The sum of terms of a G.P if `a_(1)=3,a_(n)=96` and `S_(n)=189` is

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem, we need to find the number of terms in a geometric progression (G.P.) given the first term \( a_1 = 3 \), the nth term \( a_n = 96 \), and the sum of the first n terms \( S_n = 189 \).

### Step-by-Step Solution:

1. **Identify the formulas for G.P.**:
- The nth term of a G.P. is given by:
\[ a_n = a_1 \cdot r^{n-1} \]
- The sum of the first n terms of a G.P. is given by:
\[ S_n = \frac{a_1 (r^n - 1)}{r - 1} \]
where \( r \) is the common ratio.

2. **Substitute the known values into the nth term formula**:
- We know \( a_1 = 3 \) and \( a_n = 96 \):
\[ 96 = 3 \cdot r^{n-1} \]
- Dividing both sides by 3:
\[ r^{n-1} = \frac{96}{3} = 32 \]

3. **Substitute the known values into the sum formula**:
- We know \( S_n = 189 \):
\[ 189 = \frac{3 (r^n - 1)}{r - 1} \]
- Multiplying both sides by \( r - 1 \):
\[ 189(r - 1) = 3(r^n - 1) \]
- Expanding this gives:
\[ 189r - 189 = 3r^n - 3 \]
- Rearranging:
\[ 3r^n - 189r + 186 = 0 \]

4. **Express \( r^n \) in terms of \( r \)**:
- From the equation \( r^{n-1} = 32 \), we can express \( r^n \) as:
\[ r^n = r \cdot r^{n-1} = r \cdot 32 \]
- Substitute this into the equation:
\[ 3(32r) - 189r + 186 = 0 \]
- Simplifying:
\[ 96r - 189r + 186 = 0 \]
\[ -93r + 186 = 0 \]
\[ 93r = 186 \]
\[ r = 2 \]

5. **Find the value of \( n \)**:
- Now that we have \( r = 2 \), substitute back into the equation for \( r^{n-1} \):
\[ 2^{n-1} = 32 \]
- Recognizing that \( 32 = 2^5 \):
\[ 2^{n-1} = 2^5 \]
- Therefore:
\[ n - 1 = 5 \implies n = 6 \]

### Final Answer:
The number of terms in the G.P. is \( n = 6 \).