The number of common terms in the progressions `4,9,14,19,......` ,up to `25^(" th ")` term and `3,6,9,12,......,` up to `37^(" th ")` term is :

Options

A. 7
B. 8
C. 5
D. 9

Answer

To find the number of common terms in the two given arithmetic progressions (APs), we will follow these steps:

### Step 1: Identify the first arithmetic progression (AP1)
The first progression is:
\[ 4, 9, 14, 19, \ldots \]
This is an arithmetic progression where:
- First term \( a_1 = 4 \)
- Common difference \( d_1 = 5 \)

The \( n \)-th term of an AP can be calculated using the formula:
\[ a_n = a + (n-1)d \]
Thus, the 25th term of this progression is:
\[ a_{25} = 4 + (25-1) \cdot 5 = 4 + 120 = 124 \]

### Step 2: Identify the second arithmetic progression (AP2)
The second progression is:
\[ 3, 6, 9, 12, \ldots \]
This is also an arithmetic progression where:
- First term \( a_2 = 3 \)
- Common difference \( d_2 = 3 \)

The 37th term of this progression is:
\[ a_{37} = 3 + (37-1) \cdot 3 = 3 + 108 = 111 \]

### Step 3: Determine the common terms
To find the common terms in both progressions, we need to find terms that exist in both sequences. The common terms will also form an arithmetic progression.

The first common term can be found by checking the terms of the first progression:
- The first term is \( 4 \)
- The second term is \( 9 \) (which is also in the second progression)

Next, we need to find the common difference for the common terms. The common difference of the common terms will be the least common multiple (LCM) of the two common differences \( d_1 \) and \( d_2 \):
\[ \text{LCM}(d_1, d_2) = \text{LCM}(5, 3) = 15 \]

### Step 4: Formulate the common terms AP
The common terms will start from \( 9 \) and will have a common difference of \( 15 \):
\[ 9, 24, 39, 54, \ldots \]

### Step 5: Find the last common term
The last common term must be less than or equal to the last term of the first progression (124) and the last term of the second progression (111). Therefore, we need to find the largest term in the sequence \( 9 + (n-1) \cdot 15 \) that is less than or equal to \( 111 \):
\[ 9 + (n-1) \cdot 15 \leq 111 \]
Subtracting 9 from both sides:
\[ (n-1) \cdot 15 \leq 102 \]
Dividing by 15:
\[ n-1 \leq \frac{102}{15} \]
\[ n-1 \leq 6.8 \]
Thus,
\[ n \leq 7.8 \]
Since \( n \) must be a whole number, the maximum value of \( n \) is \( 7 \).

### Conclusion
The number of common terms in the two progressions is \( 7 \).