# The sum of the series \[\dfrac{5}{13}+\dfrac{55}{{{13}^{2}}}+\dfrac{555}{{{13}^{3}}}+.....\infty \] is

(a) \[\dfrac{65}{36}\]

(b) \[\dfrac{65}{32}\]

(c) \[\dfrac{25}{36}\]

(d) None of these

Last updated date: 18th Mar 2023

•

Total views: 303.3k

•

Views today: 7.82k

Answer

Verified

303.3k+ views

Hint: In this series, first of all, we will multiply \[\dfrac{1}{13}\] on both sides and then subtract this series from the original series. Then G.P would be obtained. Use the formula for the sum of G.P that is \[{{S}_{\infty }}=\dfrac{a}{1-r}\] to get the desired sum.

Complete step-by-step answer:

Here, we have to find the sum of the series \[\dfrac{5}{13}+\dfrac{55}{{{13}^{2}}}+\dfrac{555}{{{13}^{3}}}+.....\infty \].

Let us consider the series given in the question.

\[S=\dfrac{5}{13}+\dfrac{55}{{{13}^{2}}}+\dfrac{555}{{{13}^{3}}}+.....\infty ....\left( i \right)\]

By multiplying both the sides of the above expression by \[\dfrac{1}{13}\], we get,

\[\dfrac{S}{13}=\dfrac{5}{{{\left( 13 \right)}^{2}}}+\dfrac{55}{{{13}^{3}}}+\dfrac{555}{{{13}^{4}}}+.....\infty ....\left( ii \right)\]

Now, by subtracting equation (ii) from equation (i), we get,

\[S-\dfrac{S}{13}=\dfrac{5}{13}+\left( \dfrac{55}{{{13}^{2}}}-\dfrac{5}{{{\left( 13 \right)}^{2}}} \right)+\left( \dfrac{555}{{{13}^{3}}}-\dfrac{55}{{{13}^{3}}} \right)+\left( \dfrac{5555}{{{13}^{4}}}-\dfrac{555}{{{13}^{4}}} \right)+.....\infty \]

By simplifying the above equation, we get,

\[\dfrac{12S}{13}=\dfrac{5}{13}+\left( \dfrac{50}{{{\left( 13 \right)}^{2}}} \right)+\left( \dfrac{500}{{{13}^{3}}} \right)+\left( \dfrac{5000}{{{13}^{4}}} \right)+.....\infty \]

Now by dividing both the sides of the above equation by 5, we get,

\[\dfrac{1}{5}.\dfrac{12S}{13}=\dfrac{1}{13}+\dfrac{10}{{{\left( 13 \right)}^{2}}}+\dfrac{100}{{{13}^{3}}}+\dfrac{1000}{{{13}^{4}}}+.....\infty \]

Or we can also write the above equation as,

\[\dfrac{12S}{65}=\dfrac{1}{13}+\dfrac{10}{{{\left( 13 \right)}^{2}}}+\dfrac{{{\left( 10 \right)}^{2}}}{{{13}^{3}}}+\dfrac{{{\left( 10 \right)}^{3}}}{{{13}^{4}}}+.....\infty ....\left( iii \right)\]

Now, we know the terms of the G.P are of the form \[a,ar,a{{r}^{2}},a{{r}^{3}}.....\]

By comparing the terms of RHS of the equation (iii), with the general terms of the G.P, we get, \[a=\dfrac{1}{13}\] and \[r=\dfrac{10}{13}\].

We know that the sum of the terms of the infinite G.P \[=\dfrac{a}{1-r}\]. By using this in the RHS of equation (iii), we get,

\[\dfrac{12S}{65}=\dfrac{\dfrac{1}{13}}{1-\dfrac{10}{13}}\]

By simplifying the above equation, we get,

\[\dfrac{12S}{65}=\dfrac{\dfrac{1}{13}}{\dfrac{13-10}{13}}\]

Or, \[\dfrac{12S}{65}=\dfrac{1}{3}\]

By cross multiplying the above equation, we get,

\[S=\dfrac{65}{3\times 12}=\dfrac{65}{36}\]

So, we get the sum of the series as \[\dfrac{65}{36}\].

