Chapter 11
Sequences and Series
Submitted to: Rajeev
Sir
Submitted by : Mukul
Sharma
Class : 11th
– A
Section 11-1
Arithmetic
Sequences
Arithmetic Sequences
Every day a radio station
asks a question for a prize
of $150. If the 5th caller
does not answer correctly,
the prize money increased
by $150 each day until
someone correctly answers
their question.
Arithmetic Sequences
Make a list of the prize
amounts for a week
(Mon - Fri) if the contest
starts on Monday and no one
answers correctly all week.
Arithmetic Sequences
• Monday : $150
• Tuesday: $300
• Wednesday: $450
• Thursday: $600
• Friday: $750
Arithmetic Sequences
• These prize amounts form a
sequence, more specifically
each amount is a term in an
arithmetic sequence. To
find the next term we just
add $150.
Definitions
• Sequence: a list of numbers
in a specific order.
• Term: each number in a
sequence
Definitions
• Arithmetic Sequence: a
sequence in which each term
after the first term is
found by adding a constant,
called the common
difference (d), to the
previous term.
Explanations
• 150, 300, 450, 600, 750…
• The first term of our
sequence is 150, we denote
the first term as a1.
• What is a2?
• a2 : 300 (a2 represents the
2nd term in our sequence)
Explanations
• a3 = ? a4 = ? a5 = ?
• a3 : 450 a4 : 600 a5 : 750
• an represents a general term
(nth term) where n can be
any number.
Explanations
• Sequences can continue
forever. We can calculate as
many terms as we want as
long as we know the common
difference in the sequence.
Explanations
• Find the next three terms in
the sequence:
2, 5, 8, 11, 14, __, __, __
• 2, 5, 8, 11, 14, 17, 20, 23
• The common difference is?
• 3!!!
Explanations
• To find the common
difference (d), just subtract
any term from the term that
follows it.
• FYI: Common differences
can be negative.
Formula
• What if I wanted to find the
50th (a50) term of the
sequence 2, 5, 8, 11, 14, …?
Do I really want to add 3
continually until I get there?
• There is a formula for
finding the nth term.
Formula
• Let’s see if we can figure
the formula out on our own.
• a1 = 2, to get a2 I just add 3
once. To get a3 I add 3 to a1
twice. To get a4 I add 3 to a1
three times.
Formula
• What is the relationship
between the term we are
finding and the number of
times I have to add d?
• The number of times I had
to add is one less then the
term I am looking for.
Formula
• So if I wanted to find a50
then how many times would I
have to add 3?
• 49
• If I wanted to find a193 how
many times would I add 3?
• 192
Formula
• So to find a50 I need to take
d, which is 3, and add it to
my a1, which is 2, 49 times.
That’s a lot of adding.
• But if we think back to
elementary school, repetitive
adding is just multiplication.
Formula
• 3 + 3 + 3 + 3 + 3 = 15
• We added five terms of
three, that is the same as
multiplying 5 and 3.
• So to add three forty-nine
times we just multiply 3 and
49.
Formula
• So back to our formula, to
find a50 we start with 2 (a1)
and add 3•49. (3 is d and 49
is one less than the term we
are looking for) So…
• a50 = 2 + 3(49) = 149
Formula
• a50 = 2 + 3(49) using this
formula we can create a
general formula.
• a50 will become an so we can
use it for any term.
• 2 is our a1 and 3 is our d.
Formula
• a50 = 2 + 3(49)
• 49 is one less than the term
we are looking for. So if I
am using n as the term I am
looking for, I multiply d by
n - 1.
Formula
• Thus my formula for finding
any term in an arithmetic
sequence is an = a1 + d(n-1).
• All you need to know to find
any term is the first term in
the sequence (a1) and the
common difference.
Example
• Let’s go back to our first
example about the radio
contest. Suppose no one
correctly answered the
question for 15 days. What
would the prize be on day
16?
Example
• an = a1 + d(n-1)
• We want to find a16. What is
a1? What is d? What is n-1?
• a1 = 150, d = 150, n
-1 = 16 - 1 = 15
• So a16 = 150 + 150(15) =
• $2400
Example
• 17, 10, 3, -4, -11, -18, …
• What is the common
difference?
