 39

) 2d + a1 = -3

6d = 42

d = 7

Put d = 7 into 2d + a1 = -3,

2(7) + a1 = -3


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


8d + a1 = 39

) 2d + a1 = -3

6d = 42

d = 7

Put d = 7 into 2d + a1 = -3,

2(7) + a1 = -3

14 + a1 = -3


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


8d + a1 = 39

) 2d + a1 = -3

6d = 42

d = 7

Put d = 7 into 2d + a1 = -3,

2(7) + a1 = -3

14 + a1 = -3

a1 = -17


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


8d + a1 = 39

) 2d + a1 = -3

6d = 42

d = 7

Put d = 7 into 2d + a1 = -3,

2(7) + a1 = -3

14 + a1 = -3

a1 = -17

Hence the specific formula is an = 7(n – 1) – 17


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39




an = d(n – 1) + a1, we get

a3 = d(3 – 1) + a1 = -3

2d + a1 = -3


8d + a1 = 39

) 2d + a1 = -3

6d = 42

d = 7

Put d = 7 into 2d + a1 = -3,

2(7) + a1 = -3

14 + a1 = -3

a1 = -17

Hence the specific formula is an = 7(n – 1) – 17

or an = 7n – 24.


a9 = d(9 – 1) + a1 = 39

8d + a1 = 39

Given that a1, a2 , a3 , …an an arithmetic sequence, then

a1+ a2 + a3 + … + an = nTailHead +

2( )

Sums of Arithmetic Sequences



2( )

Head Tail




2( )

ana1 +

2( )= n

Head Tail




2( )

ana1 +

2( )= n

Head Tail

Example E.

a. Given the arithmetic sequence a1= 4, 7, 10, … , and

an = 67. What is n?




2( )

ana1 +

2( )= n

Head Tail

Example E.


an = 67. What is n?

We need the specific formula.




2( )

ana1 +

2( )= n

Head Tail

Example E.


an = 67. What is n?

We need the specific formula. Find d = 7 – 4 = 3.




2( )

ana1 +

2( )= n

Head Tail

Example E.


an = 67. What is n?


Therefore the specific formula is

an = 3(n – 1) + 4




2( )

ana1 +

2( )= n

Head Tail

Example E.


an = 67. What is n?



an = 3(n – 1) + 4

an = 3n + 1.




2( )

ana1 +

2( )= n

Head Tail

Example E.


an = 67. What is n?



an = 3(n – 1) + 4

an = 3n + 1.


If an = 67 = 3n + 1,



2( )

ana1 +

2( )= n

Head Tail

Example E.


an = 67. What is n?



an = 3(n – 1) + 4

an = 3n + 1.


If an = 67 = 3n + 1, then

66 = 3n



2( )

ana1 +

2( )= n

Head Tail

Example E.


an = 67. What is n?



an = 3(n – 1) + 4

an = 3n + 1.


If an = 67 = 3n + 1, then

66 = 3n

or 22 = n

b. Find the sum 4 + 7 + 10 +…+ 67


b. Find the sum 4 + 7 + 10 +…+ 67

a1 = 4, and a22 = 67 with n = 22,


b. Find the sum 4 + 7 + 10 +…+ 67

a1 = 4, and a22 = 67 with n = 22, so the sum

4 + 7 + 10 +…+ 67 = 22 4 + 67

2( )


b. Find the sum 4 + 7 + 10 +…+ 67


4 + 7 + 10 +…+ 67 = 22 4 + 67

2( )

11


b. Find the sum 4 + 7 + 10 +…+ 67


4 + 7 + 10 +…+ 67 = 22 = 11(71) = 7814 + 67

2( )

11


b. Find the sum 4 + 7 + 10 +…+ 67


4 + 7 + 10 +…+ 67 = 22 = 11(71) = 7814 + 67

2( )

11


ana1 +

2( ) =

Formulas for the Arithmetic Sums

The sum Sn of the first n terms of an arithmetic sequence

a1, a2 , a3 , …an, i.e.

a1+ a2 + a3 + … + an = Sn= n2a1 + (n –1)d

2( ) n

b. Find the sum 4 + 7 + 10 +…+ 67


4 + 7 + 10 +…+ 67 = 22 = 11(71) = 7814 + 67

2( )

11


ana1 +

2( ) =



a1, a2 , a3 , …an, i.e.

