C2: Geometric Series
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 24th September 2013
Types of series
common difference
𝑑
?
+3
+3
+3
2, 5, 8, 11, 14, …
This is a:
Arithmetic
? Series
common?ratio 𝑟
×2 ×2
×2
3, 6, 12, 24, 48, …
? Series
Geometric
Common Ratio
Identify the common ratio 𝑟:
1
1, 2, 4, 8, 16, 32, …
𝑟=2 ?
2
24, 18, 12, 8, …
𝑟 = 2/3?
3
10, 5, 2.5, 1.25, …
𝑟 = 1/2?
4
5, −5, 5, −5, 5, −5, …
𝑟 = −1?
5
𝑥, −2𝑥 2 , 4𝑥 3
𝑟 = −2𝑥
?
6
1, 𝑝, 𝑝2 , 𝑝3 , …
𝑟=𝑝 ?
7
4, −1, 0.25, −0.0625, …
?
𝑟 = −0.25
Common Ratio Exam Question
May 2013 (Retracted)
Hint for (a): the common ratio between the first and second terms, and the second and
third terms, is the same.
a
3𝑝 + 15 5𝑝 + 20
=
4𝑝
3𝑝 + 15
2
?
3𝑝 + 15 = 4𝑝 5𝑝 + 20
11𝑝2 − 10𝑝 − 225
b
3𝑝 + 15 30 3
𝑟=
=
=
4𝑝?
20 2
𝑛th term
Arithmetic Series
𝑈𝑛 = 𝑎 + 𝑛 −? 1 𝑑
Geometric Series
𝑈𝑛 = 𝑎𝑟 𝑛−1?
Determine the following:
3, 6, 12, 24, …
40, -20, 10, -5, …
?
𝑈10 = 1536
𝑈10
𝑈𝑛 = −1
5
=− ?
64
𝑛−1
?×
5
2𝑛−4
Another Common Ratio Example
The numbers 3, 𝑥 and 𝑥 + 6 form the first three terms of a
positive geometric sequence. Find:
a) The possible values of 𝑥.
b) The 10th term in the sequence.
𝑥 = 6 𝑜𝑟 − 3
But there are no negative terms so 𝑥 = 6
?
𝑥 6
𝑟= = =2
𝑎=3
𝑛 = 10
3 3
𝑈10 = 3 × 29 = 1536
Missing information
The second term of a geometric sequence is 4 and the 4th term is 8.
The common ratio is positive. Find the exact values of:
a) The common ratio.
b) The first term.
c) The 10th term.
𝑈2 = 4
𝑈4 = 8
𝑠𝑜 𝑎𝑟 = 4 (1)
𝑠𝑜 𝑎𝑟 3 = 8 (2)
? 𝑟 2 = 2, 𝑠𝑜 𝑟 = 2
a) Dividing (1) by (2) gives
4
4
b) Substituting, 𝑎 = = = 2 2
c) 𝑈10 = 𝑎𝑟 9 = 64
𝑟
2
Bro Tip: Explicitly writing 𝑈2 = 4 first helps you avoid confusing the 𝑛th term with
the ‘sum of the first 𝑛 terms’ (the latter of which we’ll get onto).
𝑛th term with inequalities
What is the first term in the geometric progression 3, 6, 12, 24, …
to exceed 1 million?
𝑈𝑛 > 1000000
𝑎 = 3, 𝑟 = 2
?
Exam Question
Edexcel June 2010
?
25000 × 1.03 = 25750
?
𝑟 = 1.03
𝑈𝑁 > 40000
25000 × 1.03𝑁−1 > 40000
1.03𝑁−1 > 1.6
log 1.03𝑁−1 > log 1.6
𝑁 − 1 log 1.03 > log 1.6
?
𝑁−1>
log 1.6
log 1.03
?
= 15.9,
𝑆10
𝑁 > 16.9,
𝑁 = 17
25000 1 − 1.0310
=
= £287,000
1 − 1.03
?
Sum of the first 𝑛 terms
Arithmetic Series
𝑛
𝑆𝑛 = 2𝑎 + ?𝑛 − 1 𝑑
2
Geometric Series
𝑎 1 − 𝑟𝑛
?
𝑆𝑛 =
1−𝑟
Technically you could be asked in an exam the proof of
the sum of a geometric series (it once came up!)
So let’s prove it…
Sum of the first 𝑛 terms
Geometric Series
𝑎 1 − 𝑟𝑛
𝑆𝑛 =
1−𝑟
Find the sum of the first 10 terms.
3, 6, 12, 24, 48, …
? 𝑟 = 2,
?
? 𝑛 = 10
𝑎 = 3,
3 1 − 210
? = 3069
𝑆𝑛 =
1−2
1 1 1
4, 2,1, , , , …
2 4 8
1
?
?
𝑎 = ?4, 𝑟 = , 𝑛 = 10
2
1
4 1−
2
𝑆𝑛 =
1
1−
2
10
?= 1023
128
Summation Notation
Find
𝑎 = 6,?
10
𝑟=1
3×
𝑟 = 2,?
𝑟
2
𝑛 = 10?
10
𝑆10
6 1 − 2?
=
1−2
= 6138
Harder Questions: Type 1
Find the least value of 𝑛 such that the sum of 1 + 2 + 4 + 8 + ⋯
to 𝑛 terms would exceed 2 000 000.
𝒏 > 𝟐𝟎. 𝟗
?
𝑺𝒐 𝒏 = 𝟐𝟏
An investor invests £2000 on January 1st every year in a savings
account that guarantees him 4% per annum for life. If interest is
calculated on the 31st of December each year, how much will be
in the account at the end of the 10th year?
£𝟐𝟒 𝟗𝟕𝟐. 𝟕𝟎
?
