1. Dr J Frost (jfrost@tiffin.kingston.sch.uk)
GCSE :: Term-to-term Sequences and Arithmetic vs Geometric Progressions
Dr J Frost
Objectives: (From Edexcel specification)
Last modified: 15th April 2018

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3. STARTER :: Types of sequences
Determine the next two terms in each of these sequences:
! An arithmetic progression/sequence is one where we add/subtract the same number from term to term.
3, 5, 7, 9, 11, ____, _____ ? 13 15 ?
3, 6, 12, 24, ____, ____ ? 48 96 ?
! A geometric progression is where we multiply/divide by the same number.
3, 5, 8, 13, 21, ____, ____ ? 34 ? 55
A Fibonacci-type sequence is one where each term is the sum of the previous two terms.
3, 5, 8, 12, 17, ____, ____ ? 23 ? 30
! A quadratic sequence is one where the second difference is constant.
(Quadratic sequences are covered in a separate resource)

4. Arithmetic Progressions
You should already know how to find the 𝑛th term formula for a linear sequence/arithmetic progress. For example, the 𝑛th term formula of 2,5,8,… is 3𝑛−1.
But questions might specifically test your understanding that the difference of an arithmetic progression is constant:
The first three terms of an arithmetic progression are: 5,𝑥,11.
Determine the value of 𝑥. ?
11−5=6
So difference between each term is 3.
𝑥=5+3=8
The first three terms of an arithmetic progression are: 𝑥,2𝑥+1, 𝑥 2 +4.
Determine the possible values of 𝑥. ?
2𝑥+1 −𝑥=𝑥+1
𝑥 2 +4 − 2𝑥+1 = 𝑥 2 −2𝑥+3
𝑥 2 −2𝑥+3=𝑥+1 𝑥 2 −3𝑥+2=0 𝑥−1 𝑥−2 =0 𝑥=1 or 𝑥=2
These correspond to the sequences:
1, 3, 5 and 2, 5, 8
The difference between the first and second term.
The difference between the second and third term.
Since we have an arithmetic progression, these differences are the same.

5. Test Your Understanding1
The first four terms of an arithmetic progression are: 6,𝑥, 𝑦,18.
Determine the values of 𝑥 and 𝑦.
Hence determine the 100th term of the sequence.
Difference: 18−6 3 =4
𝑛th term formula: 4𝑛+2
100th term: 4×100 +2=402 a ? b ? 2
The first three terms of an arithmetic progression are:
𝑥, 3𝑥−5, 𝑥 2 −6
Determine the possible values of 𝑥, and hence the possible values for the fourth term of the sequence. ?
𝟑𝒙−𝟓 −𝒙= 𝒙 𝟐 −𝟔 − 𝟑𝒙−𝟓 𝟐𝒙−𝟓= 𝒙 𝟐 −𝟑𝒙−𝟏 𝒙 𝟐 −𝟓𝒙+𝟒=𝟎 𝒙−𝟏 𝒙−𝟒 =𝟎 𝒙=𝟏 𝒐𝒓 𝒙=𝟒
These give the sequences: 1, -2, -5 and 4, 7, 10
Continuing each, we get the possible 4th terms -8 and 13

6. Geometric Progressions
The first three terms of a geometric progression are: 2,𝑥,5, where 𝑥 is positive.
Determine the value of 𝑥.
We can similarly use the main property of a geometric progression, that the ratio (i.e. the number we’re multiplying by each time) is constant. ?
If the sequence were say 3, 6, 12, …, we know the common ratio is 6 3 =2, by dividing the 2nd term by the 1st. But notice also that =2 also, using the 2nd and 3rd terms. This allows us to get an equation.
𝑥 2 = 5 𝑥
𝑥 2 =10 𝑥= 10
Cross multiply, i.e. if 𝑎 𝑏 = 𝑐 𝑑 then 𝑎𝑑=𝑏𝑐
The first three terms of a geometric progression are: 𝑥, 𝑥+3, 2𝑥+6.
Determine the possible values of 𝑥. ?
Important note: This is strongly connected with the topic of exponential graphs, where you’re often given two points on a graph with equation 𝑦=𝑎× 𝑏 𝑛 , and have to determine the values of 𝑎 and 𝑏.
𝑥+3 𝑥 = 2𝑥+6 𝑥+3
𝑥+3 2 =𝑥(2𝑥+6) 𝑥 2 +6𝑥+9=2 𝑥 2 +6𝑥 𝑥 2 −9=0 𝑥=3 or 𝑥=−3
although 𝑥=−3 would give the sequence −3,0,0,… where the common ratio is 0 (which is typically not allowed for a geometric progression)