• Subtract any term from the
term after it.
• -4 - 3 = -7
• d = - 7
Definition
• 17, 10, 3, -4, -11, -18, …
• Arithmetic Means: the
terms between any two
nonconsecutive terms of an
arithmetic sequence.
Arithmetic Means
• 17, 10, 3, -4, -11, -18, …
• Between 10 and -18 there
are three arithmetic means
3, -4, -11.
• Find three arithmetic means
between 8 and 14.
Arithmetic Means
• So our sequence must look
like 8, __, __, __, 14.
• In order to find the means
we need to know the common
difference. We can use our
formula to find it.
Arithmetic Means
• 8, __, __, __, 14
• a1 = 8, a5 = 14, & n = 5
• 14 = 8 + d(5 - 1)
• 14 = 8 + d(4) subtract 8
• 6 = 4d divide by 4
• 1.5 = d
Arithmetic Means
• 8, __, __, __, 14 so to find
our means we just add 1.5
starting with 8.
• 8, 9.5, 11, 12.5, 14
Additional Example
• 72 is the __ term of the
sequence -5, 2, 9, …
• We need to find ‘n’ which is
the term number.
• 72 is an, -5 is a1, and 7 is d.
Plug it in.
Additional Example
• 72 = -5 + 7(n - 1)
• 72 = -5 + 7n - 7
• 72 = -12 + 7n
• 84 = 7n
• n = 12
• 72 is the 12th term.
Section 11-2
Arithmetic
Series
Arithmetic Series
• The African-American
celebration of Kwanzaa
involves the lighting of
candles every night for
seven nights. The first night
one candle is lit and blown
out.
Arithmetic Series
• The second night a new
candle and the candle from
the first night are lit and
blown out. The third night a
new candle and the two
candles from the second
night are lit and blown out.
Arithmetic Series
• This process continues for
the seven nights.
• We want to know the total
number of lightings during
the seven nights of
celebration.
Arithmetic Series
• The first night one candle
was lit, the 2nd night two
candles were lit, the 3rd
night 3 candles were lit, etc.
• So to find the total number
of lightings we would add:
1 + 2 + 3 + 4 + 5 + 6 + 7
Arithmetic Series
• 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28
• Series: the sum of the terms
in a sequence.
• Arithmetic Series: the sum
of the terms in an
arithmetic sequence.
Arithmetic Series
• Arithmetic sequence:
2, 4, 6, 8, 10
• Corresponding arith. series:
2 + 4 + 6 + 8 + 10
• Arith. Sequence: -8, -3, 2, 7
• Arith. Series: -8 + -3 + 2 + 7
Arithmetic Series
• Sn is the symbol used to
represent the first ‘n’ terms
of a series.
• Given the sequence 1, 11, 21,
31, 41, 51, 61, 71, … find S4
• We add the first four terms
1 + 11 + 21 + 31 = 64
Arithmetic Series
• Find S8 of the arithmetic
sequence 1, 2, 3, 4, 5, 6, 7, 8,
9, 10, …
• 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 =
• 36
Arithmetic Series
• What if we wanted to find
S100 for the sequence in the
last example. It would be a
pain to have to list all the
terms and try to add them
up.
• Let’s figure out a formula!! :)
Sum of Arithmetic Series
• Let’s find S7 of the sequence
1, 2, 3, 4, 5, 6, 7, 8, 9, …
• If we add S7 in too different
orders we get:
S7 = 1 + 2 + 3 + 4 + 5 + 6 + 7
S7 = 7 + 6 + 5 + 4 + 3 + 2 + 1
2S7 = 8 + 8 + 8 + 8 + 8 + 8 + 8
Sum of Arithmetic Series
S7 = 1 + 2 + 3 + 4 + 5 + 6 + 7
S7 = 7 + 6 + 5 + 4 + 3 + 2 + 1
2S7 = 8 + 8 + 8 + 8 + 8 + 8 + 8
2S7 = 7(8)
S7 =7
/2(8)
7 sums of 8
Sum of Arithmetic Series
• S7 =7
/2(8)
• What do these numbers
mean?
