The sum of $n$terms of the sequence $\log a,\log ar,\log a{r^2},...$ is:
A.$\dfrac{n}{2}\log {a^2}{r^{n - 1}}$
B.$n\log {a^2}{r^{n - 1}}$
C.$\dfrac{{3n}}{2}\log {a^2}{r^{n - 1}}$
D.None of these
Answer
644.7k+ views
Hint: We first simplify the given sequence and then find its first term and common difference. To find the sum of $n$terms of the given sequence, use the formula , ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$. To get the final answer, use the properties of log to simplify the answer.
Complete step by step answer:
The first term of the sequence is $\log a$
We will first simplify the given sequence.
We know that, $\log \left( {xy} \right) = \log x + \log y$
Therefore, we write the given sequence as, $\log a,\log a + \log r,\log a + \log {r^2},...$
Also, $\log \left( {{x^m}} \right) = m\log x$
Hence, we have, $\log a,\log a + \log r,\log a + 2\log r,...$
Now, we can observe that the given sequence is an AP as $\log r$ is added to each term.
Now, we will find the common difference by subtracting the second tern from the first one.
Hence, we get common difference $d$ as,
$
d = \log a + \log r - \log a \\
d = \log r \\
$
We have to find the sum of $n$terms of the given sequence.
We know that the sum of $n$ terms of a sequence is $\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term and $d$ is the common difference.
Substitute the values of the first term and the common difference to find the sum of $n$terms of the sequence.
Hence, sum of $n$ terms is
$\dfrac{n}{2}\left( {2\left( {\log a} \right) + \left( {n - 1} \right)\log r} \right)$
We can simplify the above expression using the properties of log, $\log \left( {{x^m}} \right) = m\log x$
$\Rightarrow$$\dfrac{n}{2}\left( {\log {a^2} + \log {r^{n - 1}}} \right)$
Hence, option A is correct.
Note: We have to use properties of log to simplify the given sequence such as, $\log \left( {xy} \right) = \log x + \log y$ and $\log \left( {{x^m}} \right) = m\log x$. Many students make mistakes by considering the given sequence as GP. It is after applying the properties of log, we will observe that the given sequence is an AP.
Complete step by step answer:
The first term of the sequence is $\log a$
We will first simplify the given sequence.
We know that, $\log \left( {xy} \right) = \log x + \log y$
Therefore, we write the given sequence as, $\log a,\log a + \log r,\log a + \log {r^2},...$
Also, $\log \left( {{x^m}} \right) = m\log x$
Hence, we have, $\log a,\log a + \log r,\log a + 2\log r,...$
Now, we can observe that the given sequence is an AP as $\log r$ is added to each term.
Now, we will find the common difference by subtracting the second tern from the first one.
Hence, we get common difference $d$ as,
$
d = \log a + \log r - \log a \\
d = \log r \\
$
We have to find the sum of $n$terms of the given sequence.
We know that the sum of $n$ terms of a sequence is $\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term and $d$ is the common difference.
Substitute the values of the first term and the common difference to find the sum of $n$terms of the sequence.
Hence, sum of $n$ terms is
$\dfrac{n}{2}\left( {2\left( {\log a} \right) + \left( {n - 1} \right)\log r} \right)$
We can simplify the above expression using the properties of log, $\log \left( {{x^m}} \right) = m\log x$
$\Rightarrow$$\dfrac{n}{2}\left( {\log {a^2} + \log {r^{n - 1}}} \right)$
Hence, option A is correct.
Note: We have to use properties of log to simplify the given sequence such as, $\log \left( {xy} \right) = \log x + \log y$ and $\log \left( {{x^m}} \right) = m\log x$. Many students make mistakes by considering the given sequence as GP. It is after applying the properties of log, we will observe that the given sequence is an AP.
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