Question 4 Exercise 4.5

Solutions of Question 4 of Exercise 4.5 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.

Convert each decimal to common fraction $0 . \overline{8}$

We can write $$0 . \overline{8}=0.888888 \ldots$$
That can be written in the form
$$\begin{align}0 . \overline{8}&=0.8+0.08+0.008 \div 0.0008+ \ldots\\ \text { or } 0 . \overline{8}&=0.8+(0.1)(0.8) +(0.1)^2(0.8)+\ldots \ldots \ldots \ldots .(\mathrm{i})\end{align}$$
It is geometric series with $$a_1=0.8, \quad r=0.1$$
We can find the infinite sum as:
$$S_{\infty}=\dfrac{a_1}{1-r}=\dfrac{0.8}{1-0.1}=\dfrac{0.8}{0.9}=\dfrac{8}{9}$$
Hence putting $S_{\infty}$ in (i), we get $$0 . \overline{8}=\dfrac{8}{9}$$ .

Convert each decimal to common fraction $1 . \overline{63}$

Since $$\begin{align}1 . \overline{63}&=1+0.63+0.0063+0.000063 +\ldots \\ \text { or } 1 . \overline{63}&=1+[0.63+ (0.01)(0.63)-(0.01)^2 0.63+\ldots \ldots \text { (i) }\end{align}$$ The serics in braces is infinite gcometric series with $a_1=0.63, \quad r=0.01<1$.
Therefore the sum exists and is given by
$$\begin{align}S_{\infty}&=\dfrac{a_1}{1-r}\\ &=\dfrac{0.63}{1-0.01}\\ &=\dfrac{0.63}{0.99} \\ \Rightarrow S_{\infty}&=\dfrac{7}{11} \ldots \ldots \ldots \ldots . . . \text { (ii) }\end{align}$$ Putting (ii) in (i), we get
$$1.63=1+\dfrac{7}{11}=\dfrac{18}{11} \text {. }$$

Convert each decimal to common fraction $2 . \overline{15}$

Since $$\begin{align}2 . \overline{15}&=2+0.15+0.0015+0.000015+\ldots\\ \text{or we can write}\\ 2 . \overline{15}&=2+[0.15+(0.01) 0.15+ .(0.01)^2 0.15+\ldots ..]\end{align}$$
The sequence in bracket is infinite geometric series with $a_1=0.15, r=0.01<1$.
Thus the sum of the given series exists and given by $$S_{\infty}=\dfrac{a_1}{1-r}$$ ,
putting $a_1, r$ we get
$$S_{\infty}=\dfrac{0.15}{1-0.01}=\dfrac{0.15}{0.99}=\dfrac{5}{33}$$ putting in (i) then
$$2 . \overline{15}=2+\dfrac{5}{33}=\dfrac{66+5}{33}=\dfrac{71}{33}$$
is the desired common fraction.

Convert each decimal to common fraction $0 . \overline{123}$

Since $$0 . \overline{123}=0.123+0.000123 +0.000000123+\ldots$$ $$\begin{align}\text{or}\quad 0 . \overline{123}&=0.123+(0.001)(0.123)+(0.001)^2(0.123)+\ldots\end{align}$$ The series on the R.H.S is geometric series with
$$\begin{align}a_1&=0.123, \quad r=0.001<1 \end{align}$$ Thus the sum exists and is given by
$$\begin{align}S_{\infty}&=\dfrac{a_1}{1-r}\end{align}$$ putting $a_1, r$ we get
$$\begin{align}S_{\infty}&=\dfrac{0.123}{1-0.001}\\ &=\dfrac{0.123}{0.999} \\ S_{\infty}&=\dfrac{123}{999}\\ &=\dfrac{41}{333}\end{align}$$ Thus the required common fraction is: $\quad 0 . \overline{123}=\dfrac{41}{333}$

< Question 3 Question 5 & 6 >