• 7 is n, 8 is the sum of the
first and last term (a1 + an)
• So Sn = n
/2(a1 + an)
Examples
• Sn = n
/2(a1 + an)
• Find the sum of the first 10
terms of the arithmetic
series with a1 = 6 and a10 =51
• S10 = 10/2(6 + 51) = 5(57) =
285
Examples
• Find the sum of the first 50
terms of an arithmetic
series with a1 = 28 and d = -4
• We need to know n, a1, and
a50.
• n= 50, a1 = 28, a50 = ?? We
have to find it.
Examples
• a50 = 28 + -4(50 - 1) =
28 + -4(49) = 28 + -196 =
-168
• So n = 50, a1 = 28, & an =-168
• S50 = (50/2)(28 + -168) = 25(-
140) = -3500
Examples
• To write out a series and
compute a sum can
sometimes be very tedious.
Mathematicians often use
the greek letter sigma &
summation notation to
simplify this task.
Examples
• This means to find the sum
of the sums n + 1 where we
plug in the values 1 - 5 for n
n + 1
n = 1
5
!
last value of n
First value of n
formula used to
find sequence
Examples
• Basically we want to find
(1 + 1) + (2 + 1) + (3 + 1) +
(4 + 1) + (5 + 1) =
• 2 + 3 + 4 + 5 + 6 =
• 20
n + 1
n = 1
5
!
Examples
• So
• Try:
• First we need to plug in the
numbers 2 - 7 for x.
n + 1 = 20
n = 1
5
!
3x - 2
x = 2
7
!
Examples
• [3(2)-2]+[3(3)-2]+[3(4)-2]+
[3(5)-2]+[3(6)-2]+[3(7)-2] =
• (6-2)+(9-2)+(12-2)+(15-2)+
(18-2)+ (21-2) =
• 4 + 7 + 10 + 13 + 17 + 19 = 70
3x - 2
x = 2
7
!
Section 11-3
Geometric
Sequences
GeometricSequence
• What if your pay check
started at $100 a week and
doubled every week. What
would your salary be after
four weeks?
GeometricSequence
• Starting $100.
• After one week - $200
• After two weeks - $400
• After three weeks - $800
• After four weeks - $1600.
• These values form a
geometric sequence.
Geometric Sequence
• Geometric Sequence: a
sequence in which each term
after the first is found by
multiplying the previous term
by a constant value called
the common ratio.
Geometric Sequence
• Find the first five terms of
the geometric sequence with
a1 = -3 and common ratio (r)
of 5.
• -3, -15, -75, -375, -1875
Geometric Sequence
• Find the common ratio of
the sequence 2, -4, 8, -16,
32, …
• To find the common ratio,
divide any term by the
previous term.
• 8 ÷ -4 = -2
• r = -2
Geometric Sequence
• Just like arithmetic
sequences, there is a
formula for finding any given
term in a geometric
sequence. Let’s figure it out
using the pay check example.
Geometric Sequence
• To find the 5th term we look
100 and multiplied it by two
four times.
• Repeated multiplication is
represented using
exponents.
Geometric Sequence
• Basically we will take $100
and multiply it by 24
• a5 = 100•24
= 1600
• A5 is the term we are looking
for, 100 was our a1, 2 is our
common ratio, and 4 is n-1.
Examples
• Thus our formula for finding
any term of a geometric
sequence is an = a1•rn-1
• Find the 10th term of the
geometric sequence with a1 =
2000 and a common ratio of
1
/2.
Examples
• a10 = 2000• (1
/2)9
=
• 2000 • 1
/512 =
• 2000
/512 = 500
/128 = 250
/64 = 125
/32
• Find the next two terms in
the sequence -64, -16, -4 ...
Examples
• -64, -16, -4, __, __
• We need to find the common
ratio so we divide any term
by the previous term.
• -16/-64 = 1/4
• So we multiply by 1/4 to find
the next two terms.
Examples
• -64, -16, -4, -1, -1/4
Geometric Means
• Just like with arithmetic
sequences, the missing terms
between two nonconsecutive
terms in a geometric
sequence are called
geometric means.
Geometric Means
• Looking at the geometric
sequence 3, 12, 48, 192, 768
the geometric means
between 3 and 768 are 12,
48, and 192.
• Find two geometric means
between -5 and 625.
Geometric Means
• -5, __, __, 625
• We need to know the
common ratio. Since we only
know nonconsecutive terms
we will have to use the
formula and work backwards.
Geometric Means
• -5, __, __, 625
• 625 is a4, -5 is a1.
• 625 = -5•r4-1
divide by -5
• -125 = r3
take the cube root
of both sides
• -5 = r
Geometric Means
• -5, __, __, 625
• Now we just need to multiply
by -5 to find the means.
• -5 • -5 = 25
• -5, 25, __, 625
• 25 • -5 = -125
• -5, 25, -125, 625
Section 11-4
Geometric
Series
Geometric Series
• Geometric Series - the sum
of the terms of a geometric
sequence.
• Geo. Sequence: 1, 3, 9, 27, 81
• Geo. Series: 1+3 + 9 + 27 + 81
• What is the sum of the
geometric series?
Geometric Series
• 1 + 3 + 9 + 27 + 81 = 121
• The formula for the sum Sn
of the first n terms of a
geometric series is given by
Sn= 1 - r
a1- a1rn
or Sn= 1 - r
a1(1 - rn
)
Geometric Series
• Find
• You can actually do it two
ways. Let’s use the old way.
• Plug in the numbers 1 - 4 for
n and add.
• [-3(2)1-1
]+[-3(2)2-1
]+[-3(2)3-1
]+ [-
3(2)4-1
]
- 3 (2)n - 1
n = 1
4
!
Geometric Series
• [-3(1)] + [-3(2)] + [-3(4)] +
[-3(8)] =
• -3 + -6 + -12 + -24 = -45
• The other method is to use
the sum of geometric series
formula.
Geometric Series
• use
• a1 = -3, r = 2, n = 4
Sn=
1 - r
a1(1 - rn
)
- 3 (2)n - 1
n = 1
4
!
Geometric Series
• use
• a1 = -3, r = 2, n = 4
•
Sn=
1 - r
a1(1 - rn
)
- 3 (2)n - 1
n = 1
4
!
S4= 1 - 2
- 3 (1 - 24
)
Geometric Series
•
S4= 1 - 2
- 3 (1 - 24
)
S4=
- 1
- 3 (1 - 16)
S4=
- 1
- 3 (- 15)
=
- 1
45 =- 45
THE END

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Ppt on sequences and series by mukul sharma

  • 1. Chapter 11 Sequences and Series Submitted to: Rajeev Sir Submitted by : Mukul Sharma Class : 11th – A
  • 3. Arithmetic Sequences Every day a radio station asks a question for a prize of $150. If the 5th caller does not answer correctly, the prize money increased by $150 each day until someone correctly answers their question.
  • 4. Arithmetic Sequences Make a list of the prize amounts for a week (Mon - Fri) if the contest starts on Monday and no one answers correctly all week.
  • 5. Arithmetic Sequences • Monday : $150 • Tuesday: $300 • Wednesday: $450 • Thursday: $600 • Friday: $750
  • 6. Arithmetic Sequences • These prize amounts form a sequence, more specifically each amount is a term in an arithmetic sequence. To find the next term we just add $150.
  • 7. Definitions • Sequence: a list of numbers in a specific order. • Term: each number in a sequence
  • 8. Definitions • Arithmetic Sequence: a sequence in which each term after the first term is found by adding a constant, called the common difference (d), to the previous term.
  • 9. Explanations • 150, 300, 450, 600, 750… • The first term of our sequence is 150, we denote the first term as a1. • What is a2? • a2 : 300 (a2 represents the 2nd term in our sequence)
  • 10. Explanations • a3 = ? a4 = ? a5 = ? • a3 : 450 a4 : 600 a5 : 750 • an represents a general term (nth term) where n can be any number.
  • 11. Explanations • Sequences can continue forever. We can calculate as many terms as we want as long as we know the common difference in the sequence.
  • 12. Explanations • Find the next three terms in the sequence: 2, 5, 8, 11, 14, __, __, __ • 2, 5, 8, 11, 14, 17, 20, 23 • The common difference is? • 3!!!
  • 13. Explanations • To find the common difference (d), just subtract any term from the term that follows it. • FYI: Common differences can be negative.
