92 arithmetic sequences

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.


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.


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.Fact: If a1, a2 , a3 , …is an arithmetic sequence and that an = d*n + c then the difference between any two terms is d, i.e. ak+1 – ak = d.




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



The following theorem gives the converse of the above factand the main formula for arithmetic sequences.

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



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.



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 and the formula for the sequence is an = d(n – 1) + a1.


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



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 and the formula for the sequence is an = d(n – 1) + a1. This is the general formula of arithemetic sequences.


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

Arithmetic Sequences Given the description of a 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 represents this sequence.





Plug a1 = 2 and d = 3, into the general formula an = d(n – 1) + a1






we get an = 3(n – 1) + 2






we get an = 3(n – 1) + 2 an = 3n – 3 + 2






we get an = 3(n – 1) + 2 an = 3n – 3 + 2 an = 3n – 1 the specific formula.







c. Find a1000.







c. Find a1000. Set n = 1000 in the specific formula,







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.

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,

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, we get an = –4(n – 1) + a1.



Set d = –4 in the general formula an = d(n – 1) + a1, we get an = –4(n – 1) + a1. Set n = 6 in this formula,



Set d = –4 in the general formula an = d(n – 1) + a1, we get an = –4(n – 1) + a1. Set n = 6 in this formula, we geta6 = -4(6 – 1) + a1 = 5



Set d = –4 in the general formula an = d(n – 1) + a1, we get an = –4(n – 1) + a1. Set n = 6 in this formula, we geta6 = -4(6 – 1) + a1 = 5 -20 + a1 = 5



Set d = –4 in the general formula an = d(n – 1) + a1, we get an = –4(n – 1) + a1. Set n = 6 in this formula, we geta6 = -4(6 – 1) + a1 = 5 -20 + a1 = 5 a1 = 25



Set d = –4 in the general formula an = d(n – 1) + a1, we get an = –4(n – 1) + a1. Set n = 6 in this formula, we geta6 = -4(6 – 1) + a1 = 5 -20 + a1 = 5 a1 = 25To find the specific formula , set 25 for a1 in an = -4(n – 1) + a1




an = -4(n – 1) + 25




an = -4(n – 1) + 25 an = -4n + 4 + 25




an = -4(n – 1) + 25 an = -4n + 4 + 25 an = -4n + 29




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) + 25 an = -4n + 4 + 25 an = -4n + 29


To find a1000, set n = 1000 in the specific formulaa1000 = –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 formulaan = d(n – 1) + a1,



Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3



Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3 2d + a1 = -3





a9 = d(9 – 1) + a1 = 39




a9 = d(9 – 1) + a1 = 39 8d + a1 = 39


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


Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3 2d + a1 = -3 Subtract these equations: 8d + a1 = 39 ) 2d + a1 = -3 6d = 42


a9 = d(9 – 1) + a1 = 39 8d + a1 = 39


Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3 2d + a1 = -3 Subtract these equations: 8d + a1 = 39 ) 2d + a1 = -3 6d = 42 d = 7


a9 = d(9 – 1) + a1 = 39 8d + a1 = 39


Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3 2d + a1 = -3 Subtract these equations: 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


Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3 2d + a1 = -3 Subtract these equations: 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


Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3 2d + a1 = -3 Subtract these equations: 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


Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3 2d + a1 = -3 Subtract these equations: 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


Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3 2d + a1 = -3 Subtract these equations: 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 = -17Hence the specific formula is an = 7(n – 1) – 17


a9 = d(9 – 1) + a1 = 39 8d + a1 = 39


Set n = 3 and n = 9 in the general arithmetic formulaan = d(n – 1) + a1, we get a3 = d(3 – 1) + a1 = -3 2d + a1 = -3 Subtract these equations: 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 = -17Hence 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 = n

TailHead + 2( )

Sum of Arithmetic Sequences


a1+ a2 + a3 + … + an = n

TailHead + 2( )

Head Tail



a1+ a2 + a3 + … + an = n

TailHead + 2( )

an a1 + 2( )= n

Head Tail



a1+ a2 + a3 + … + an = n

TailHead + 2( )

an a1 + 2( )= n

Head Tail

Example E. a. Given the arithmetic sequence a1= 4, 7, 10, … , and an = 67. What is n?



a1+ a2 + a3 + … + an = n

TailHead + 2( )

an a1 + 2( )= n

Head Tail


We need the specific formula.



a1+ a2 + a3 + … + an = n

TailHead + 2( )

an a1 + 2( )= n

Head Tail


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



a1+ a2 + a3 + … + an = n

TailHead + 2( )

an a1 + 2( )= n

Head Tail


We need the specific formula. Find d = 7 – 4 = 3. Therefore the specific formula is an = 3(n – 1) + 4



a1+ a2 + a3 + … + an = n

TailHead + 2( )

an a1 + 2( )= n

Head Tail


We need the specific formula. Find d = 7 – 4 = 3. Therefore the specific formula is an = 3(n – 1) + 4 an = 3n + 1.



a1+ a2 + a3 + … + an = n

TailHead + 2( )

an a1 + 2( )= n

Head Tail




If an = 67 = 3n + 1,


a1+ a2 + a3 + … + an = n

TailHead + 2( )

an a1 + 2( )= n

Head Tail




If an = 67 = 3n + 1, then 66 = 3n


a1+ a2 + a3 + … + an = n

TailHead + 2( )

an a1 + 2( )= n

Head Tail




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) 4 + 67

2( )11


b. Find the sum 4 + 7 + 10 +…+ 67


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

2( )11


92 arithmetic sequences

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