====== Question 3 and 4 Exercise 4.1 ======
Solutions of Question 3 and 4 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 3(i)====
Write down the nth term of the sequence as suggested by the pattern.
$\dfrac{1}{2}, \dfrac{2}{3} \dfrac{3}{4}, \dfrac{4}{5}, \ldots$
====Solution====
We can reform the given sequence to pick the pattern of the sequence as:
$$\dfrac{1}{1+1}, \dfrac{2}{2+1}, \dfrac{3}{3+1}, \dfrac{4}{4+1},...$$
Hence the general term of the sequence is $\dfrac{n}{n+1}$.

=====Question 3(ii)====
Write down the nth term of the sequence as suggested by the pattern.
$2,-4,6,-8,10, \ldots$
====Solution====
We can reform the given sequence to pick the pattern of the sequence as:
\begin{align}
&(-1)^2 \cdot 2 \cdot 1, (-1)^3 \cdot 2 \cdot 2, (-1)^4 \cdot 2 \cdot 3, (-1)^5 \cdot 2 \cdot 4, \ldots \\
&(-1)^{1+1} \cdot 2 \cdot 1, (-1)^{2+1} \cdot 2 \cdot 2, (-1)^{3+1} \cdot 2 \cdot 3, (-1)^{4+1} \cdot 2 \cdot 4, \ldots
\end{align}
Hence the general term of the sequence is $(-1)^{n+1} 2 n$.

=====Question 3(iii)====
Write down the nth term of the sequence as suggested by the pattern.
$1,-1,1,-1, \ldots$
====Solution====
We can reform the give sequence to pick the pattern of the sequence as:
\begin{align}(-1)^2,(-1)^3,(-1)^4,(-1)^5, \ldots, (-1)^{n+1}, \ldots \end{align}
Hence the general term of the sequence is $(-1)^{n+1}$.

=====Question 4(i)=====
Write down the first five terms of each sequence defined recursively.
$a_1=3$, $a_{n+1}=5-a_n$.
====Solution====
Given
$$a_1=3, a_{n+1}=5-a_n.$$
For $n=1$
\begin{align}a_{1+1}&=5-a_1\\ 
\Rightarrow a_2&=5-3=2\end{align}
For $n=2$
\begin{align}a_{2+1}&=5-a_2\\
\Rightarrow a_3&=5-2=3\end{align}
For $n=3$
\begin{align}a_{3+1}&=5-a_3\\
\Rightarrow a_4&=5-3=2\end{align}
For $n=4$
\begin{align}a_{4+1}&=5-a_4\\
\Rightarrow a_5&=5-2=3\end{align}
Hence the first five terms are $3,2,3,2,3$.

=====Question 4(ii)=====
Write down the first five terms of each sequence detined recursively.
$a_1=3, a_{n+1}=\dfrac{a_n}{n}$
====Solution====
Given
$$a_1=3, a_{n+1}=\frac{a_n}{n}$$
For $n=1$
\begin{align}&a_{1+1}=\dfrac{a_1}{1} \\
\implies &a_2=\dfrac{3}{1}=3.\end{align}
For $n=2$
\begin{align}&a_{2+1}=\dfrac{a_2}{2} \\
\implies &a_3=\dfrac{3}{2}.\end{align}
For $n=3$
\begin{align}&a_{3+1}=\dfrac{a_3}{3} \\
\implies &a_4=\dfrac{\dfrac{3}{2}}{3}=\dfrac{1}{2}.\end{align}
For $n=4$
\begin{align}&a_{4+1}=\dfrac{a_4}{4} \\
\implies &a_5=\dfrac{\dfrac{1}{2}}{4}=\dfrac{1}{8}.\end{align}
Hence the first five terms are $3, 3, \dfrac{3}{2}, \dfrac{1}{2}, \dfrac{1}{8}$.




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