  • 14. Formula • What if I wanted to find the 50th (a50) term of the sequence 2, 5, 8, 11, 14, …? Do I really want to add 3 continually until I get there? • There is a formula for finding the nth term.
  • 15. Formula • Let’s see if we can figure the formula out on our own. • a1 = 2, to get a2 I just add 3 once. To get a3 I add 3 to a1 twice. To get a4 I add 3 to a1 three times.
  • 16. Formula • What is the relationship between the term we are finding and the number of times I have to add d? • The number of times I had to add is one less then the term I am looking for.
  • 17. Formula • So if I wanted to find a50 then how many times would I have to add 3? • 49 • If I wanted to find a193 how many times would I add 3? • 192
  • 18. Formula • So to find a50 I need to take d, which is 3, and add it to my a1, which is 2, 49 times. That’s a lot of adding. • But if we think back to elementary school, repetitive adding is just multiplication.
  • 19. Formula • 3 + 3 + 3 + 3 + 3 = 15 • We added five terms of three, that is the same as multiplying 5 and 3. • So to add three forty-nine times we just multiply 3 and 49.
  • 20. Formula • So back to our formula, to find a50 we start with 2 (a1) and add 3•49. (3 is d and 49 is one less than the term we are looking for) So… • a50 = 2 + 3(49) = 149
  • 21. Formula • a50 = 2 + 3(49) using this formula we can create a general formula. • a50 will become an so we can use it for any term. • 2 is our a1 and 3 is our d.
  • 22. Formula • a50 = 2 + 3(49) • 49 is one less than the term we are looking for. So if I am using n as the term I am looking for, I multiply d by n - 1.
  • 23. Formula • Thus my formula for finding any term in an arithmetic sequence is an = a1 + d(n-1). • All you need to know to find any term is the first term in the sequence (a1) and the common difference.
  • 24. Example • Let’s go back to our first example about the radio contest. Suppose no one correctly answered the question for 15 days. What would the prize be on day 16?
  • 25. Example • an = a1 + d(n-1) • We want to find a16. What is a1? What is d? What is n-1? • a1 = 150, d = 150, n -1 = 16 - 1 = 15 • So a16 = 150 + 150(15) = • $2400
  • 26. Example • 17, 10, 3, -4, -11, -18, … • What is the common difference? • Subtract any term from the term after it. • -4 - 3 = -7 • d = - 7
  • 27. Definition • 17, 10, 3, -4, -11, -18, … • Arithmetic Means: the terms between any two nonconsecutive terms of an arithmetic sequence.
  • 28. Arithmetic Means • 17, 10, 3, -4, -11, -18, … • Between 10 and -18 there are three arithmetic means 3, -4, -11. • Find three arithmetic means between 8 and 14.
  • 29. Arithmetic Means • So our sequence must look like 8, __, __, __, 14. • In order to find the means we need to know the common difference. We can use our formula to find it.
  • 30. Arithmetic Means • 8, __, __, __, 14 • a1 = 8, a5 = 14, & n = 5 • 14 = 8 + d(5 - 1) • 14 = 8 + d(4) subtract 8 • 6 = 4d divide by 4 • 1.5 = d
  • 31. Arithmetic Means • 8, __, __, __, 14 so to find our means we just add 1.5 starting with 8. • 8, 9.5, 11, 12.5, 14
  • 32. Additional Example • 72 is the __ term of the sequence -5, 2, 9, … • We need to find ‘n’ which is the term number. • 72 is an, -5 is a1, and 7 is d. Plug it in.
  • 33. Additional Example • 72 = -5 + 7(n - 1) • 72 = -5 + 7n - 7 • 72 = -12 + 7n • 84 = 7n • n = 12 • 72 is the 12th term.
  • 35. Arithmetic Series • The African-American celebration of Kwanzaa involves the lighting of candles every night for seven nights. The first night one candle is lit and blown out.
  • 36. Arithmetic Series • The second night a new candle and the candle from the first night are lit and blown out. The third night a new candle and the two candles from the second night are lit and blown out.
  • 37. Arithmetic Series • This process continues for the seven nights. • We want to know the total number of lightings during the seven nights of celebration.
