Lesson 4 - Geometric Sequencesasasasa.pdf

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1. MATH 10
WELCOME TO OUR ONLINE CLASS !

2. Geometric
Sequence
B

3. What is Geometric Sequence?
qIs a sequence of numbers where each term
after the first is found by multiplying the
previous one by a fixed, non – zero number
called the common ratio.

4. EXAMPLE:
Since we multiply 2 to every
term to get the next term, then 2
is our common ratio.
2, 4, 8, 16, 32, …
x2x2 x2 x2
t1t2t3 t4 t5
tn

5. Formula of Geometric Sequence
Where:
tn is the last term r is the common ratio
t1 is the first term n is the number of terms
tn = t1???
???−1

6. qWhat is the 10
th
term of the sequence?
2, 4, 8, 16, 32, …
tn = t1???
???−1
t1 = 2
r = 2
n = 10
tn =?
t10 = 2(2)
10−1
t10 = 2(2)
9
t10 = 2(512)
t10 = 1,024

7. qWhat is the 12
th
term of the sequence?
1, 3, 9, 27, 81, …
tn = t1???
???−1
t1 = 1
r = 3
n = 12
tn =?
t12 = 1(3)
12−1
t12 = 1(3)
11
t12 = 1(177,147)
t12 = 177, 147

8. qWhat is the 16
th
term of the sequence?
6, 12, 24, 48, 96, …
tn = t1???
???−1
t1 = 6
r = 2
n = 16
tn =?
t16 = 6(2)
16−1
t16 = 6(2)
15
t16 = 6(32, 768)
t16 = 196, 608

9. qWhat is the 13
th
term of the sequence?
21,42,84,168,336, …
tn = t1???
???−1
t1 = 21
r = 2
n = 13
tn =?
t13 = 21(2)
13−1
t13 = 21(2)
12
t13 = 21(4,096)
t13 = 86, 016
Challenge:

10. Word Problems Involving Geometric Sequence
qA virus reproduces by dividing into two, and after a certain growth
period, it divides into two again. As the virus continues to reproduce,
it will continue to divide in two. How many viruses will be in a system
starting with a single virus AFTER 10 divisions?
DIVISION1st2nd3rd4th5th6th7th8th9th10th
NUMBER
OF VIRUS
1

11. Illustration:
1
2
2
3
3
3
3
DIVISION1st2nd3rd4th5th6th7th8th9th10th
NUMBER
OF VIRUS
1248163264128256512

12. Word Problems Involving Geometric Sequence
qA brand new machine cost 75,000.00. It’s value depreciates each year at the
rate of 6% of the original value. Find its value at the end of the fourth year.
YEAR Initial
Value
1st2nd3rd4th
AMOUNT75, 000
(75,000)*(1 - .06)
(75,000)*(0.94)
= 70,500
(70,500)*(1 - .06)
(70,500)*(0.94)
= 66,270
(66,270)*(1 - .06)
(66,270)*(0.94)
= 62,293.8
END OF 1
ST
YEAR END OF 2
ND
YEAR
END OF 3
RD
YEAR

13. Word Problems Involving Geometric Sequence
qA brand new machine cost 75,000.00. It’s value depreciates each year at the
rate of 6% of the original value. Find its value at the end of the fourth year.
YEAR Initial
Value
1st2nd3rd4th
AMOUNT75, 000
t4 = (70,500)(???,??????)
??? −???
t4 = (70,500)(???,??????)
???
t4 = (70,500)(0.830584)
t4 = 58, 556.172
tn = t1???
???−1
t1 = 70,500
r = 0.94
n = 4
t4 =?
62,293.8
End of 3
rd Year
(62,293.8)*(1 - .06)
(62,293.8)*(0.94)
= 58, 556.172

14. Word Problems Involving Geometric Sequence
qA brand new machine cost 75,000.00. It’s value depreciates each year at the
rate of 6% of the original value. Find its value at the end of the fourth year.
YEAR 1st 2nd 3rd 4th
AMOUNT70, 50066, 27062, 293.858,556.172
0.940.940.94