Exercise Set D
These questions are considered Ordinary Level.
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?
(a)Â The answer is already in the question!
(b)Â \(T_n = 4n+1\)
(c) \(1{,}890\)
(a)Â 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}
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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?
(a)Â The answer is already in the question.
(b)Â \(T_n = \dfrac{n}{8}\)
(c)Â \(\dfrac{2{,}525}{4}\)
(a)Â 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]\\&=Â \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}
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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 \(\dfrac{3n^2+13n}{2}\).
The answer is already in the question!
\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.
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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?
(a) The answer is already in the question!
(b)Â \(T_n =2(3^{n-1})\)
(c)Â \(59{,}048\)
(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}
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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?
(a) The answer is already in the question!
(b)Â \(T_n =\dfrac{1}{2}^n\)
(c)Â \(\dfrac{1{,}023}{1{,}024}\)
(d)Â \(1\)
(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}
