In the following two A.P. how many terms are identical ?
`2,5,8,11…` to 60 terms and `3,5,7,…..50` terms

Answer

To solve the problem of finding how many terms are identical in the two given arithmetic progressions (A.P.), we will follow these steps:

### Step 1: Identify the first A.P.
The first A.P. is given as:
\[ 2, 5, 8, 11, \ldots \]
This A.P. has:
- First term \( a_1 = 2 \)
- Common difference \( d_1 = 5 - 2 = 3 \)

The \( n \)-th term of the first A.P. can be expressed as:
\[ T_n = a_1 + (n - 1) \cdot d_1 = 2 + (n - 1) \cdot 3 = 3n - 1 \]

### Step 2: Identify the second A.P.
The second A.P. is given as:
\[ 3, 5, 7, \ldots \]
This A.P. has:
- First term \( a_2 = 3 \)
- Common difference \( d_2 = 5 - 3 = 2 \)

The \( m \)-th term of the second A.P. can be expressed as:
\[ T_m = a_2 + (m - 1) \cdot d_2 = 3 + (m - 1) \cdot 2 = 2m + 1 \]

### Step 3: Set the terms equal to find common terms
We need to find the values of \( n \) and \( m \) such that:
\[ T_n = T_m \]
This gives us the equation:
\[ 3n - 1 = 2m + 1 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ 3n = 2m + 2 \]
\[ n = \frac{2m + 2}{3} \]

### Step 5: Ensure \( n \) is a natural number
For \( n \) to be a natural number, \( 2m + 2 \) must be divisible by 3. This can be expressed as:
\[ 2m + 2 \equiv 0 \mod 3 \]
Simplifying gives:
\[ 2m \equiv 1 \mod 3 \]
This means:
\[ m \equiv 2 \mod 3 \]
Thus, \( m \) can take the values:
\[ m = 3k + 2 \]
for \( k = 0, 1, 2, \ldots \)

### Step 6: Determine the maximum value of \( m \)
Given that the second A.P. has 50 terms, we have:
\[ 3k + 2 \leq 50 \]
This gives:
\[ 3k \leq 48 \]
\[ k \leq 16 \]

### Step 7: Calculate the corresponding \( n \) values
Now substituting \( k = 0, 1, 2, \ldots, 16 \) into \( m = 3k + 2 \):
- For \( k = 0 \), \( m = 2 \) → \( n = \frac{2(2) + 2}{3} = \frac{6}{3} = 2 \)
- For \( k = 1 \), \( m = 5 \) → \( n = \frac{2(5) + 2}{3} = \frac{12}{3} = 4 \)
- For \( k = 2 \), \( m = 8 \) → \( n = \frac{2(8) + 2}{3} = \frac{18}{3} = 6 \)
- Continue this until \( k = 16 \), where \( m = 50 \) → \( n = \frac{2(50) + 2}{3} = \frac{102}{3} = 34 \)

### Step 8: Count the values of \( k \)
The values of \( k \) range from 0 to 16, giving us a total of:
\[ 17 \text{ identical terms} \]

### Final Answer
Thus, the number of identical terms in the two A.P.s is:
**17**