The first term and the last term of a GP are a and k respectively. If the number of terms be n, then n is equal to (r `to` common ratio)

Options

A. `1-(logk-loga)/(logr)`
B. `1+(loga+logk)/(logr)`
C. `1+(logk-loga)/(logr)`
D. `logr=logk-loga`

Answer

To solve the problem, we need to find the number of terms \( n \) in a geometric progression (GP) where the first term is \( a \) and the last term is \( k \), with a common ratio \( r \).

### Step-by-Step Solution:

1. **Understanding the nth Term of a GP**:
The nth term of a geometric progression can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( T_n \) is the nth term, \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

2. **Setting Up the Equation**:
Since the last term of the GP is given as \( k \), we can set up the equation:
\[ k = a \cdot r^{n-1} \]

3. **Rearranging the Equation**:
To isolate \( r^{n-1} \), we can rearrange the equation:
\[ r^{n-1} = \frac{k}{a} \]

4. **Taking Logarithms**:
To solve for \( n \), we take the logarithm of both sides. We can use logarithm base \( r \):
\[ \log_r(r^{n-1}) = \log_r\left(\frac{k}{a}\right) \]

5. **Applying Logarithm Properties**:
Using the property of logarithms, we simplify the left side:
\[ n-1 = \log_r\left(\frac{k}{a}\right) \]

6. **Solving for \( n \)**:
Now, we can solve for \( n \):
\[ n = 1 + \log_r\left(\frac{k}{a}\right) \]

7. **Using Change of Base Formula**:
We can express the logarithm in terms of natural logarithms or logarithms of any other base using the change of base formula:
\[ n = 1 + \frac{\log(k) - \log(a)}{\log(r)} \]

### Final Result:
Thus, the number of terms \( n \) in the geometric progression is given by:
\[ n = 1 + \frac{\log(k) - \log(a)}{\log(r)} \]