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Digital Lessons

# Exercise Set D

## Question 1

Consider the following sequence:

\begin{align}5,9,13,17,21…\end{align}

(a) Show that this sequence is linear.

(b) What is the general term for this sequence?

(c) What is the sum of the first $$30$$ terms of this sequence?

(b) $$T_n = 4n+1$$

(c) $$1{,}890$$

Solution

(a) As the difference between successive terms is a constant value of $$4$$, the sequence is linear.

(b)

\begin{align}T_n = pn+q\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5=p(1)+q\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p+q=5\end{align}

and

\begin{align}9=p(2)+q\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2p+q=9\end{align}

We now have two equations for two unknowns.

\begin{align}p+q=5\end{align}

\begin{align}2p+q=9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p&=9-5\\&=4\end{align}

and therefore

\begin{align}4+q=5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q&=5-4\\&=1\end{align}

Therefore, the general term is

\begin{align}T_n = 4n+1\end{align}

(c)

\begin{align}S_n &= \frac{n}{2}[2a+(n-1)d]\\&=\frac{30}{2}[2(5)+(30-1)(4)]\\&=15(10+116)\\&=1{,}890\end{align}

Video Walkthrough

## Question 2

Consider the following sequence:

\begin{align}\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2},…\end{align}

(a) Show that this sequence is linear.

(b) What is the general term for this sequence?

(c) What is the sum of the first $$100$$ terms of this sequence?

(b) $$T_n = \dfrac{n}{8}$$

(c) $$\dfrac{2{,}525}{4}$$

Solution

(a) As the difference between successive terms is a constant value of $$\dfrac{1}{8}$$, the sequence is linear.

(b)

\begin{align}T_n = pn+q\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{8}=p(1)+q\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p+q=\frac{1}{8}\end{align}

and

\begin{align}\frac{1}{4}=p(2)+q\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2p+q=\frac{1}{4}\end{align}

We now have two equations for two unknowns.

\begin{align}p+q=\frac{1}{8}\end{align}

\begin{align}2p+q=\frac{1}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p&=\frac{1}{4}-\frac{1}{8}\\&=\frac{1}{8}\end{align}

and therefore

\begin{align}\frac{1}{8}+q=\frac{1}{8}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q&=\frac{1}{8}-\frac{1}{8}\\&=0\end{align}

Therefore, the general term is

\begin{align}T_n = \frac{n}{8}\end{align}

(c)

\begin{align}S_n &= \frac{n}{2}[2a+(n-1)d]\\&= \frac{100}{2}\left[2\left(\dfrac{1}{8}\right)+(100-1)\left(\dfrac{1}{8}\right)\right]\\&=50\left(\frac{2}{8}+\frac{99}{8}\right)\\&=\frac{2{,}525}{4}\end{align}

Video Walkthrough

## Question 3

The general term $$T_n$$ of a particular sequence is given by

\begin{align}T_n = 3n+5\end{align}

Show that that the sum of the first $$n$$ terms of this sequence is $$\dfrac{3n^2+13n}{2}$$.

Solution

\begin{align}a&=T_1\\&=3(1)+5\\&=8\end{align}

and

\begin{align}T_2&=3(2)+5\\&=11\end{align}

\begin{align}\downarrow\end{align}

\begin{align}d&=T_2-T_1\\&=11-8\\&=3\end{align}

and therefore

\begin{align}S_n &= \frac{n}{2}[2a+(n-1)d]\\&=\frac{n}{2}[2(8)+(n-1)(3)]\\&=\frac{n}{2}(16+3n-3)\\&=\frac{n}{2}(13+3n)\\&=\frac{3n^2+13n}{2}\end{align}

as required.

Video Walkthrough

## Question 4

Consider the following sequence:

\begin{align}2,6,18,54,…\end{align}

(a) Show that this sequence is geometric.

(b) What is the general term for this sequence?

(c) What is the sum of the first $$10$$ terms of this sequence?

(b) $$T_n =2(3^{n-1})$$

(c) $$59{,}048$$

Solution

(a) As the ratio between each successive term is a constant value of $$3$$, the sequence is geometric.

(b)

\begin{align}T_n &= ar^{n-1}\\&=2(3^{n-1})\end{align}

(c)

\begin{align}S_n = \frac{a(1-r^n)}{1-r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S_{10} &= \frac{2(1-3^{10})}{1-3}\\&=59{,}048\end{align}

Video Walkthrough

## Question 5

Consider the following sequence:

\begin{align}\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},…\end{align}

(a) Show that this sequence is geometric.

(b) What is the general term for this sequence?

(c) What is the sum of the first $$10$$ terms of this sequence?

(d) What does the corresponding series converge to?

(b) $$T_n =\dfrac{1}{2}^n$$

(c) $$\dfrac{1{,}023}{1{,}024}$$

(d) $$1$$

Solution

(a) As the ratio between each successive term is a constant value of $$\dfrac{1}{2}$$, the sequence is geometric.

(b)

\begin{align}T_n &= ar^{n-1}\\&=\frac{1}{2}\left(\frac{1}{2}\right)^{n-1}\\&=\frac{1}{2}^n\end{align}

(c)

\begin{align}S_n = \frac{a(1-r^n)}{1-r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S_{10} &= \frac{\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^{10}\right]}{1-\frac{1}{2}}\\&=\frac{1{,}023}{1{,}024}\end{align}

(d)

\begin{align}S_\infty &= \frac{a}{1-r}\\&= \frac{\frac{1}{2}}{1-\frac{1}{2}}\\&=1\end{align}

Video Walkthrough