Question

Use Geometric progression to express \[0.\bar 3\] and $1.2\bar 3$ as $\dfrac{p}{q}$ form.

Answer 1

Hint: First express the given expression in expanded form and then change the decimals in fraction form. After that, you can find a GP series. Apply the formula given below for the sum of GP series to convert the obtained result in $\dfrac{p}{q}$ form.
Sum of the infinite term of GP series $ = \dfrac{a}{{1 - r}}$
Here, $a$ is the first term of the series and $r$is the common ratio of the series, which is the ratio of consecutive terms.

Complete step by step solution:
We have given \[0.\bar 3\] and we have to express it in the form of fraction, $\dfrac{p}{q}$ form.
Let us assume that $x = 0.\bar 3$, then we can also express \[0.\bar 3\]as:
\[0.\bar 3 = 0.333333 \cdot \cdot \cdot \cdot \cdot \]
Now, we express in expanded form:
\[0.\bar 3 = 0.3 + 0.03 + 0.003 + \cdot \cdot \cdot \cdot \cdot \]
We can express the decimals in the form of fractions.
\[0.\bar 3 = \dfrac{3}{{10}} + \dfrac{3}{{100}} + \dfrac{3}{{1000}} + \cdot \cdot \cdot \cdot \cdot \]
Now, we have the first term of the series as:
$a = \dfrac{3}{{10}}$
Now, we find the common ratio of the series by taking the ratio of the consecutive terms.
$r = \dfrac{{\dfrac{3}{{100}}}}{{\dfrac{3}{{10}}}}$
$r = \dfrac{1}{{10}}$
Now, we have the first term and the common ratio of the obtained series, which are given below:
$a = \dfrac{3}{{10}}$ and $r = \dfrac{1}{{10}}$
We now that the series in the Geometric Progression and the sum of infinite terms of the geometric series is given as:
${S_\infty } = \dfrac{a}{{1 - r}}$
Substitute $\left( {\dfrac{3}{{10}}} \right)$for $a$and $\left( {\dfrac{1}{{10}}} \right)$for $d$ into the above expression:
${S_\infty } = \dfrac{{\left( {\dfrac{3}{{10}}} \right)}}{{1 - \left( {\dfrac{1}{{10}}} \right)}}$
Now, we simplify the expression to find the sum up to infinite terms:
${S_\infty } = \dfrac{{\left( {\dfrac{3}{{10}}} \right)}}{{\left( {\dfrac{{10 - 1}}{{10}}} \right)}} = \dfrac{{\left( {\dfrac{3}{{10}}} \right)}}{{\left( {\dfrac{9}{{10}}} \right)}}$
${S_\infty } = \dfrac{3}{10} \times \dfrac{10}{9} = \dfrac{1}{3}$
Hence,\[0.\bar 3\] can be expressed in $\dfrac{p}{q}$ form as $\dfrac{1}{3}$ using the Geometric Progression.
Let us assume that $x = 1.2\bar 3$, then we can also express \[1.2\bar 3\] as:
\[1.2\bar 3 = 1.2333333 \cdot \cdot \cdot \cdot \cdot \]
Now, we express in expanded form:
\[1.2\bar 3 = 1.2 + + 0.03 + 0.003 + 0.0003 \cdot \cdot \cdot \cdot \cdot \]
We can express the decimals in the form of fractions.
\[1.2\bar 3 = 1.2 + \dfrac{3}{{100}} + \dfrac{3}{{1000}} + \dfrac{3}{{10000}} \cdot \cdot \cdot \cdot \cdot \]
Break the expanded form as:
\[1.2\bar 3 = 1.2 + \left( {\dfrac{3}{{100}} + \dfrac{3}{{1000}} + \dfrac{3}{{10000}} \cdot \cdot \cdot \cdot \cdot } \right)\]
We can notice that \[\left( {\dfrac{3}{{100}} + \dfrac{3}{{1000}} + \dfrac{3}{{10000}} \cdot \cdot \cdot \cdot \cdot } \right)\] is in Geometric progression.
Now, we have the first term of the series as:
$a = \dfrac{3}{{100}}$
Now, we find the common ratio of the series by taking the ratio of the consecutive terms.
$r = \dfrac{{\dfrac{3}{{1000}}}}{{\dfrac{3}{{100}}}}$
$r = \dfrac{1}{{10}}$
Now, we have the first term and the common ratio of the obtained series, which are given below:
$a = \dfrac{3}{{100}}$and$r = \dfrac{1}{{10}}$
We now that the series in the Geometric Progression and the sum of infinite terms of the geometric series is given as:
${S_\infty } = \dfrac{a}{{1 - r}}$
Then the equation becomes:
\[1.2\bar 3 = 1.2 + \dfrac{a}{{1 - r}}\]
Substitute $\left( {\dfrac{3}{{100}}} \right)$for $a$and $\left( {\dfrac{1}{{10}}} \right)$ for $d$ into the above expression:
$1.2\bar 3 = 1.2 + \dfrac{{\left( {\dfrac{3}{{100}}} \right)}}{{1 - \left( {\dfrac{1}{{10}}} \right)}}$
Now, we simplify the expression to find the sum up to infinite terms:
$1.2\bar 3 = 1.2 + \dfrac{{\left( {\dfrac{3}{{100}}} \right)}}{{\left( {\dfrac{{10 - 1}}{{10}}} \right)}} = 1.2 + \dfrac{{\left( {\dfrac{3}{{100}}} \right)}}{{\left( {\dfrac{9}{{10}}} \right)}}$
$1.2\bar 3 = 1.2 + \left( {\dfrac{3}{{100}} \times \dfrac{{10}}{9}} \right) = 1.2 + \dfrac{1}{{30}}$
$1.2\bar 3 = \dfrac{{36 + 1}}{{30}}$
$1.2\bar 3 = \dfrac{{37}}{{30}}$

Hence, $1.2\bar 3$ can be expressed in $\dfrac{p}{q}$ form as $\dfrac{{37}}{{30}}$ using the Geometric Progression.

Note: In the second part of the problem,$1.2\bar 3$. When we express it in the expanded form it contains the term$1.2$in the summation but this is not taken as the part of the geometric series.
\[1.2\bar 3 = 1.2 + \dfrac{3}{{100}} + \dfrac{3}{{1000}} + \dfrac{3}{{10000}} \cdot \cdot \cdot \cdot \cdot \]
Here, \[\left( {\dfrac{3}{{100}} + \dfrac{3}{{1000}} + \dfrac{3}{{10000}} \cdot \cdot \cdot \cdot \cdot } \right)\]is in geometric progression, because the ratio of each consecutive term is same.