There are several types of series that are studied at school. Words used are progression, sequence and series.
There is the progression, which is a sequence of numbers that may (or may not) have a pattern you can discern. Recommended technique (recommended by me) is to write down differences (and differences of the differences) until you see a pattern. You will find this explained in Formulae in the Lower School section.
Usually at school, mathematical progressions are either arithmetic or geometric. Pronunciation of the adjective arithmetic is arith-MEH-tick, where the noun is ArITH-mu-tick [Addition is an arithmetic (adj) process, one of the four base functions of arithmetic (noun)].
In music you might find progressions whose adjectives are omnibus, ragtime or backdoor; elsewhere in school it refers to any intended design of education, such as spiralling in teaching maths. What makes maths ‘boring’ for many is the revisiting of stuff you thought you’d already covered; this is called both progress and progression; they have their place and I am in two or more minds about the appropriateness of my (lack of) enthusiasm for the process of repetition. I think this stems from it being very hard to know where to stop without going too far for several (too many) members of a class. Maybe that’s a moan about streaming?
Sequence is the technical term (oops) well defined in mathematics; progression by comparison is vague. One difference is that terms of a sequence can usually be referenced (the 4th term, the 50th term).
A series is, technically, the sum of the terms of a sequence. In practice these words are often fudged in class so that the fine difference (a nicety) is lost. When handling arithmetic and geometric series at A-level it is understood that the sum of the series (which is wrong English) is what will be sought, when correct English would be to say “the series sum of the sequence”. We will let convention lie, i.e. stay as it is, largely wrong. It seems to me that ‘the sum of a series’ is made a tautology by that definition of series. In practice, we use the word series to mean ‘a sequence of numbers whose sum we will be expecting to find’.
An arithmetic series (or arithmetic progression, an AP) is a sequence of numbers that go up by a constant addition: 2 5 8 11 14 17 goes up by three. The theoretical version call the first term a and the difference d, so a a+d a+2d a+3d and so on.
This makes the nth term a + (n-1)d [and wouldn’t you have preferred to start counting at zero?]. To add up a series (a common event) up to and including the nth term, called Sn, is done by adding the end pair to make 2a+(n-1)d and then the next outermost pair does the same, and so on until either you finish or you have a middle term. the number of pairs is then n/2 and so
Sn = n/2 {2a+(n-1)d} where the curly brackets are used only for historic reasons.
Examples: Find the next two terms, and S10 and S50
1 1 4 7 10 13
2 11 14 17 20 23 I added ten to case1
3 4 16 28 40 52 I multiplied case1 by four
4 14 26 38 50 63 I multiplied case1 by four and added 10
5 44 56 68 80 86 I added 10 to case1 and then multiplied by four.
A geometric series is a sequence of numbers that go up by a constant multiplication: 2 6 18 54 162 starts at two and multiplies by three. The theoretical version calls the first term a and the multiplying ratio r so the sequence underlying the series is a ar ar² ar³ ar⁴ and so on.
This makes the nth term a rn-1 [and again you might prefer to start counting at zero]. To add up this series up to and including the nth term, again called Sn, write the infinite series (and that requires an appreciation of the Binomial series expansions). Suffice it here to simply observe that
because (1+x)-1 = 1 + x + x² + x³ + x⁴ + .........
so Sn = a + ar + ar² + ar³ + ar⁴ ..+ arn-1 = a (1 + r + r² + r³ + r⁴ ..+ rn-1)
= a ((1 + r + r² + r³ + r⁴ ..+......) - rⁿ (1 + r + r²+ r³ + r⁴ ..+......))
= a (1-rⁿ ) / (1-r) = a (rn -1) / (r-1)
Traditionally, we pick the version of the last line that is positive, the first one for r<1, obviously. If r is small enough, less than one, then the series converges to a value, rⁿ tends to zero as n tends to infinity, and the infinite sum S∞ has a value, S∞ = a/(1-r) |r|<1
Examples: Find the next two terms, and S10 and S∞ if that is appropriate.
6 1 4 16 64
7 1.1 4.4 17.6 70.4 Q6 x 1.1. S∞ is still not appropriate, r>1
8 100 20 4 0.8 This time you can find S∞
9 4 3.6 3.24 2.916
10 128 64 32 16 8
Questions then become inventive. These are FPM, Y11 standard. For more, see here.
11 The third term of an arithmetic series is 50 and the sum of the first 3 terms of the series is 51. Find the values of a & d for this series. Find the first value over 1000.
12 The third term of an geometric series is 16 and the sum of the first 3 terms of the series is 5. Find the values of a & r for this series. Find the value of n that first makes Sn over 1000. Find the different solution that gives a value for S∞
Questions also get more academic and abstract, like this one, of mine own, around scholarship standard, meaning it uses standard learning in a hard way, which is what extension papers should do, e.g. STEP and what was called S level.
13 The format of Q12 requires integer values. Let the third term be K and the sum of the first three terms be k+λ. Form a quadratic in r, k and λ. Establish the condition for this to factorise. By completing the square find integer solutions [including the one in Q12].
DJS 20130214
edits 20151210
Formatting of Question numbers has been odd on this page. I quite like it, but I don’t understand what is occurring and I can’t undo it. DJS 20160602.
Updates and repeated editing may have repaired some of that formatting. DJS 20171116
11 a+2d=50, 3/2 (2a+2d)=51 => a+d=17 => d=33, a=-16.
33n-16>1000 => n>1016/33=30.8 so 31st term, 1007.
12 ar2=16, a(1+r+r2)=5 => 5r2=16(1+r) =>(5r+4)(r-4)=0 so two solutions: (a,r)=(1,4) and (a,r)=(25, -4/5). When r=4, 4n>3001=> n>5.77 =>first n is 6. When r=-4/5 the series converges, so S∞ = a/(1-r) = 125/9.
13 λr2-kr-k=0 will factorise if the discriminant is a perfect square, say p2. Then
k2=4λk=p2=(k+2λ)2-4λ2 => (k+2λ)2- p2 = 4λ2 e.g. 4λ2 = 4x25, λ=5, k=16. Other solutions are found when 4λ2 = 4x(odd)2 = a2-b2 where a-b=2 and a,b= λ2±1. There are other solution sets, such as λ=10, k=9 where r=3/5 or -3/2 and λ=14, k=25 where r=5/2 or -5/7.