Hence, option (a) is the right answer.

Note: In these types of questions, always try to transform the series into geometric progression by multiplying a suitable number with it. This approach should be there in mind. Also, take special care whether the given series is finite or infinite and use the formula for the sum of the finite series as \[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\] and for the sum of the infinite series as \[{{S}_{\infty }}=\dfrac{a}{\left( 1-r \right)}\].

Complete step-by-step answer:

Here, we have to find the sum of the series \[\dfrac{5}{13}+\dfrac{55}{{{13}^{2}}}+\dfrac{555}{{{13}^{3}}}+.....\infty \].

Let us consider the series given in the question.

\[S=\dfrac{5}{13}+\dfrac{55}{{{13}^{2}}}+\dfrac{555}{{{13}^{3}}}+.....\infty ....\left( i \right)\]

By multiplying both the sides of the above expression by \[\dfrac{1}{13}\], we get,

\[\dfrac{S}{13}=\dfrac{5}{{{\left( 13 \right)}^{2}}}+\dfrac{55}{{{13}^{3}}}+\dfrac{555}{{{13}^{4}}}+.....\infty ....\left( ii \right)\]

Now, by subtracting equation (ii) from equation (i), we get,

\[S-\dfrac{S}{13}=\dfrac{5}{13}+\left( \dfrac{55}{{{13}^{2}}}-\dfrac{5}{{{\left( 13 \right)}^{2}}} \right)+\left( \dfrac{555}{{{13}^{3}}}-\dfrac{55}{{{13}^{3}}} \right)+\left( \dfrac{5555}{{{13}^{4}}}-\dfrac{555}{{{13}^{4}}} \right)+.....\infty \]

By simplifying the above equation, we get,

\[\dfrac{12S}{13}=\dfrac{5}{13}+\left( \dfrac{50}{{{\left( 13 \right)}^{2}}} \right)+\left( \dfrac{500}{{{13}^{3}}} \right)+\left( \dfrac{5000}{{{13}^{4}}} \right)+.....\infty \]

Now by dividing both the sides of the above equation by 5, we get,

\[\dfrac{1}{5}.\dfrac{12S}{13}=\dfrac{1}{13}+\dfrac{10}{{{\left( 13 \right)}^{2}}}+\dfrac{100}{{{13}^{3}}}+\dfrac{1000}{{{13}^{4}}}+.....\infty \]

Or we can also write the above equation as,

\[\dfrac{12S}{65}=\dfrac{1}{13}+\dfrac{10}{{{\left( 13 \right)}^{2}}}+\dfrac{{{\left( 10 \right)}^{2}}}{{{13}^{3}}}+\dfrac{{{\left( 10 \right)}^{3}}}{{{13}^{4}}}+.....\infty ....\left( iii \right)\]

Now, we know the terms of the G.P are of the form \[a,ar,a{{r}^{2}},a{{r}^{3}}.....\]

By comparing the terms of RHS of the equation (iii), with the general terms of the G.P, we get, \[a=\dfrac{1}{13}\] and \[r=\dfrac{10}{13}\].

We know that the sum of the terms of the infinite G.P \[=\dfrac{a}{1-r}\]. By using this in the RHS of equation (iii), we get,

\[\dfrac{12S}{65}=\dfrac{\dfrac{1}{13}}{1-\dfrac{10}{13}}\]

By simplifying the above equation, we get,

\[\dfrac{12S}{65}=\dfrac{\dfrac{1}{13}}{\dfrac{13-10}{13}}\]

Or, \[\dfrac{12S}{65}=\dfrac{1}{3}\]

By cross multiplying the above equation, we get,

\[S=\dfrac{65}{3\times 12}=\dfrac{65}{36}\]

So, we get the sum of the series as \[\dfrac{65}{36}\].

Hence, option (a) is the right answer.

Note: In these types of questions, always try to transform the series into geometric progression by multiplying a suitable number with it. This approach should be there in mind. Also, take special care whether the given series is finite or infinite and use the formula for the sum of the finite series as \[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\] and for the sum of the infinite series as \[{{S}_{\infty }}=\dfrac{a}{\left( 1-r \right)}\].

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE