Unformatted text preview: Section 11.3: Geometric Sequences Instructor: Ms. Hoa Nguyen (nguyenescs.fsu.edu) 1 Geometric Sequences A geometric sequence {an} satisﬁes the following:
. (1,1
. an = art—17' Where a1 is the ﬁrst term, 1" 7E 0 is the common ratio. SO, 2. Find the nth term of a geometric sequence 01
(12 = 0417‘
as = a2?" = alrz a," = an_1r = awn—1
So, the n—th term of a geometric sequence is: [an = alrn‘l. 
Example 1:
[11.3.1aP'fIThe 6“ term of a geometric sequence with ﬁrst term :11, =
—4 and common ratio r = a: is \ n —
0 ml 6 I
2 0i : 0\. I"
_..1. Q 1
o a n 5
e m: (A) (i _ , J.
2  8 Example 2: [11.3.1bP’i‘]Then"‘ term. of a, geometric sequence with ﬁrst term a] = 6
and common ratio r 2: w2 is o __,(—12)n1 , 0 ‘None ofthese a”! : 1,31 r‘  Z
a 6(—2)““1 :1;  o 12»! 2 an: al F‘” 0 “mar” 3‘“'
o —6(2)"~1 2 02 ( 2) Example 3: [11.3.2aPT]If a geometric sequence has a“; = ‘13 and a.” a 12, what is thegcomsénon. ratio? CL‘H Que . r» 12 => Y“:
Q. 13'
$1
46 _
None of these 1:
12 @960 3 Summing the terms of geometric sequence Let 8,609 = '1, ..., n) be the sum of the ﬁrst 1:: terms of a geometric sequence {an}. Then 81 = an S; = a1+a2=a1+(a1r)=a1(1+r) ,. S3 = a1 + a2 + a3 = a1+(a1r)+(a1r2) = a1(1 +r+ r2) 3,, = _ a1 + a2_+ a3 + + an =, a1 +(a1r)+(a17‘2) +1.. +[017‘n—1]
1— n = a1(1+r+r2+...+r"‘1)=a1 T
1 — 'r
if r 7E 1.
So, the sum of the ﬁrst n terms of a geometric sequence is Sn = a1 111’: if r 7E 1. 4 Geometric series A geometric series is a series of the form: 00
z» 00
a1+a1r+a1r2+... = E alrk=a1
16:0 16:0 In other words, the geometric series is a sum of inﬁnite number of terms of a geometric sequence.
If r < 1, then .900 : alx+ a1?" + alr2 + = inﬁ
Example 4: at 04
u u
[11.3.3bP’I‘]Find the sum of the inﬁnite geometric series 1 + § + 33 +
‘."+(§)n—l,+,,, a
0 :14, a : l i" ; j = £ ‘7 i =“’
e 6 ‘  — r ».«:ql 3 '
G 5 w, km 1: y i
00 ‘ i A__ 4_ Z 
Example 5 — ._
3 3 Alan 0,,— 11.3.3cPT1Find the sum of the alternating inﬁnite geometric series er, dz \ ‘ a1: '. “all: all" .2) _——%—:—.\.“_Ll'_
0‘41 0 5% _ 4 ‘ ‘ 3 4 8 i
6 l1? a J'— 1 : ,1. l = l _ 2 Li
0 § "L  r" l“ L} é I. 3‘ g
9  \
5 Write Repeating Decimal as a Fraction ,6 Example. Repeating Decimal Write 0.898989 «  as a fraction.
Solution. 0.898989 ~   = $693 + + + . .. =§§5u+ﬁﬁ+~f§g+mi
=¥&r:§g
=3—3— So,a1=%’5,r=ﬁ,5m=g.
Example 6:
[Il.3.3tiPT]If the repeating decimal 0.135135135 '  is written as {if in Where m and. n age integers, then m =
a $15 0. :55 135+.2»;122,5 s. 5 ‘ a1: E5. r»: ) goo: mtg : 5 e 139 ‘ :9 , [000 ’ " tmg 3 3 3 37 k 9“
Example 7: => W1 3 [11.3.3ePTllf the repeating decimal 0.444444  » i is written as in re
‘duced form where. m and n are integers, then m = 0‘ 44 e 13 OuﬁliLHHLQ? 5 =‘\_ _ m _ Lt . 4 4&1 7 l~' _3,‘ gou— T‘ _. E.
=7 m= Li ...
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 Spring '11
 Nuegyen

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