7. 𝑛th term formula of a Geometric Progression
Determine the 𝑛th term formula for the geometric progressions:
2, 4, 8, 16, 32, …
6, 18, 54, …
The common ratio is 2 so try 2 𝑛 . This works, as 1st term is 2 1 =2, 2nd term is 2 2 =4, … a ?
Fro Tip: If the terms ×𝑎 each time, then try the nth term formula 𝑎 𝑛
2 𝑛 ?
Common ratio is =3 so try 3 𝑛 . This gives the sequence 3, 9, 27, … We can see we need to double it to get the correct sequence. b
2× 3 𝑛

8. Test Your Understanding1
Determine the 5th term of the geometric progression:
3, , 6, … 2
Determine the 8th term of the geometric progression 3, 6, 12, … ?
Common ratio = 2. We need to advantage 7 terms from the 1st:
𝟑× 𝟐 𝟕 =𝟑𝟖𝟒 ?
Common ratio: 𝟑 𝟐 𝟑 = 𝟐
𝟔× 𝟐 × 𝟐 =𝟏𝟐 3
The first three terms of a geometric progression are:
𝑥−4, 𝑥, 2𝑥+6
Determine the possible values of 𝑥.
𝒙 𝒙−𝟒 = 𝟐𝒙+𝟔 𝒙
𝒙 𝟐 = 𝟐𝒙+𝟔 𝒙−𝟒 𝒙 𝟐 =𝟐 𝒙 𝟐 −𝟐𝒙−𝟐𝟒 𝒙 𝟐 −𝟐𝒙−𝟐𝟒=𝟎 𝒙+𝟒 𝒙−𝟔 =𝟎 𝒙=−𝟒 𝒐𝒓 𝒙=𝟔 ?

9. Exercise 1 ? ? ? ? ? ? ? ? ? (On provided worksheet)
Find the 8th term of the following geometric progressions:
2, 4, 8, …
400, 200, 100, …
3, 3 2 , 6, … 𝟐𝟒 𝟐
The first three terms of an arithmetic progression are
3, 2𝑥+1, 𝑥 2 +3
Determine the value of 𝑥.
𝟐𝒙+𝟏 −𝟑= 𝒙 𝟐 +𝟑 −(𝟐𝒙+𝟏)
𝟐𝒙−𝟐= 𝒙 𝟐 −𝟐𝒙+𝟐 𝒙 𝟐 −𝟒𝒙+𝟒=𝟎 𝒙−𝟐 𝟐 =𝟎 𝒙=𝟐
The first three terms of a geometric progression are:
𝑥, 2𝑥, 𝑥+9
Find the value of 𝑥.
𝟐𝒙 𝒙 = 𝒙+𝟗 𝟐𝒙 𝟒𝒙=𝒙+𝟗 → 𝒙=𝟑
[OCR GCSE(9-1) Practice set 1 5H Q14b]
Here is a sequence.
2, , 14,           
Find the 𝑛th term.
𝟐 𝟕 𝒏−𝟏 or 𝟐 𝟕 𝟕 𝒏
[Edexcel GCSE(9-1) Nov H Q23a]
𝑆 is a geometric sequence.
(a) Given that  𝑥 −1, 1 and 𝑥 +1 are the first three terms of 𝑆, find the value of 𝑥.
𝟏 𝒙 −𝟏 = 𝒙 +𝟏 𝟏 𝟏= 𝒙 +𝟏 𝒙 −𝟏 𝟏=𝒙−𝟏 𝒙=𝟐
(b) Find the 5th term of 𝑆.
𝟐 +𝟏 × 𝟐 +𝟏 𝟐 =𝟕+𝟓 𝟐
Prove that the first three terms of an arithmetic progression (where the difference is not 0) cannot be the same as the first three terms of a geometric progression.
Let the first three terms be: 𝒙−𝒂, 𝒙, 𝒙+𝒂
If geometric: 𝒙 𝒙−𝒂 = 𝒙+𝒂 𝒙 → 𝒙 𝟐 = 𝒙 𝟐 − 𝒂 𝟐
∴𝒂=𝟎. This would mean the terms have to be the same. 1 4 ? ? ? ? 5 2 ? ? ? 3 N1 ? ?

