Two sequences `lta_(n)gtandltb_(n)gt` are defined by
`a_(n)=log((5^(n+1))/(3^(n-1))),b_(n)={log((5)/(3))}^(n)`, then

Options

A. `lta_(n)gt ` is an A.P. and `ltb_(n)gt` is a G.P
B. `lta_(n)gt` and `ltb_(n)gt` both are G.P.
C. `lta_(n)gt` and `ltb_(n)gt` both are A.P.
D. `lta_(n)gt` is a G.P. and `ltb_(n)gt` is neither an A.P. nor a G.P.

Answer

To solve the problem, we need to analyze the sequences \( a_n \) and \( b_n \) given by:

\[ a_n = \log\left(\frac{5^{n+1}}{3^{n-1}}\right) \]
\[ b_n = \left(\log\left(\frac{5}{3}\right)\right)^n \]

We will determine if these sequences are in Arithmetic Progression (AP) or Geometric Progression (GP).

### Step 1: Simplifying \( a_n \)

Using the properties of logarithms, we can simplify \( a_n \):

\[ a_n = \log(5^{n+1}) - \log(3^{n-1}) \]
Using the property \( \log(a^b) = b \log(a) \):

\[ a_n = (n+1) \log(5) - (n-1) \log(3) \]

Expanding this gives:

\[ a_n = n \log(5) + \log(5) - n \log(3) + \log(3) \]
\[ a_n = n(\log(5) - \log(3)) + \log(15) \]

### Step 2: Finding the first few terms of \( a_n \)

Now we can find the first few terms:

- For \( n = 1 \):
\[ a_1 = 1(\log(5) - \log(3)) + \log(15) = \log(15) + \log\left(\frac{5}{3}\right) \]

- For \( n = 2 \):
\[ a_2 = 2(\log(5) - \log(3)) + \log(15) = \log(15) + 2\log\left(\frac{5}{3}\right) \]

- For \( n = 3 \):
\[ a_3 = 3(\log(5) - \log(3)) + \log(15) = \log(15) + 3\log\left(\frac{5}{3}\right) \]

### Step 3: Checking if \( a_n \) is in AP

To check if \( a_n \) is in AP, we need to find the common difference:

\[ a_2 - a_1 = \left(\log(15) + 2\log\left(\frac{5}{3}\right)\right) - \left(\log(15) + \log\left(\frac{5}{3}\right)\right) = \log\left(\frac{5}{3}\right) \]

\[ a_3 - a_2 = \left(\log(15) + 3\log\left(\frac{5}{3}\right)\right) - \left(\log(15) + 2\log\left(\frac{5}{3}\right)\right) = \log\left(\frac{5}{3}\right) \]

Since the common difference is the same, \( a_n \) is in AP.

### Step 4: Simplifying \( b_n \)

Now let's simplify \( b_n \):

\[ b_n = \left(\log\left(\frac{5}{3}\right)\right)^n \]

### Step 5: Finding the first few terms of \( b_n \)

- For \( n = 1 \):
\[ b_1 = \log\left(\frac{5}{3}\right) \]

- For \( n = 2 \):
\[ b_2 = \left(\log\left(\frac{5}{3}\right)\right)^2 \]

- For \( n = 3 \):
\[ b_3 = \left(\log\left(\frac{5}{3}\right)\right)^3 \]

### Step 6: Checking if \( b_n \) is in GP

To check if \( b_n \) is in GP, we need to find the common ratio:

\[ \frac{b_2}{b_1} = \frac{\left(\log\left(\frac{5}{3}\right)\right)^2}{\log\left(\frac{5}{3}\right)} = \log\left(\frac{5}{3}\right) \]

\[ \frac{b_3}{b_2} = \frac{\left(\log\left(\frac{5}{3}\right)\right)^3}{\left(\log\left(\frac{5}{3}\right)\right)^2} = \log\left(\frac{5}{3}\right) \]

Since the common ratio is the same, \( b_n \) is in GP.

### Conclusion

Thus, we conclude that:
- The sequence \( a_n \) is in Arithmetic Progression (AP).
- The sequence \( b_n \) is in Geometric Progression (GP).