a1+ a2 + a3 + … + an = Sn= n2a1 + (n –1)d

2( ) n

Example F.

a. How many bricks are

there as shown

if there are 100

layers of bricks

continuing in the same pattern?

b. Find the sum 4 + 7 + 10 +…+ 67


4 + 7 + 10 +…+ 67 = 22 = 11(71) = 7814 + 67

2( )

11


ana1 +

2( ) =



a1, a2 , a3 , …an, i.e.

a1+ a2 + a3 + … + an = Sn= n2a1 + (n –1)d

2( ) n

Example F.


there as shown

if there are 100

layers of bricks


The 1st layer has 3 = 1 x 3 bricks the 2nd layer has 6 = 2 x 3

bricks, etc..,

b. Find the sum 4 + 7 + 10 +…+ 67


4 + 7 + 10 +…+ 67 = 22 = 11(71) = 7814 + 67

2( )

11


ana1 +

2( ) =



a1, a2 , a3 , …an, i.e.

a1+ a2 + a3 + … + an = Sn= n2a1 + (n –1)d

2( ) n

Example F.


there as shown

if there are 100

layers of bricks


The 1st layer has 3 = 1 x 3 bricks the 2nd layer has 6 = 2 x 3

bricks, etc.., hence the 100th layer has 100 x 3 = 300 bricks.


The 1st layer has 3 bricks



The last layer has 300 bricks



n = 100 layers



The sum 3 + 6 + 9 + .. + 300 is arithmetic.


n = 100 layers



3 + 3002

( )


Hence the total number of bricks is


n = 100 layers

100



3 + 3002

( )


Hence the total number of bricks is


n = 100 layers

100


= 50 x 303

= 15150


2. –2, –5, –8, –11,..1. 2, 5, 8, 11,..

4. –12, –5, 2, 9,..3. 6, 2, –2, –6,..

6. 23, 37, 51,..5. –12, –25, –38,..

8. –17, .., a7 = 13, .. 7. 18, .., a4 = –12, ..

10. a12 = 43, d = 59. a4 = –12, d = 6

12. a42 = 125, d = –511. a8 = 21.3, d = –0.4

14. a22 = 25, a42 = 12513. a6 = 21, a17 = 54

16. a17 = 25, a42 = 12515. a3 = –4, a17 = –11,

Exercise A. For each arithmetic sequence below

a. find the first term a1 and the difference d

b. find a specific formula for an and a100

c. find the sum ann=1

100

B. For each sum below, find the specific formula of

the terms, write the sum in the notation,

then find the sum.

1. – 4 – 1 + 2 +…+ 302

Sum of Arithmetic Sequences

2. – 4 – 9 – 14 … – 1999

3. 27 + 24 + 21 … – 1992

4. 3 + 9 + 15 … + 111,111,111

5. We see that it’s possible to add infinitely many

numbers and obtain a finite sum.

For example ½ + ¼ + 1/8 + 1/16... = 1.

Give a reason why the sum of infinitely many terms

of an arithmetic sequence is never finite,

except for 0 + 0 + 0 + 0..= 0.


1. a1 = 2

d = 3

an = 3n – 1

a100 = 299

an = 15 050

(Answers to the odd problems) Exercise A.

n=1

100

3. a1 = 6

d = – 4

an = – 4n + 10

a100 = – 390

an = – 19 200n=1

100

5. a1 = – 12

d = –13

an = – 13n + 1

a100 = – 129

an = – 65 550n=1

100

7. a1 = 18

d = – 10

an = – 10n +28

a100 = – 972

an = – 47 700n=1

100

9. a1 = –30

d = 6

an = 6n – 36

a100 = 564

an = 26 700n=1

100

11. a1 = 24.1

d = –0.4

an = –0.4n + 24.5

a100 = –15.5

an = 430n=1

100


13. a1 = 6

d = 3

an = 3n + 3

a100 = 303

an = 15 450n=1

100

15. a1 = –3

d = – 0.5

an = – 0.5n – 2.5

a100 = –52.5

an = –2 775n=1

100

Exercise B.

1. – 4 – 1 + 2 +…+ 302 = 3n – 7 = 15 347

3. 27 + 24 + 21 … – 1992 = –3n + 30 = –662 205

n=1

103

n=1

674

5.1.5 arithmetic sequences and sums

Education