  • 38. Arithmetic Series • The first night one candle was lit, the 2nd night two candles were lit, the 3rd night 3 candles were lit, etc. • So to find the total number of lightings we would add: 1 + 2 + 3 + 4 + 5 + 6 + 7
  • 39. Arithmetic Series • 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 • Series: the sum of the terms in a sequence. • Arithmetic Series: the sum of the terms in an arithmetic sequence.
  • 40. Arithmetic Series • Arithmetic sequence: 2, 4, 6, 8, 10 • Corresponding arith. series: 2 + 4 + 6 + 8 + 10 • Arith. Sequence: -8, -3, 2, 7 • Arith. Series: -8 + -3 + 2 + 7
  • 41. Arithmetic Series • Sn is the symbol used to represent the first ‘n’ terms of a series. • Given the sequence 1, 11, 21, 31, 41, 51, 61, 71, … find S4 • We add the first four terms 1 + 11 + 21 + 31 = 64
  • 42. Arithmetic Series • Find S8 of the arithmetic sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, … • 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = • 36
  • 43. Arithmetic Series • What if we wanted to find S100 for the sequence in the last example. It would be a pain to have to list all the terms and try to add them up. • Let’s figure out a formula!! :)
  • 44. Sum of Arithmetic Series • Let’s find S7 of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, … • If we add S7 in too different orders we get: S7 = 1 + 2 + 3 + 4 + 5 + 6 + 7 S7 = 7 + 6 + 5 + 4 + 3 + 2 + 1 2S7 = 8 + 8 + 8 + 8 + 8 + 8 + 8
  • 45. Sum of Arithmetic Series S7 = 1 + 2 + 3 + 4 + 5 + 6 + 7 S7 = 7 + 6 + 5 + 4 + 3 + 2 + 1 2S7 = 8 + 8 + 8 + 8 + 8 + 8 + 8 2S7 = 7(8) S7 =7 /2(8) 7 sums of 8
  • 46. Sum of Arithmetic Series • S7 =7 /2(8) • What do these numbers mean? • 7 is n, 8 is the sum of the first and last term (a1 + an) • So Sn = n /2(a1 + an)
  • 47. Examples • Sn = n /2(a1 + an) • Find the sum of the first 10 terms of the arithmetic series with a1 = 6 and a10 =51 • S10 = 10/2(6 + 51) = 5(57) = 285
  • 48. Examples • Find the sum of the first 50 terms of an arithmetic series with a1 = 28 and d = -4 • We need to know n, a1, and a50. • n= 50, a1 = 28, a50 = ?? We have to find it.
  • 49. Examples • a50 = 28 + -4(50 - 1) = 28 + -4(49) = 28 + -196 = -168 • So n = 50, a1 = 28, & an =-168 • S50 = (50/2)(28 + -168) = 25(- 140) = -3500
  • 50. Examples • To write out a series and compute a sum can sometimes be very tedious. Mathematicians often use the greek letter sigma & summation notation to simplify this task.
  • 51. Examples • This means to find the sum of the sums n + 1 where we plug in the values 1 - 5 for n n + 1 n = 1 5 ! last value of n First value of n formula used to find sequence
  • 52. Examples • Basically we want to find (1 + 1) + (2 + 1) + (3 + 1) + (4 + 1) + (5 + 1) = • 2 + 3 + 4 + 5 + 6 = • 20 n + 1 n = 1 5 !
  • 53. Examples • So • Try: • First we need to plug in the numbers 2 - 7 for x. n + 1 = 20 n = 1 5 ! 3x - 2 x = 2 7 !
  • 54. Examples • [3(2)-2]+[3(3)-2]+[3(4)-2]+ [3(5)-2]+[3(6)-2]+[3(7)-2] = • (6-2)+(9-2)+(12-2)+(15-2)+ (18-2)+ (21-2) = • 4 + 7 + 10 + 13 + 17 + 19 = 70 3x - 2 x = 2 7 !
  • 56. GeometricSequence • What if your pay check started at $100 a week and doubled every week. What would your salary be after four weeks?
  • 57. GeometricSequence • Starting $100. • After one week - $200 • After two weeks - $400 • After three weeks - $800 • After four weeks - $1600. • These values form a geometric sequence.
  • 58. Geometric Sequence • Geometric Sequence: a sequence in which each term after the first is found by multiplying the previous term by a constant value called the common ratio.
