# Calculate the sum of the following sequence:

0.45, 0.015, 0.0005,…

Last updated date: 25th Mar 2023

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Answer

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Hint: Try to find a pattern. What kind of series is it? A geometric series, but take care that the given series is infinite, so don’t use the finite series formula. So use the sum of terms of the infinite series formula.

According to the question, we have to find the value of S, i.e.,

S = 0.45 + 0.015 + 0.005………..

This is an infinite sum.

Let, $S={{a}_{1}}+{{a}_{2}}+{{a}_{3}}........$

Where, ${{a}_{1}}=0.45,{{a}_{2}}=0.015,{{a}_{3}}=0.005$ and so on

Now, divide the first two term, we get

$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{0.45}{0.015}=30$

Now divide the second and the third term, we get

$\dfrac{{{a}_{2}}}{{{a}_{3}}}=\dfrac{0.015}{0.005}=30$

Here we can observe that the series has a common ratio, that is $r=\dfrac{1}{30}$ .

Hence the given series is infinite geometric progression.

So the given infinite series can be written in the form of,

$S=a+ar+a{{r}^{2}}.......$

Where ‘a’ is the first term and ‘r’ is the common ratio, i.e,

$S=0.45+\dfrac{0.45}{\left( 30 \right)}+\dfrac{0.45}{{{\left( 30 \right)}^{2}}}........$

We know the sum of infinite geometric progression is given by the formula,

$S=\dfrac{a}{1-r}$

Substituting the values of ‘a’ and ‘r’, we get

$S=\dfrac{0.45}{1-\dfrac{1}{30}}$

Taking the LCM in the denominator, we get

$\begin{align}

& S=\dfrac{0.45}{\dfrac{30-1}{30}} \\

& \Rightarrow S=\dfrac{0.45}{\dfrac{29}{30}} \\

\end{align}$

We know division is also written as the multiplication of the inverse of the number, so the above equation can be written as,

\[\begin{align}

& S=0.45\times \dfrac{30}{29} \\

& \Rightarrow S=0.45\times \dfrac{30}{29} \\

& S=\dfrac{13.5}{29}=0.4655 \\

\end{align}\]

So the sum of 0.45, 0.015, 0.0005,… is 0.4655.

Note: There is no need to get confused by decimals. Also, students often think the infinite sum is infinite. This is not true, students also take $\dfrac{{{a}_{1}}}{{{a}_{2}}}$ to be r, when $r=\dfrac{{{a}_{2}}}{{{a}_{1}}}.$

Sometimes student get confused with the infinite series and apply the formula of finite geometric progression formula, i.e.

$S=\dfrac{a(1-{{r}^{n}})}{1-r}$

This will give the wrong answer.

According to the question, we have to find the value of S, i.e.,

S = 0.45 + 0.015 + 0.005………..

This is an infinite sum.

Let, $S={{a}_{1}}+{{a}_{2}}+{{a}_{3}}........$

Where, ${{a}_{1}}=0.45,{{a}_{2}}=0.015,{{a}_{3}}=0.005$ and so on

Now, divide the first two term, we get

$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{0.45}{0.015}=30$

Now divide the second and the third term, we get

$\dfrac{{{a}_{2}}}{{{a}_{3}}}=\dfrac{0.015}{0.005}=30$

Here we can observe that the series has a common ratio, that is $r=\dfrac{1}{30}$ .

Hence the given series is infinite geometric progression.

So the given infinite series can be written in the form of,

$S=a+ar+a{{r}^{2}}.......$

Where ‘a’ is the first term and ‘r’ is the common ratio, i.e,

$S=0.45+\dfrac{0.45}{\left( 30 \right)}+\dfrac{0.45}{{{\left( 30 \right)}^{2}}}........$

We know the sum of infinite geometric progression is given by the formula,

$S=\dfrac{a}{1-r}$

Substituting the values of ‘a’ and ‘r’, we get

$S=\dfrac{0.45}{1-\dfrac{1}{30}}$

Taking the LCM in the denominator, we get

$\begin{align}

& S=\dfrac{0.45}{\dfrac{30-1}{30}} \\

& \Rightarrow S=\dfrac{0.45}{\dfrac{29}{30}} \\

\end{align}$

We know division is also written as the multiplication of the inverse of the number, so the above equation can be written as,

\[\begin{align}

& S=0.45\times \dfrac{30}{29} \\

& \Rightarrow S=0.45\times \dfrac{30}{29} \\

& S=\dfrac{13.5}{29}=0.4655 \\

\end{align}\]

So the sum of 0.45, 0.015, 0.0005,… is 0.4655.

Note: There is no need to get confused by decimals. Also, students often think the infinite sum is infinite. This is not true, students also take $\dfrac{{{a}_{1}}}{{{a}_{2}}}$ to be r, when $r=\dfrac{{{a}_{2}}}{{{a}_{1}}}.$

Sometimes student get confused with the infinite series and apply the formula of finite geometric progression formula, i.e.

$S=\dfrac{a(1-{{r}^{n}})}{1-r}$

This will give the wrong answer.

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