A G.P. has first term `a=3`, last term `l=96` and sum of n terms `S=189`. Find the number of terms in it.

Answer

To find the number of terms in the given geometric progression (G.P.) with the first term \( a = 3 \), last term \( l = 96 \), and sum of \( n \) terms \( S = 189 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- First term \( a = 3 \)
- Last term \( l = 96 \)
- Sum of \( n \) terms \( S_n = 189 \)

2. **Assume the Number of Terms**:
- Let the number of terms in the G.P. be \( n \).

3. **Use the Formula for the Last Term**:
- The last term of a G.P. can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
- Here, \( T_n = 96 \), so we can write:
\[ 3 \cdot r^{n-1} = 96 \]
- Dividing both sides by 3 gives:
\[ r^{n-1} = \frac{96}{3} = 32 \]

4. **Use the Formula for the Sum of n Terms**:
- The sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
- Substituting the known values, we have:
\[ 189 = \frac{3(r^n - 1)}{r - 1} \]
- Dividing both sides by 3 gives:
\[ 63 = \frac{r^n - 1}{r - 1} \]
- Rearranging gives:
\[ r^n - 1 = 63(r - 1) \]

5. **Substituting for \( r^{n-1} \)**:
- From step 3, we know \( r^{n-1} = 32 \). Thus, we can express \( r^n \) as:
\[ r^n = r \cdot r^{n-1} = r \cdot 32 \]
- Substituting this into the equation gives:
\[ r \cdot 32 - 1 = 63(r - 1) \]

6. **Expanding and Rearranging**:
- Expanding the right side:
\[ 32r - 1 = 63r - 63 \]
- Rearranging gives:
\[ 32r - 63r = -63 + 1 \]
\[ -31r = -62 \]
- Dividing by -31 gives:
\[ r = \frac{62}{31} = 2 \]

7. **Finding the Number of Terms \( n \)**:
- Now that we have \( r = 2 \), we can substitute back to find \( n \):
\[ r^{n-1} = 32 \implies 2^{n-1} = 32 \]
- Since \( 32 = 2^5 \), we equate the exponents:
\[ n - 1 = 5 \implies n = 6 \]

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