In a lot of private schools, and even some state schools in the UK, the students will do the following:
This makes a lot of sense to do, because a lot of the Further Maths has assumed knowledge from the A-Level maths course. In my school though, due to the way Maths is taught in my area, you have to take both Further Maths and A-Level Maths at the same time. But then a problem appears, if you want to take some of the University Admissions Tests (i.e. the MAT), they are taken at the end of your Year 12 (with the exception of the STEP paper, which is taken at the end of Year 13). So the questions in these papers assume you have learnt the full mathematics course, but you haven't yet. So, this here is going to try and make sense of sequences, because I too am trying to learn them by myself.
We are going to start with some knowledge from the GCSEs. I recommend you read pages 2 & 4 from BBC Bitesize. But I will also explain it for you here:
1st 2nd 3rd 4th 5th
+10 +10
┌──┐ ┌──┐
│ │ │ │
20 30 40 50 60
│ │ │ │
└──┘ └──┘
+10 +10
And then if you look at the 10 times table, you can see that this has the formula: \(10n + 10\)
Now that you have remembered how to do this, we can look at an alternative way to do this.
Can you see how the starting term is \(20\). And from that you add \(10\) each time. So, could we not say instead, that this is:
\(20+10 \times 0, 20+10 \times 1, 20+10 \times 2, 20+10 \times 3, 20+10 \times 4\)
This can be generalised as: \(20 + 10(n-1)\). (If you expand and simplify, that is \(10n + 10\), the same formula from earlier)
To differentiate this from the \(n^{th} \text{ term}\) formula, we use the following notation: $$ \Large a_n = 20 + 10(n-1) $$
Or, more commonly, in its general form:
$$ \Large a_n = a_1 + (n - 1)d $$
In english that means: The nth term is equal to the first term plus "n - 1" lots of the multiplier d.
Now that we have this formula, we need to look at another important formula, which is the summation formula. Most of the questions related to Arithmetic Series use this formula so I recommend you learn it off by heart.
Let's take the above example of the series: 20 30 40 50 60. The summation of this is:
200. Therefore, we can say that the summation of the first 5 times of the sequence \(a_n = 20 +
10(n-1)\) is 200.
But how could we do this algebraically? Let's define a new variable \(S_k\) to be the summation of the first \(k\) terms of our formula. Using the sigma notation that we've learnt from the Further Maths Core Pure, we can say:
$$ \Large S_k = \sum_{n=1}^{k} a_n $$
We can then substitute our formula for \(a_n\):
$$ \Large S_k = \sum_{n=1}^{k} a_1 + (n - 1)d $$
We can then create two separate equations:
$$ \Large S_k = \sum_{n=1}^{k} a_1 + \sum_{n=1}^{k} (n - 1)d $$
Simplifying using the summation rules, this can be rewritten as:
$$ \Large \begin{align} S_k &= k \times a_1 + d \times \sum_{n=1}^{k} (n - 1) \\ S_k &= k \times a_1 + d \times \left ( \sum_{n=1}^{k} n - \sum_{n=1}^{k} 1 \right) \\ S_k &= k \times a_1 + d \times \left ( \frac{k(k+1)}{2} - k \right) \\ S_k &= k \times a_1 + \frac{1}{2}d \left ( k(k+1) - 2k \right) \\ S_k &= k \times a_1 + \frac{1}{2}dk \left ( k - 1 \right) \\ S_k &= k \times \left [a_1 + \frac{1}{2}d \left ( k - 1 \right) \right] \\ S_k &= \frac{1}{2} k \times \left [2a_1 + d \left ( k - 1 \right) \right] \\ \end{align} $$
Voila! We have a new formula! Let's input the values we had earlier. We wanted the first 5 terms, so \(k = 5\). Our starting value was \(20\), so \(a_1 = 20\). And our "increaser" (the bit we were adding on) was 10. So \(d = 10\).
$$ \Large \begin{align} S_5 &= \frac{1}{2} 5 \times \left [2(20) + 10 \left ( 5 - 1 \right) \right] \\ S_5 &= \frac{1}{2} 5 \times \left [40 + 40 \right] \\ S_5 &= \frac{1}{2} 5 \times \left [80 \right] \\ S_5 &= 5 \times 40 \\ S_5 &= 200 \\ \end{align} $$
It works!
First we'll start off with GCSE knowledge (BBC Bitesize.) Let's take the following geometric series: 3, 9, 27, 81. We can find out the common multiplier (if you read the bitesize page) by dividing each term by the term next to it. $$ \frac{9}{3} = \frac{27}{9} = \frac{81}{27} = 3 $$
Therefore, we know the multiplier is 3 and we know the first term of the sequence is 3. Let's write out what is actually happening here:
$$ 3 \times 1, 3 \times 3, 3 \times 3 \times 3, 3 \times 3 \times 3 \times 3 $$
Let's convert these into indices and see what we get:
$$ 3 \times 3^0, 3 \times 3^1, 3 \times 3^2, 3 \times 3^3 $$
Can you see how there is a general pattern here? This is often written as:
$$ \Large a_n = a_1 \times r ^ {n - 1} $$
So for our pattern, the multiplier is 3, so \(r = 3 \) and the first term is 3, so \(a_1 = 3\):
$$ \Large a_n = 3 \times 3^{n-1} $$
Let's look at Q7 from D513/11:
$$ \begin{align} \text{The first three terms of an arithmetic progression are } p, q \text{ and } p^2 \text{ respectively, where } p < 0. \\ \text{The first three terms of a geometric progression are } p, p^2 \text{ and } q \text{ respectively.} \\ \text{Find the sum of the first 10 terms of the arithmetic progression} \end{align} $$
Let's rewrite the arithmetic progression first:
$$ a_n = p + (n - 1)d $$
$$ \begin{align} p &= p + 0d \\ q &= p + d \\ p^2 &= p + 2d \end{align} $$
Now let's look at the geometric sequence:
$$ a_n = p \times r ^ {n - 1} $$
$$ \begin{align} p &= p \times r^0 \\ p^2 &= p \times r^1 \\ q &= p \times r^2 \end{align} $$
Because we know for a fact, that p can't be 0 (it's stated in the question). We now know the value of r, because: \(p^2 = p \times r\). So the only value r can be, is p.
This means we now know that \(q = p \times p^2 = p^3\)
Substituting that back into the equation \(q = p + d\). We now have a set of simultaneous equations:
$$ \begin{align} p^2 = p + 2d \\ p^3 = p + d \end{align} $$
We can then remove the d from this equation:
$$ \begin{align} p^2 = p + 2d \\ - 2\times [p^3 = p + d] \\ \\ p^2 - p^3 &= -p \\ 0 &=p^3 -p^2 -p \\ p^3 -p^2 -p &= 0\\ p(p^2 -p^1 - 1) &= 0\\ p(2p+1)(p-1) &= 0\\ p = 0, p = -\frac{1}{2}, p = 1 \end{align} $$
We are told that \(p < 0 \), so we can say \(p=-\frac{1}{2}\)
Our next step is to work out what \(d\) is, given that we have the value of p, we can simply substitute it.
$$ \begin{align} p^2 &= p + 2d \\ (-\frac{1}{2})^2 &= -\frac{1}{2} + 2d \\ \frac{1}{4} + \frac{1}{2} &= 2d \\ \frac{3}{4} &= 2d \\ \frac{3}{8} &= d \end{align} $$
We can now write the first algorithmic progression as:
$$ \Large \begin{align} a_n = a_1 + (n-1)d \\ a_n = p + (n-1) \times \frac{3}{8} \end{align} $$
We can then use the \(S_k\) rule
$$ \Large \begin{align} S_k &= \frac{1}{2}k(a_1 + a_k) \\ S_k &= \frac{1}{2}k \left(p + p + (k-1)\frac{3}{8} \right) \\ S_{10} &= \frac{1}{2}10 \left(2 \times \frac{-1}{2} + 9\times\frac{3}{8} \right) \\ S_{10} &= 5\left(-1 + \frac{27}{8}\right) \\ S_{10} &= 5\left (\frac{19}{8} \right) \\ S_{10} &= \frac{95}{8} \\ \end{align} $$