# 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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