  • 59. Geometric Sequence • Find the first five terms of the geometric sequence with a1 = -3 and common ratio (r) of 5. • -3, -15, -75, -375, -1875
  • 60. Geometric Sequence • Find the common ratio of the sequence 2, -4, 8, -16, 32, … • To find the common ratio, divide any term by the previous term. • 8 ÷ -4 = -2 • r = -2
  • 61. Geometric Sequence • Just like arithmetic sequences, there is a formula for finding any given term in a geometric sequence. Let’s figure it out using the pay check example.
  • 62. Geometric Sequence • To find the 5th term we look 100 and multiplied it by two four times. • Repeated multiplication is represented using exponents.
  • 63. Geometric Sequence • Basically we will take $100 and multiply it by 24 • a5 = 100•24 = 1600 • A5 is the term we are looking for, 100 was our a1, 2 is our common ratio, and 4 is n-1.
  • 64. Examples • Thus our formula for finding any term of a geometric sequence is an = a1•rn-1 • Find the 10th term of the geometric sequence with a1 = 2000 and a common ratio of 1 /2.
  • 65. Examples • a10 = 2000• (1 /2)9 = • 2000 • 1 /512 = • 2000 /512 = 500 /128 = 250 /64 = 125 /32 • Find the next two terms in the sequence -64, -16, -4 ...
  • 66. Examples • -64, -16, -4, __, __ • We need to find the common ratio so we divide any term by the previous term. • -16/-64 = 1/4 • So we multiply by 1/4 to find the next two terms.
  • 67. Examples • -64, -16, -4, -1, -1/4
  • 68. Geometric Means • Just like with arithmetic sequences, the missing terms between two nonconsecutive terms in a geometric sequence are called geometric means.
  • 69. Geometric Means • Looking at the geometric sequence 3, 12, 48, 192, 768 the geometric means between 3 and 768 are 12, 48, and 192. • Find two geometric means between -5 and 625.
  • 70. Geometric Means • -5, __, __, 625 • We need to know the common ratio. Since we only know nonconsecutive terms we will have to use the formula and work backwards.
  • 71. Geometric Means • -5, __, __, 625 • 625 is a4, -5 is a1. • 625 = -5•r4-1 divide by -5 • -125 = r3 take the cube root of both sides • -5 = r
  • 72. Geometric Means • -5, __, __, 625 • Now we just need to multiply by -5 to find the means. • -5 • -5 = 25 • -5, 25, __, 625 • 25 • -5 = -125 • -5, 25, -125, 625
  • 74. Geometric Series • Geometric Series - the sum of the terms of a geometric sequence. • Geo. Sequence: 1, 3, 9, 27, 81 • Geo. Series: 1+3 + 9 + 27 + 81 • What is the sum of the geometric series?
  • 75. Geometric Series • 1 + 3 + 9 + 27 + 81 = 121 • The formula for the sum Sn of the first n terms of a geometric series is given by Sn= 1 - r a1- a1rn or Sn= 1 - r a1(1 - rn )
  • 76. Geometric Series • Find • You can actually do it two ways. Let’s use the old way. • Plug in the numbers 1 - 4 for n and add. • [-3(2)1-1 ]+[-3(2)2-1 ]+[-3(2)3-1 ]+ [- 3(2)4-1 ] - 3 (2)n - 1 n = 1 4 !
  • 77. Geometric Series • [-3(1)] + [-3(2)] + [-3(4)] + [-3(8)] = • -3 + -6 + -12 + -24 = -45 • The other method is to use the sum of geometric series formula.
  • 78. Geometric Series • use • a1 = -3, r = 2, n = 4 Sn= 1 - r a1(1 - rn ) - 3 (2)n - 1 n = 1 4 !
  • 79. Geometric Series • use • a1 = -3, r = 2, n = 4 • Sn= 1 - r a1(1 - rn ) - 3 (2)n - 1 n = 1 4 ! S4= 1 - 2 - 3 (1 - 24 )
  • 80. Geometric Series • S4= 1 - 2 - 3 (1 - 24 ) S4= - 1 - 3 (1 - 16) S4= - 1 - 3 (- 15) = - 1 45 =- 45