m A.M. and 3 G.M. are inserted between 3 and 243 such that `2^(nd)` GM=`4^(th)`AM then `m=`

Answer

To solve the problem, we need to find the value of \( m \) such that \( m \) Arithmetic Means (A.M.) and 3 Geometric Means (G.M.) are inserted between 3 and 243, with the condition that the second G.M. is equal to the fourth A.M.

### Step-by-Step Solution:

1. **Identify the Terms**:
- The first term \( a_1 = 3 \)
- The last term \( a_{m+2} = 243 \)
- We need to insert \( m \) A.M. between these two terms.

2. **Formulate the A.M.**:
- The A.M. sequence can be expressed as:
\[ a_1, a_2, a_3, \ldots, a_{m+2} \]
- The \( n \)-th term of an A.M. can be given by:
\[ a_n = a_1 + (n-1)d \]
- Here, \( d \) is the common difference.

3. **Calculate the Common Difference**:
- The last term can be expressed as:
\[ a_{m+2} = a_1 + (m+1)d \]
- Substituting the known values:
\[ 243 = 3 + (m+1)d \]
- Rearranging gives:
\[ (m+1)d = 240 \quad \text{(1)} \]

4. **Formulate the G.M.**:
- The G.M. sequence can be expressed as:
\[ g_1, g_2, g_3 \]
- The first term \( g_1 = 3 \) and the last term \( g_4 = 243 \).
- The \( n \)-th term of a G.M. can be given by:
\[ g_n = g_1 \cdot r^{n-1} \]
- Here, \( r \) is the common ratio.

5. **Calculate the Common Ratio**:
- The fourth term can be expressed as:
\[ g_4 = g_1 \cdot r^3 \]
- Substituting the known values:
\[ 243 = 3 \cdot r^3 \]
- Rearranging gives:
\[ r^3 = 81 \implies r = 3 \quad \text{(2)} \]

6. **Identify the Condition**:
- The second G.M. is equal to the fourth A.M.:
\[ g_2 = a_4 \]
- The second G.M. can be expressed as:
\[ g_2 = g_1 \cdot r^{2-1} = 3 \cdot r = 3 \cdot 3 = 9 \]
- The fourth A.M. can be expressed as:
\[ a_4 = a_1 + 3d = 3 + 3d \]
- Setting these equal gives:
\[ 9 = 3 + 3d \]
- Rearranging gives:
\[ 3d = 6 \implies d = 2 \quad \text{(3)} \]

7. **Substituting Back to Find \( m \)**:
- Substitute \( d = 2 \) into equation (1):
\[ (m+1) \cdot 2 = 240 \]
- Simplifying gives:
\[ m + 1 = 120 \implies m = 119 \]

### Final Answer:
The value of \( m \) is \( 119 \).