Answer

Verified

483.9k+ views

Hint: We have to change the base of the logarithm so that the terms will get simplified and form a geometric progression. Then apply the formula of sum of infinite terms of a G.P. which is $S=\dfrac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio of the G.P.

Complete step-by-step answer:

There are different types of series but mainly we have to study three of them namely Arithmetic progression (A.P.), Geometric Progression (G.P.) and Harmonic Progression (H.P.). In this question geometric progression will be used.

In Mathematics, a G.P. also known as geometric sequence is a sequence of numbers where each term after its predecessor is obtained by multiplying the previous term with a fixed non-zero number known as the common ratio of the G.P.

For example: \[2,\text{ }4,\text{ }8,\text{ }16,\text{ }32,\text{ }.........\] is a G.P. with common ratio $2$.

Generally, G.P. is represented by $a,\text{ }a{{r}^{2}},\text{ }a{{r}^{3}},\text{ }\ldots \ldots \ldots $ where \[a\] is the first term and \[r\] is the common ratio.

Now, we have been given:

${{S}_{\infty }}=\text{Sum of infinite series}=\dfrac{1}{{{\log }_{3}}e}+\dfrac{1}{{{\log }_{3}}{{e}^{2}}}+\dfrac{1}{{{\log }_{3}}{{e}^{4}}}+.........\infty $

We can change the base of the logarithm by using the following formula:

\[{{\log }_{m}}n=\dfrac{1}{{{\log }_{n}}m}\]

For example; \[{{\log }_{3}}e=\dfrac{1}{{{\log }_{e}}3}\].

$\therefore {{S}_{\infty }}={{\log }_{e}}3+{{\log }_{{{e}^{2}}}}3+{{\log }_{{{e}^{3}}}}3+.........\infty $

Now, using the following property of logarithm: ${{\log }_{{{m}^{a}}}}n=\dfrac{1}{a}{{\log }_{m}}n$, we get

$\therefore {{S}_{\infty }}={{\log }_{e}}3+\dfrac{1}{2}{{\log }_{e}}3+\dfrac{1}{3}{{\log }_{e}}3+.........\infty $

We can clearly see that this a G.P. with the first term $a={{\log }_{e}}3$ and $r=\dfrac{1}{2}$.

$\therefore {{S}_{\infty }}=\dfrac{a}{1-r}=\dfrac{{{\log }_{e}}3}{1-\dfrac{1}{2}}=2{{\log }_{e}}3$

Now, using the rule that: \[a\times {{\log }_{m}}n={{\log }_{n}}({{n}^{a}})\], we get

${{S}_{\infty }}={{\log }_{e}}({{3}^{2}})={{\log }_{e}}9$

Hence the correct option is (A).

Note: Always try to convert the question in simplified form just as we did here. We first simplified the terms of the given sequence by converting the fractions into simple terms and then applied the formula for summation of infinite terms of a G.P. Here, we have to sum infinite terms so the formula for summation of $n$ terms cannot be used.

Complete step-by-step answer:

There are different types of series but mainly we have to study three of them namely Arithmetic progression (A.P.), Geometric Progression (G.P.) and Harmonic Progression (H.P.). In this question geometric progression will be used.

In Mathematics, a G.P. also known as geometric sequence is a sequence of numbers where each term after its predecessor is obtained by multiplying the previous term with a fixed non-zero number known as the common ratio of the G.P.

For example: \[2,\text{ }4,\text{ }8,\text{ }16,\text{ }32,\text{ }.........\] is a G.P. with common ratio $2$.

Generally, G.P. is represented by $a,\text{ }a{{r}^{2}},\text{ }a{{r}^{3}},\text{ }\ldots \ldots \ldots $ where \[a\] is the first term and \[r\] is the common ratio.

Now, we have been given:

${{S}_{\infty }}=\text{Sum of infinite series}=\dfrac{1}{{{\log }_{3}}e}+\dfrac{1}{{{\log }_{3}}{{e}^{2}}}+\dfrac{1}{{{\log }_{3}}{{e}^{4}}}+.........\infty $

We can change the base of the logarithm by using the following formula:

\[{{\log }_{m}}n=\dfrac{1}{{{\log }_{n}}m}\]

For example; \[{{\log }_{3}}e=\dfrac{1}{{{\log }_{e}}3}\].

$\therefore {{S}_{\infty }}={{\log }_{e}}3+{{\log }_{{{e}^{2}}}}3+{{\log }_{{{e}^{3}}}}3+.........\infty $

Now, using the following property of logarithm: ${{\log }_{{{m}^{a}}}}n=\dfrac{1}{a}{{\log }_{m}}n$, we get

$\therefore {{S}_{\infty }}={{\log }_{e}}3+\dfrac{1}{2}{{\log }_{e}}3+\dfrac{1}{3}{{\log }_{e}}3+.........\infty $

We can clearly see that this a G.P. with the first term $a={{\log }_{e}}3$ and $r=\dfrac{1}{2}$.

$\therefore {{S}_{\infty }}=\dfrac{a}{1-r}=\dfrac{{{\log }_{e}}3}{1-\dfrac{1}{2}}=2{{\log }_{e}}3$

Now, using the rule that: \[a\times {{\log }_{m}}n={{\log }_{n}}({{n}^{a}})\], we get

${{S}_{\infty }}={{\log }_{e}}({{3}^{2}})={{\log }_{e}}9$

Hence the correct option is (A).

Note: Always try to convert the question in simplified form just as we did here. We first simplified the terms of the given sequence by converting the fractions into simple terms and then applied the formula for summation of infinite terms of a G.P. Here, we have to sum infinite terms so the formula for summation of $n$ terms cannot be used.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which places in India experience sunrise first and class 9 social science CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE