APGP: Discussion and examples

Worksheet 1 Q2c

We will encounter the word “within” in a few different topics (very commonly in normal distribution), so it will be useful to know how to translate that into mathematical notation.

$x$ is within 5 units of $y$ $\Rightarrow$ ${|x-y| < 5}$
Alternatively, ${-5 < x-y < 5}$

So the translation for this part of the question, "$S_n$ to be within 0.1% of the sum to infinity", would be $\boxed{|S_n - S_\infty| < \frac{0.1}{100} S_\infty}$.

We will then apply our formulas and use either algebra or the GC (the table method is useful here as $n$ is an integer) to get the answer.

Algebraic type questions: worksheet 2 Q1

Algebraic type questions involve making use of the $u_n$, $S_n$ and $S_\infty$ formulas from our topic, forming equations and manipulating them. We will use the following example to illustrate a few common techniques:

Question A geometric series has common ratio $r$, and an arithmetic series has first term $a$ and common difference $d$, where $a$ and $d$ are non-zero. The first three terms of the geometric series are equal to the first, fourth and sixth terms respectively of the arithmetic series.

We set up our equations based on what the question says:

$$\begin{align} a &= a \\ a + 3d &= ar \\ a + 5d &= ar^2 \end{align}$$

Part (i)

(i) Show that ${3r^2 - 5r + 2 = 0}$

A very useful technique for AP questions is to subtract equations (because this helps us find the common difference $d$). As an aside, if we will like to find an expression for $r$, it is then useful to divide equations.

Let’s subtract our equations for this question:

$$\begin{align*} (2) - (1): && 3d &= ar-a \\ (3) - (1): && 5d &= ar^2-a \\ \end{align*}$$

We can then make $d$ the subject from each of the two and equate:
$$\frac{ar-a}{3} = \frac{ar^2-a}{5}$$

Rearrangement will then get us the answer ($a$ will cancel out in the process).

Part (ii)

(ii) Deduce that the geometric series is convergent and find, in terms of $a$, the sum to infinity.

Recall that

GP is convergent $\Rightarrow$ ${-1 < r < 1}$

We thus want to find $r$: which we do by solving the equation in (i). ${r=1}$ (as well as other cases like ${r=0}$ and ${d=0}$) are almost always rejected (do you know why?) so we have ${r=\frac{2}{3}}$ and we can make our conclusion.

Part (iii)

(iii) The sum of the first $n$ terms of the arithmetic series is denoted by $S$. Given that ${a > 0}$, find the set of possible values of $n$ for which $S$ exceeds $4a$.

Since we are interested in the AP, we will need the common difference $d$.
Recall in the derivation in part (i) we have found that ${d = \frac{ar-a}{3}}$. Substituting ${r=\frac{2}{3}}$ in gives us ${d = -\frac{1}{9}a}$.

We can then plug this into the $S_n$ formula to get

$$\frac{n}{2} \Big( 2a + (n-1) ( {\textstyle -\frac{1}{9}a} ) \Big) > 4a$$

which we can solve for $n$ (once again, $a$ will cancel out).

Showing type questions: worksheet 1 Q6

To show that a sequence is arithmetic, we will have to show that
$\displaystyle {u_n - u_{n-1} = \textrm{constant}}$

To show that a sequence is geometric, we will have to show that
$\displaystyle {\frac{u_n}{u_{n-1}}= \textrm{constant}}$

Being a “constant” typically means that “$n$” in our expression cancels out in the process.

Part (i)

(i) Given that $T_n$ represents the $n$th term of an arithmetic progression, show that the terms of a new sequence where the $n$th term is represented by ${U_n = \mathrm{e}^{T_n}}$ form as geometric progression.

Since $T_n$ is arithmetic, ${T_n = a + (n-1)d}$. Hence

$$\begin{align*} \frac{U_n}{U_{n-1}} &= \frac{\mathrm{e}^{T_n}}{\mathrm{e}^{T_{n-1}}} \\ &= \frac{\mathrm{e}^{a+(n-1)d}}{\mathrm{e}^{a+(n-2)d}} \\ &= e^d \quad \textrm{which is a constant} \end{align*}$$

so $U_n$ forms a geometric progression.

Part (ii)

(ii) Given that $X_n$ represents the $n$th term of a geometric progression, show that the terms of a new sequence where the $n$th term is represented by ${Y_n = \ln X_n}$ form as geometric progression.

Since $X_n$ is geometric, ${X_n = ar^{n-1}}$. Hence

$$\begin{align*} Y_n - Y_{n-1} &= \ln (X_n) - \ln(X_{n-1}) \\ &= \ln(ar^{n-1}) - \ln (ar^{n-2}) \\ &= \ln \left( \frac{ar^{n-1}}{ar^{n-2}} \right) \\ &= \ln r \quad \textrm{which is a constant} \end{align*}$$

so $Y_n$ forms a arithmetic progression.