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

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