To find the number of common terms in the two sequences 3, 7, 11, ..., 407 and 2, 9, 16, ..., 905, we can follow these steps:
### Step 1: Identify the sequences
The first sequence is an arithmetic progression (AP) with:
- First term (a1) = 3
- Common difference (d1) = 4
- Last term (l1) = 407
The second sequence is also an AP with:
- First term (a2) = 2
- Common difference (d2) = 7
- Last term (l2) = 905
### Step 2: Write the general terms for both sequences
The general term of the first sequence can be expressed as:
\[ T_n = a1 + (n - 1) \cdot d1 = 3 + (n - 1) \cdot 4 = 4n - 1 \]
The general term of the second sequence can be expressed as:
\[ S_m = a2 + (m - 1) \cdot d2 = 2 + (m - 1) \cdot 7 = 7m - 5 \]
### Step 3: Set the general terms equal to find common terms
To find the common terms, we set the two general terms equal to each other:
\[ 4n - 1 = 7m - 5 \]
Rearranging gives:
\[ 4n - 7m = -4 \]
This is a linear Diophantine equation.
### Step 4: Find integer solutions
We can express \( n \) in terms of \( m \):
\[ 4n = 7m - 4 \]
\[ n = \frac{7m - 4}{4} \]
For \( n \) to be an integer, \( 7m - 4 \) must be divisible by 4.
### Step 5: Determine the values of \( m \)
We can check the congruence:
\[ 7m - 4 \equiv 0 \mod 4 \]
Since \( 7 \equiv 3 \mod 4 \), we have:
\[ 3m - 4 \equiv 0 \mod 4 \]
This simplifies to:
\[ 3m \equiv 4 \mod 4 \]
\[ 3m \equiv 0 \mod 4 \]
Thus, \( m \) must be a multiple of 4. Let \( m = 4k \) for some integer \( k \).
### Step 6: Substitute back to find \( n \)
Substituting \( m = 4k \) into the equation for \( n \):
\[ n = \frac{7(4k) - 4}{4} = 7k - 1 \]
### Step 7: Determine the limits for \( n \) and \( m \)
Now we need to find the limits for \( n \) and \( m \):
- For the first sequence, the last term is 407:
\[ 4n - 1 \leq 407 \]
\[ 4n \leq 408 \]
\[ n \leq 102 \]
- For the second sequence, the last term is 905:
\[ 7m - 5 \leq 905 \]
\[ 7m \leq 910 \]
\[ m \leq 130 \]
### Step 8: Find the maximum value of \( k \)
From \( n \leq 102 \):
\[ 7k - 1 \leq 102 \]
\[ 7k \leq 103 \]
\[ k \leq 14.71 \]
Thus, \( k \) can take integer values from 0 to 14 (14 values).
From \( m \leq 130 \):
\[ 4k \leq 130 \]
\[ k \leq 32.5 \]
Thus, \( k \) can also take integer values from 0 to 32 (32 values).
### Step 9: Conclusion
The limiting factor is \( k \leq 14 \), so the number of common terms is:
\[ k = 0, 1, 2, ..., 14 \]
This gives us a total of 15 common terms.
### Final Answer
The number of common terms in both sequences is **15**.
