Find the number of common terms to the two sequences 17,21,25,...,417 and 16,21,26,...,466.

Explanation

To find the number of common terms in the sequences 17, 21, 25,..., 417 and 16, 21, 26,..., 466, we will follow these steps:

### Step 1: Identify the first sequence
The first sequence is an arithmetic progression (AP) with:
- First term \( a_1 = 17 \)
- Common difference \( d_1 = 4 \)
- Last term \( l_1 = 417 \)

### Step 2: Find the number of terms in the first sequence
To find the number of terms \( n_1 \) in the first sequence, we use the formula for the \( n \)-th term of an AP:
\[ l_n = a + (n-1)d \]
Setting \( l_n = 417 \), we have:
\[ 417 = 17 + (n_1 - 1) \cdot 4 \]
Subtracting 17 from both sides:
\[ 400 = (n_1 - 1) \cdot 4 \]
Dividing by 4:
\[ 100 = n_1 - 1 \]
Adding 1:
\[ n_1 = 101 \]

### Step 3: Identify the second sequence
The second sequence is also an AP with:
- First term \( a_2 = 16 \)
- Common difference \( d_2 = 5 \)
- Last term \( l_2 = 466 \)

### Step 4: Find the number of terms in the second sequence
Using the same formula for the \( n \)-th term:
\[ l_n = a + (n-1)d \]
Setting \( l_n = 466 \):
\[ 466 = 16 + (n_2 - 1) \cdot 5 \]
Subtracting 16 from both sides:
\[ 450 = (n_2 - 1) \cdot 5 \]
Dividing by 5:
\[ 90 = n_2 - 1 \]
Adding 1:
\[ n_2 = 91 \]

### Step 5: Formulate the common terms
The general term of the first sequence can be expressed as:
\[ T_n = 17 + (n - 1) \cdot 4 = 4n + 13 \]
The general term of the second sequence can be expressed as:
\[ S_m = 16 + (m - 1) \cdot 5 = 5m + 11 \]

### Step 6: Set the two general terms equal to find common terms
To find the common terms, we set:
\[ 4n + 13 = 5m + 11 \]
Rearranging gives:
\[ 5m - 4n = 2 \]

### Step 7: Solve for integer solutions
We can express \( m \) in terms of \( n \):
\[ m = \frac{4n + 2}{5} \]
For \( m \) to be an integer, \( 4n + 2 \) must be divisible by 5. This can be rewritten as:
\[ 4n + 2 \equiv 0 \mod 5 \]
Simplifying gives:
\[ 4n \equiv -2 \mod 5 \quad \Rightarrow \quad 4n \equiv 3 \mod 5 \]
Multiplying both sides by the modular inverse of 4 modulo 5, which is 4:
\[ n \equiv 12 \mod 5 \quad \Rightarrow \quad n \equiv 2 \mod 5 \]
Thus, \( n \) can take values of the form:
\[ n = 5k + 2 \quad \text{for integers } k \]

### Step 8: Determine the range of \( n \)
Since \( n \) must be between 1 and 101:
\[ 1 \leq 5k + 2 \leq 101 \]
Subtracting 2:
\[ -1 \leq 5k \leq 99 \]
Dividing by 5:
\[ 0 \leq k \leq 19.8 \]
Thus, \( k \) can take integer values from 0 to 19, giving us 20 possible values for \( n \).

### Conclusion
The number of common terms in the two sequences is **20**.

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