Mathematics10 Quarter1 Module4 Week4
Mathematics10 Quarter1 Module4 Week4
Mathematics10 Quarter1 Module4 Week4
Objectives
The learner…
a. enumerates the next few terms and the 𝑛𝑡ℎ term of a geometric
sequence.
b. inserts means between two given terms of a geometric sequence.
c. finds the sum of the terms of the given finite or infinite geometric
sequence.
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MATHEMATICS10-QUARTER1-WEEK4
EXAMPLES
1) Let 𝑎1 , 𝑎2 , 𝑎3 , … , 𝑎𝑛 be the geometric sequence. Fill in the blanks with the
correct expressions.
𝑎1 = 𝑎1
𝑎2 = 𝑎1 𝑟
𝑎3 = 𝑎2 𝑟 = (𝑎1 𝑟)𝑟 = __________ answer: 𝑎1 𝑟 2
𝑎4 = 𝑎3 𝑟 = (𝑎1 𝑟 2 )𝑟 = __________ answer: 𝑎1 𝑟 3
𝑎5 = 𝑎4 𝑟 = __________ = _________ answer: (𝑎1 𝑟 3 )𝑟 = 𝑎1 𝑟 4
a. Can you identify the pattern? Write down your observations below.
__________________________________________________________
b. If the pattern continues, what expressions will be for the 𝑛𝑡ℎ term of a
geometric sequence?
answer: 𝑎𝑛 = 𝑎1 𝑟 𝑛−1 General Formula of a Geometric Sequence
2) Find the fifth term of the geometric sequence 6, 24, 96, ...
Solution: 6, 24, 96, 384, 𝟓𝒕𝒉 term, …
𝑎5 = 𝑎4 𝑟 𝑎𝑛 = 𝑎1 𝑟 𝑛−1 To find the common
𝑎5 = 384 (4) 𝑎5 = 6(4 )5−1
substitute the given ratio, 𝒓, simply DIVIDE
𝑎5 = 1536 𝑎5 = 6(44 ) apply PEMDAS rule any term in the
sequence by the term
𝑎5 = 6(256) simplify
that precedes it.
𝑎5 = 1536
𝒓 = 24 ÷ 6
𝒓=4
EXAMPLES
1) Find 𝑆7 of the finite geometric sequence 6, 24, 96, 384, …, 𝑎7 .
1−𝑟 𝑛 −16383
Solution: 𝑆𝑛 = 𝑎1 ( 1−𝑟 ) 𝑆7 = 6 ( )
−3
1−47
𝑆7 = 6 ( 1−4 ) 𝑆7 = 6 (5461)
1−16384
𝑆7 = 6 ( ) 𝑆7 = 32 766 answer
1−4
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MATHEMATICS10-QUARTER1-WEEK4
For an infinite geometric sequence to have sum, the common ratio, 𝒓, must
be between −1 and 1, as 𝒏 increases 𝒂𝒏 gets closer to 0.
If 𝑎1 + 𝑎1 𝑟 + 𝑎1 𝑟 2 + ⋯ is an infinite geometric series, with −1 < 𝑟 < 1, then
𝒂𝟏
the sum of 𝑺 of all terms of this series is given by: 𝑺∞ = 𝟏−𝒓
(https://upload.wikimedia.org/wikipedia/commons/thumb/a/ab/Geometric_Segment.svg/1280px-Geometric_Segment.svg.png)
1 1 1
Prove that + + + ⋯ = 1
2 4 8
1 1
Proof: Given 𝑎1 = 2 and common ratio 𝑟 = 2
𝒂𝟏
𝑺∞ = 𝟏−𝒓
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MATHEMATICS10-QUARTER1-WEEK4
𝟏 𝟏
𝟐 𝟐
Substitute the given to the formula: 𝑺∞ = 𝟏 = 𝟏 =𝟏
𝟏−
𝟐 𝟐
EXAMPLES
NOTE
Infinite geometric series may also express in sigma notation, with the
variable going from the lower limit to the positive infinity, or
∞
∑ 𝑎1 𝑟 𝑛−1
𝑛=1
where: −1 < 𝑟 < 1
EXAMPLE: Find the value of:
∞
1 𝑛−1
∑8( )
5
𝑛=1
1 𝒂𝟏
Solution: Let 𝑎1 = 8 𝑎𝑛𝑑 𝑟 = 5, 𝑢𝑠𝑖𝑛𝑔 𝑺∞ = , 𝒔𝒊𝒏𝒄𝒆 𝒓 < 1.
𝟏−𝒓
𝟖 𝟖
𝑺∞ = 𝟏 = 𝟒 = 10
𝟏−
𝟓 𝟓
Geometric Means
Geometric means are terms between two given terms of a geometric
sequence.
𝒃
In geometric sequence 𝑎, 𝑏, 𝑐 the geometric mean is 𝑏, whose ratio is 𝒓 = 𝒂 or
𝒄
𝒓 = 𝒃.
𝒃 𝒄
Since 𝒂 = 𝒃 and 𝒃𝟐 = 𝒂𝒄, 𝒃 = √𝒂𝒄.
The positive square root of the product of the given terms is called the mean
proportional.
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MATHEMATICS10-QUARTER1-WEEK4
EXAMPLE
1) Find the geometric mean between 12 and 36.
Solution: Let 𝑐 be the geometric mean between 12 and 36.
𝒄 = √𝒂𝒃 𝒄 = √𝟒𝟑𝟐
𝒄 = √𝟏𝟐 ∗ 𝟑𝟔 𝒄 = ±𝟏𝟐√𝟑
2) Insert three geometric means between 5 and 3125.
Solution:
𝐿𝑒𝑡 𝑎1 = 5, 𝑎5 = 3125 , 𝑎𝑛𝑑 𝑛 = 5
Determine the common ratio
𝑎𝑛 = 𝑎1 𝑟 𝑛−1 Therefore, 𝒂𝟐 = 𝟐𝟓, 𝒂𝟑 = 𝟏𝟐𝟓,
𝑎5 = 5(𝑟 5−1 ) 𝒂𝟒 = 𝟔𝟐𝟓 or 𝒂𝟐 = −𝟐𝟓, 𝒂𝟑 = 𝟏𝟐𝟓,
3125 = 5𝑟 4 𝒂𝟒 = −𝟔𝟐𝟓 are the geometric
means between 5 and 3125.
625 = 𝑟 4
𝑟 = ±5
Let’s Apply
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MATHEMATICS10-QUARTER1-WEEK4
Let’s Analyze