10. Term-to-term formulae
Consider the sequences we saw earlier…
𝒖 𝟏 , 𝒖 𝟐 , 𝒖 𝟑 , 𝒖 𝟒 , 𝒖 𝟓
We can use 𝑢 1 , 𝑢 2 , … to refer to the 1st and 2nd terms of the sequence, and so on.
(But any variable could be used)
3, 5, 7, 9, 11, …
3, 6, 12, 24, 48, …
3, 5, 8, 13, 21, …
How could we relate the first two terms of the sequence using the 𝑢 notation?
𝑢 2 = 𝑢 1 +2
(i.e. the 2nd term is two more than the first term)
Similarly we could connect the second and third term in this way…
𝑢 3 = 𝑢 2 +2

11. 3, 5, 7, 9, 11, … Term-to-term formulae
3, 5, 7, 9, 11, …
𝑢 2 = 𝑢 𝑢 3 = 𝑢 𝑢 4 = 𝑢 3 +2 …
Notice that the small “subscript” numbers in each equation are consecutive numbers.
When you studied ‘proofs involving integers’, how did we represent two consecutive numbers?
𝒏, 𝒏+𝟏
Therefore how could we more generically express that “each term is the previous term plus two”?
𝒖 𝒏+𝟏 = 𝒖 𝒏 +𝟐 ? ?

12. 3, 5, 7, 9, 11, … Term to term formulae 𝑢 1 =3, 𝑢 𝑛+1 = 𝑢 𝑛 +2
3, 5, 7, 9, 11, …
𝑢 1 =3, 𝑢 𝑛+1 = 𝑢 𝑛 +2
Such a formula is known as a term-to-term formula, because each term is defined using previous term(s).
We need to specify the first term ( 𝒖 𝟏 ), otherwise we don’t know what number to start the sequence with!
Contrast this with a ‘position-to-term’ formula for the same sequence, which is based on the position 𝑛 in the sequence.
𝑢 𝑛 =2𝑛+1

13. Term to term formulae
Can you come up with term-to-term formulae for the other sequences?
3, 5, 7, 9, 11, …
𝑢 1 =3, 𝑢 𝑛+1 = 𝑢 𝑛 +2
3, 6, 12, 24, 48, …
𝑢 1 =3, 𝑢 𝑛+1 =2 𝑢 𝑛 ?
3, 5, 8, 13, 21, … ?
𝑢 1 =3, 𝑢 2 =5 𝑢 𝑛+2 = 𝑢 𝑛+1 + 𝑢 𝑛
We need two starting terms because each term is based on the previous two.
We could have also used 𝑢 𝑛+1 = 𝑢 𝑛 + 𝑢 𝑛−1

14. More Examples
The amount of money 𝑚 (in pounds) in Dr Frost’s bank account after 𝑡 years is represented by:
𝑚 0 = 𝑚 𝑡+1 =1.1 𝑚 𝑡
What does the 1.1 represent in the second equation?
How much do I have after 2 years?
Show that the sequence forms a geometric progression. ?
The second line expresses that the amount the following year is 1.06 times the previous year. The 1.06 represents a 6% increase in money each year.
𝑚 1 =1.1× 𝑚 0 = 𝑚 2 =1.1× 𝑚 1 =60 500
So we have £ after 2 years.
The first three terms of the sequence are:
50 000, ,
This is a geometric progression because the terms have a constant ratio of 1.1. a b ? c ?
The key is identifying that we have a common ratio between terms.

15. Test Your Understanding1
Why do you think the sequence refers to the first term as 𝑝 0 rather than 𝑝 1 ?
Because 𝒑 𝟎 indicates the number of slugs ‘after 0 days’, i.e. the initial value. ?
119 ? 2
A sequence is defined as:
𝑥 1 =5, 𝑥 2 =2 𝑥 𝑛+2 =2 𝑥 𝑛+1 − 𝑥 𝑛
Determine 𝑥 5 .
𝒙 𝟑 =𝟐 𝒙 𝟐 − 𝒙 𝟏 =−𝟏 𝒙 𝟒 =𝟐 𝒙 𝟑 − 𝒙 𝟐 =−𝟑 𝒙 𝟓 =𝟐 𝒙 𝟒 − 𝒙 𝟑 =−𝟒 ?

