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.

Write down the nth term of the sequence as suggested by the pattern. $\dfrac{1}{2}, \dfrac{2}{3} \dfrac{3}{4}, \dfrac{4}{5}, \ldots$

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 $\dfrac{n}{n+1}$.

Write down the nth term of the sequence as suggested by the pattern. $2,-4,6,-8,10, \ldots$

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 $(-1)^{n+1} 2 n$.

Write down the nth term of the sequence as suggested by the pattern. $1,-1,1,-1, \ldots$

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 $(-1)^{n+1}$.

Write down the first five terms of each sequence defined recursively. $a_1=3$, $a_{n+1}=5-a_n$.

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 $3,2,3,2,3$.

Write down the first five terms of each sequence detined recursively. $a_1=3, a_{n+1}=\dfrac{a_n}{n}$

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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