illustrate geometric sequence, finding the nth term of a geometric sequence, finding the first term of the geometric sequence, determining the term representing the given term of a geometric sequence.
3. Think-pair-share
Materials: paper and scissors
Procedure:
a) Cut the paper in half
b) Stack the halves. Cut the stack in halves.
c) Continue stacking and cutting the paper strips about half an inch
wide.
Record the total number of pieces obtained in the given table.
4. Number of cuts 1 2 3 4 5 6 7 8
Number of pieces
New number of pieces
ππππ£πππ’π ππ’ππππππ ππππππ
QUESTIONS TO PONDER:
1. WHAT DO YOU OBSERVE ABOUT THE CHANGE IN NUMBER OF PIECES
WITH EACH CUT AFTER THE FIRST?
2. LIST THE NUMBER IN ROW 2 FROM LEAST TPO GREATEST, SEPARATE
THE NUMBERS WITH COMMA.
3. WHAT IS THE COMMON RATIO BETWEEN ANY TWO CONSECUTIVE
TERMS IN THE SEQUENCE?
6. GEOMETRIC SEQUENCE
β’ A SEQUENCE IN WHICH EACH TERM IS OBTAINED BY MULTIPLYING THE
PRECEDING TERM BY A FIXED NUMBER.
β’ COMMON RATIO (r) IS THE FIXED NUMBER MULTIPLIED IN EACH TERM
OF A GEOMETRIC RATIO.
nth TERM OF A GEOMETRIC SEQUENCE
ππ = ππβ1 β π or ππ = π1 β ππβ1
7. Example1
β’ Write a formula for the nth term of the given geometric sequence.
8, 2, Β½ , 1/8, 1/32, . . .
Solution: Find the common ratio: r =
π2
π1
=
2
8
=
1
4
ππ = π1 β ππβ1
ππ = π β (
π
π
)πβπ
Check: π1 = 8 β
1
4
1β1
= 8 π2 = 8 β
1
4
2β1
= 2
π3 = 8 β
1
4
3β1
=
1
2
π5 = 8 β
1
4
5β1
=
1
32
Note: any number raise to 0 is
equal to 1.
8. Example 2
β’ WRITE THE FIRST THREE TERMS OF A GEOMETRIC SEQUENCE WHOSE
nth TERM IS GIVEN BY ππ = 3(β5)πβ1.
SOLUTION:
π1 = 3(β5)1β1 π2 = 3(β5)2β1 π3 = 3(β5)3β1
π1 = 3(β5)0
π2 = 3(β5)1
π3 = 3(β5)2
π1 = 3 1 π2 = 3(β5) π3 = 3(25)
π1 = 3 π2 = β15 π3 = 75
*AFTER GETTING THE FIRST TERM, THE OTHER TERMS MAY BE OBTAINED
BY MULTIPLYING THE PREVIOUS TERM BY -5.
9. EXAMPLE 3
β’ FIND THE 8th TERM OF THE GEOMETRIC SEQUENCE 24, 12, 6, 3, . . .
GIVEN: π1 = 24 r =
12
24
=
1
2
π8 = ?
SOLUTION: ππ = π1ππβ1
π8 = 24(
1
2
)8β1
substitute the values in the formula
π8 = 24(
1
2
)7
π8 = 24(
1
128
)
ππ =
ππ
πππ
or ππ =
π
ππ
10. EXAMPLE 4
FIND THE FIRST TERM OF A GEOMETRIC SEQUENCE, IF THE 6th TERM IS 96
and r = 2.
GIVEN: π6 = 96 r = 2 π1 = ?
SOLUTION: π6 = π1(2)6β1
96 = π1(2)5
(2)5
= 2 β 2 β 2 β 2 β 2 = 32
96 = (32)π1
96
32
=
(32)π1
32
ππ = π
Sequence is
3, 6, 12, 24, 48, 96, . . .
11. EXAMPLE 5
β’ FIND THE COMMON RATIO OF A GEOMETRIC SEQUENCE, IF THE FIRST
TERM IS 789 AND THE FIFTH TERM IS 12 624.
GIVEN: π1 = 789 π5 = 12 624 r = ?
SOLUTION: π5 = π1π5β1
12 624 = 789(π)4
12 624
789
=
789π4
789
16 = π4
4
16 =
4
π4
π = π
12. EXAMPLE 6
β’ IN A GEOMETRIC SEQUENCE THE FIRST TERM IS 1 AND THE COMMON
RATIO IS 4, WHAT TERM DOES 1024 REPRESENTS?
SOLUTION: ππ = π1ππβ1
1 024 = 1(4)πβ1
1 024 = (4)πβ1
45 = 4πβ1
5 = π β π
π + π = π
6 = π
13. PRACTICE TASK
β’ WRITE WHETHER THE GIVEN SEQUENCE IS ARITHMETIC, GEOMETRIC,
OR NEITHER. IF IT IS ARITHMETIC, GIVE THE COMMON DIFFERENCE. IF
IT IS GEOMETRIC, GIVE THE COMMON RATIO
1) 11, 14, 17, 20, . . .
2) 4, 8, 16, 32, . . .
3) 5, 8, 12, 17, 26, . . .
4) 100, -50, 25, -12.5, . . .
5) 1, 8, 27, 64, . . .
14. PRACTICE TASK
β’WRITE THE nth TERM FORMULA OF THE FOLLOWING:
1) 4, 8, 16, 32, . . .
2) 100, -50, 25, -12.5, . . .
3) 1, 3, 9, 27, . . .
WRITE THE 6TH TO 10TH TERMS OF THE SEQUENCE
ABOVE.
15. PRACTICE TASK
β’WRITE THE nth TERM FORMULA OF THE FOLLOWING:
1) 4, 8, 16, 32, . . .
2) 100, -50, 25, -12.5, . . .
3) 1, 3, 9, 27, . . .
WRITE THE 6TH TO 8TH TERMS OF THE SEQUENCE ABOVE.
16. ASSIGNMENT
β’ Write whether the sequence is arithmetic or geometric. Write the nth
term of each sequence.
1)2, 8, 32, 128, 512, . . .
2)3, 12, 48, 192, 768, . . .
3)-35, -32, -29, -26, -23, . . .
4)-24, -14, -5, 6, 16, . . .
5)3, -9, 27, -81, 243, . . .
17. β’WHAT IS GEOMETRIC SERIES?
β’WHAT IS THE FORMULA FOR THE
GEOMETRIC SERIES?
β’GIVE 3 OF YOUR OWN EXAMPLES OF
GEOMETRIC SEQUENCE,WRITE THE nth
TERM FORMULA.
18.
19. JOURNAL ENTRY
β’ HOW DID YOU FIND THE LESSON ABOUT GEOMETRIC SEQUENCE?
β’ HOW WILL YOU FIND THE VALUE OF n IF THE GIVEN ARE
π1 = 5, π = 3, ππ = 10935
EXPLAIN YOUR SOLUTION.