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8/10/2019 Arithmetic Sequences and Examples

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8/10/2019 Arithmetic Sequences and Examples

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A sequence in which a constant (d)can be added to each term to getthe next term is called an

Arithmetic Sequence.

The constant (d) is called theCommon Difference.

To find the common difference (d),subtract any term from one thatfollows it.

2 5 8 11 14

3 3 3 3

t1 t2 t3 t4 t5

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Find the first term and the

common difference of each

arithmetic sequence.

1.) 4,9,14,19,24First term (a): 4Common difference (d): 2 1a a = 9 4 = 5

2.) 34, 2!, 2",13,#, 1, 8,.... First term (a): 34Common difference (d): $!

%& C'&F*: ALWAYSCHECK TO MAKE

SURE THE DIFFERENCE IS THE SAME

BETWEEN EACH TERM !

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Now you try!

Find the first term and the common difference of each of

these arithmetic sequences.

+) 11, 23, 35, 4!, .

a) 1, $4, $9, $14, .

d) s$4, 3s$3, 5s$2, !s$1, ..

c) 3x$8, x$8, $x$8, $3x$8

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+) 11, 23, 35, 4!, .

a) 1, $4, $9, $14, .

d) s$4, 3s$3, 5s$2, !s$1, ..

c) 3x$8, x$8, $x$8, $3x$8

Answers with solutions

a = 1 and

d = a2 a1 = 4 1 = 5

a = 11 and

d = a2 a1 = 23 11 = 12

a = 3x$8 and

d = a2 a1 = x " #3x "$ = 2x

a =s$ 4 and

d = a2 a1 = 3s 3 (s 4) = 2s % 1

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The first term of an arithmetic sequenceis (a) . We add (d) to get the next term.There is a attern, therefore there is aformula we can use to gi!e use any termthat we need without listing the wholesequence .

3, !, 11, 15, . -e no/ a = 3 and d = 4

t1= a = 3

t2= a0d = 304 = !t3= a0d0d = a02d = 302(4) = 11

t4 = a0d0d0d = a03d = 303(4) = 15

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The first term of an arithmetic sequenceis (a) . We add (d) to get the next term.There is a attern, therefore there is aformula we can use to gi!e use any termthat we need without listing the wholesequence .

The nth term of an arithmetic sequence is gi!en by"

he ast in the

sequenceor the

ou are ooin6 for

First

term

he 7osition

the term is in he common

difference

tn= a % #n 1$ d

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#ind the $%thterm of thearithmetic sequence

%, &, $', $,

(14 1)+ 44 (13)3= +

4 39= +

43=

tn= a % #n 1$ d

t14 = 3 ou areooin6 for

t&' t'()!

The 14thterm in this sequence

is the number 43!

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Now you try!Find the 1"th and 25thterm 6ien the foo/in6 information. ae sure to derie

the 6enera formua first and then ist ehat ou hae +een 7roided.

+) x%1",x%!,x%4,x01, .

a) 1, !, 13, 19 .

d) he second term is 8 and the common difference is 3

c) he first term is 3 and the common difference is $21

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+)x%1",x%!,x%4,x01,.

a) 1, !, 13, 19 . .

d) he second term is 8and the common

difference is 3

c) he first term is 3 and thecommon difference is $21

Answers with solutionsa = 1 and d = a2 a1 = * 1 = +

tn=a%#n1$d = 1 % #n1$ + = 1%+n+ S, tn= 6n-5

t1- = +#1-$ 5 = 55

t25= +#25$5 = 145

a =x+10 and d = a2 a1 = ;0!$(;01") = $3

tn=a%#n$1)d = ;01"% #n1$#3$ = ;01"$3n03 S, tn=x-3n+13

t1- = ; $3(1")013 =x - 17

t25= ; $3(25)013 =x - 62

a = 3 and d = -21

tn=a%#n$1)d = 3 % #n$1)21 = 3$21n021 S, tn= 24-21n

t1- = 2421#1-$= $18# t25= 24$21(25) = $5"1

a = 8 $ 3 = 5 and d = 3

tn=a%#n$1)d = 5 0 (n$1) 3 = 503n$3

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#ind the $%thterm of the arithmeticsequence with first term of * and

the common difference is +.

(14 1)+

tn= a % #n 1$ dt14 =

You are looin "or the

term! List which variabes

r"m the #enera term are

$r"vi%e%!

The 14thterm in this sequence

is the number -#3!

a = 5 and d = +

5 -6

= 5 + (13) * -6

= 5 + -78= -73

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-n the arithmetic sequence%,&,$',$,, which term has a!alue of '$

3"1 4 ( 1)3n= + 3"1 4 3 3n= +

3"1 1 3n= +

3"" 3n=1"" n=

tn= a % #n 1$ d

ou areooin6

for n!

The 1$$thterm in this

sequence is 3$1!

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-n an arithmetic sequence, term $' is andterm // is +. What are the first four terms ofthe sequence

The sequence is %$& 5#& 54& 51& ''(

se /hat ou no/t1"=33

t22= $3

tn= a % #n 1$ d

For term 1": 33= a 0 9d

tn= a % #n 1$ d

For term 22: $3= a 0 21d

> /o equations ou can soe

33 = a09d

$3 = a021d

% eimination $3# = 12d

$3 = d