The sum upto n terms of the sequence log a, log ar, log `a r^(2)`, …. Is

Options

A. `(n)/(2) log a^(2) r^(n-1)`
B. `(n)/(2) log a r^(n-1)`
C. `(3n)/(2) log a r^(n-1)`
D. `(5n)/(2) log a^(2) r^(n-1)`

Answer

To find the sum of the sequence \( \log a, \log ar, \log ar^2, \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Identify the terms of the sequence
The given sequence can be expressed as:
- First term: \( \log a \)
- Second term: \( \log ar = \log a + \log r \)
- Third term: \( \log ar^2 = \log a + 2\log r \)
- Continuing this way, the \( k \)-th term can be expressed as:
\[ T_k = \log a + (k-1) \log r \]

### Step 2: Recognize the sequence as an arithmetic progression (AP)
The sequence can be rewritten as:
\[ T_1 = \log a, \quad T_2 = \log a + \log r, \quad T_3 = \log a + 2\log r, \ldots \]
The first term \( a = \log a \) and the common difference \( d = \log r \).

### Step 3: Use the formula for the sum of the first \( n \) terms of an AP
The sum \( S_n \) of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
Substituting \( a = \log a \) and \( d = \log r \):
\[ S_n = \frac{n}{2} \left(2\log a + (n-1)\log r\right) \]

### Step 4: Simplify the expression
We can factor out the logarithms:
\[ S_n = \frac{n}{2} \left(2\log a + (n-1)\log r\right) = \frac{n}{2} \left(\log a^2 + \log r^{n-1}\right) \]
Using the property of logarithms \( \log m + \log n = \log(m \cdot n) \):
\[ S_n = \frac{n}{2} \log \left(a^2 \cdot r^{n-1}\right) \]

### Final Result
Thus, the sum of the first \( n \) terms of the sequence is:
\[ S_n = \frac{n}{2} \log \left(a^2 r^{n-1}\right) \]

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