16. Exercise 2
For each of the following determine:
the position-to-term formula and
the term-to-term formula, stating any initial terms required.
e.g. For 3, 5, 7, 9, …
(a) 𝑢 𝑛 =2𝑛+1,
(b) 𝑢 𝑛 = 𝑢 𝑛−1 +2, 𝑢 1 =3 3
Find the 3rd term for each of the following sequences:
𝑢 1 =4, 𝑢 𝑛+1 = 𝑢 𝑛 → 𝟏𝟓 𝑢 1 =5, 𝑢 𝑛+1 =3 𝑢 𝑛 → 𝟒𝟓 𝑢 1 =10, 𝑢 𝑛+1 = 𝑢 𝑛 → 𝟐.𝟓 𝑢 1 =3, 𝑢 𝑛+1 =2 𝑢 𝑛 → 𝟏𝟖 1 ? ? ? ?
0, 1, 2, 3, 4…
𝒖 𝒏 =𝒏−𝟏,
𝒖 𝒏 = 𝒖 𝒏−𝟏 +𝟏, 𝒖 𝟏 =𝟎
6, 11, 16, 21,…
𝒖 𝒏 =𝟓𝒏+𝟏,
𝒖 𝒏 = 𝒖 𝒏−𝟏 +𝟓, 𝒖 𝟏 =𝟔
4, 16, 64, 256…
𝒖 𝒏 = 𝟒 𝒏 ,
𝒖 𝒏 =𝟒 𝒖 𝒏−𝟏 , 𝒖 𝟏 =𝟒
3, 6, 12, 24,…
𝒖 𝒏 = 𝟑 𝟐 × 𝟐 𝒏 =𝟑 𝟐 𝒏−𝟏 ,
𝒖 𝒏 =𝟐 𝒖 𝒏−𝟏 , 𝒖 𝟏 =𝟑 i ? ?
[Edexcel Specimen Papers Set 1, Paper 2H Q21]
The number of bees in a beehive at the start of year 𝑛 is  𝑃 𝑛 . The number of bees in the beehive at the start of the following year is given by
𝑃 𝑛+1 =1.05 𝑃 𝑛 −250
At the start of 2015 there were 9500 bees in the beehive.
How may bees will there be in the beehive at the start of 2018?
10170 bees 2 ii ? ?
iii ? ? iv ? ? ?

17. Exercise 2
[Edexcel New SAMs Paper 3F Q20b, Paper 3H Q3b] Here are the first six terms of a Fibonacci sequence.
      1    1    2    3    5    8
The rule to continue a Fibonacci sequence is,
        the next term in the sequence is the sum of the two previous terms
The first three terms of a different Fibonacci sequence are
     𝑎, 𝑏, 𝑎+𝑏
Find the 6th term of this sequence, in terms of 𝑎 and 𝑏. Simplify your answer.
Answer: 𝟑𝒂+𝟓𝒃 4 N1
Find the position-to-term formula of the sequence:
1,−2,3,−4,5,…
𝒖 𝒏 =𝒏 −𝟏 𝒏−𝟏 ?
[JMO 2008 B6] In a sequence of positive integers, each term is larger than the previous term. Also, after the first two terms, each term is the sum of the previous two terms.
The eighth term of the sequence is 390. What is the ninth term?
Solution: 631 N2 ?
Find the term-to-term and position-to-term formulae.
8, 24, 72, 216, 648…
𝒖 𝒏 = 𝟖 𝟑 𝟑 𝒏 =𝟖 𝟑 𝒏−𝟏 ,
𝒖 𝒏 =𝟑 𝒖 𝒏−𝟏 , 𝒖 𝟏 =𝟖
−1, +1, −1, +1, −1…
𝒖 𝒏 = −𝟏 𝒏 ,
𝒖 𝒏 =− 𝒖 𝒏−𝟏 , 𝒖 𝟏 =−𝟏
0.5, 0.25, 0.125, …
𝒖 𝒏 = 𝟏 𝟐 𝒏 = 𝟐 −𝒏 ,
𝒖 𝒏 = 𝒖 𝒏−𝟏 𝟐 , 𝒖 𝟏 =𝟎.𝟓 5 i ? ? ii ? ? ?
iii ? ?