Exercise 7D
1 Find the sum of the following geometric series (to 3dp if necessary).
a) 1 + 2 + 4 + 8 + ⋯ (8 terms)
c) 4 − 12 + 36 − 108 + ⋯ (6 terms)
𝑟
e) 10
𝑟=1 4
h)
2
5
𝑟=0 60
×
1 𝑟
−3
𝑺𝟖 = ?𝟐𝟓𝟓
𝑺𝟔 = ?−𝟕𝟐𝟖
= 𝟏𝟑𝟗𝟖𝟏𝟎𝟎
?
= 𝟒𝟒. 𝟗𝟑𝟖?(𝒕𝒐 𝟑𝒅𝒑)
The sum of the first three terms of a geometric series is 30.5. If the first term is 8, find the
possible values of 𝑟.
𝟓
𝟗
= 𝟒,−?
𝟒
4
Jane invest £4000 at the start of every year. She negotiates a rate of interest of 4% per
annum, which is paid at the end of the year. How much is her investment worth at the
end of (a) the 10th year and (b) the 20th year.
(a) £𝟒𝟗𝟗𝟒𝟓.
? 𝟒𝟏 (b) £𝟏𝟐𝟑𝟖𝟕𝟔.
? 𝟖𝟏
5
A ball is dropped from a height of 10m. It bounces to a height of 7m and continues to
bounce. Subsequent heights to which it bounces follow a geometric sequence. Find out:
a) How high it will bounce after the fourth bounce,
= 𝟐. 𝟒𝟎𝟏
?
b) The total distance travelled until it hits the ground for a sixth time.
= 𝟒𝟖. 𝟖𝟐𝟑𝟒
?
6
Find the least value of 𝑛 such that the sum 3 + 6 + 12 + 24 + ⋯ to 𝑛 terms would first
exceed 1.5 million.
= 𝟏𝟗 𝒕𝒆𝒓𝒎𝒔
?
Different types of series
What can you say about the sum of each series up to infinity?
1 + 2 + 4 + 8 + 16 + ...
This is divergent – the
sum of the
?
values tends towards infinity.
1 + 2 + 3 + 4 + 5 + ...
This is divergent – the sum of the values
tends towards infinity.
? But arguably, the
sum of the natural numbers is −1/12.
1 + 0.5 + 0.25 + 0.125 + ...
This is convergent – the sum of
the values tends towards
? a fixed
value, in this case 2.
Just for fun...
1 1 1 1
+ + + +⋯
1 2 3 4
1 1 1 1
+ + +
+⋯
1 4 9 16
This is divergent . This is known
as the Harmonic Series
?
This is convergent . This is known
as the Basel Problem, and the
?
value is 𝜋 2 /6.
Sum to Infinity
Think about our formula for the sum of the first 𝑛 terms.
If we make 𝑛 infinity, what do we require of 𝑟 for 𝑆∞ not to be infinity (i.e. we
want to keep the series convergent). And what will the formula become?
𝑎 1 − 𝑟𝑛
𝑆𝑛 =
1−𝑟
𝑎
?
𝑆∞ =
1−𝑟
Restriction on 𝑟:
−1 < 𝑟 ?< 1
Examples
1 1 1
1, , , , …
2 4 8
𝒂 = 𝟏,?
27, −9,3, −1, …
𝒂 = 𝟐𝟕,
?
2
3
4
𝑝, 𝑝 , 𝑝 , 𝑝 , …
𝟏
𝒓= ?
𝟐
𝟏
𝒓 = −?
𝟑
𝟖𝟏
𝑺∞ = ?
𝟒
𝒂 = 𝒑,?
𝒓 = 𝒑?
𝒑
𝑺∞ = ?
𝟏−𝒑
𝒂 = 𝒑,?
𝟏
𝒓= ?
𝒑
𝒑𝟐
𝑺∞ = ?
𝒑−𝟏
𝑤ℎ𝑒𝑟𝑒 − 1 < 𝑝 < 1
1 1
𝑝, 1, , 2 , …
𝑝 𝑝
𝑺∞ = 𝟐?
A somewhat esoteric Futurama joke explained
Bender (the robot) manages to self-clone himself, where some excess is required to produce the
duplicates (e.g. alcohol), but the duplicates are smaller versions of himself. These smaller clones
also have the capacity to clone themselves. The Professor is worried that the total amount mass
𝑀 consumed by the growing population is divergent, and hence they’ll consume to Earth’s entire
resources.
A somewhat esoteric Futurama joke explained
This simplifies to
∞
𝑀 = 𝑀0
The sum
1
1
+
1
2
1
3
𝑛=1
1
𝑛
+ + ⋯ is known as the harmonic series, which is divergent.
Another Example
The sum to 4 terms of a geometric series is 15 and
the sum to infinity is 16.
a) Find the possible values of 𝑟.
𝟏
𝒓 = ±?
𝟐
b) Given that the terms are all positive, find the first
term in the series.
𝑼𝟏 = 𝟖
?
Another Example
Edexcel May 2011
𝟑
𝒓= ?
𝟒
𝒂 = 𝟐𝟓𝟔
?
𝑺∞ = 𝟏𝟎𝟐𝟒
?
𝟑
𝟐𝟓𝟔 𝟏 − 𝟒
𝟎. 𝟐𝟓
𝒏
> 𝟏𝟎𝟎𝟎
?
𝟔
𝟐𝟓𝟔
𝒏>
𝒍𝒐𝒈 𝟎. 𝟕𝟓
𝒍𝒐𝒈
⇒ 𝒏 = 𝟏𝟒
Exercises
Exercise 7D Q6, 7
Exercise 7E Q8
Exercise 7F Q10