Let`a_(1),a_(2),"...."` be in AP and `q_(1),q_(2),"...."` be in GP. If `a_(1)=q_(1)=2 " and "a_(10)=q_(10)=3`, then

Options

A. `a_(7)q_(19)` is not an integer
B. `a_(19)q_(7)`is an integer
C. `a_(7)q_(19)=a_(19)q_(10)`
D. None of these

Explanation

To solve the problem, we need to analyze the sequences given in the question.

### Step 1: Define the sequences
Let \( a_n \) be the \( n \)-th term of the arithmetic progression (AP) and \( q_n \) be the \( n \)-th term of the geometric progression (GP).

For the AP:
- The first term \( a_1 = a = 2 \)
- The common difference \( d \) can be found using the 10th term:
\[ a_{10} = a + 9d = 3 \]

For the GP:
- The first term \( q_1 = a = 2 \)
- The common ratio \( r \) can be found using the 10th term:
\[ q_{10} = ar^9 = 3 \]

### Step 2: Find the common difference \( d \)
From the equation for the AP:
\[ 2 + 9d = 3 \]
Subtracting 2 from both sides:
\[ 9d = 1 \]
Dividing by 9:
\[ d = \frac{1}{9} \]

### Step 3: Find the common ratio \( r \)
From the equation for the GP:
\[ 2r^9 = 3 \]
Dividing both sides by 2:
\[ r^9 = \frac{3}{2} \]
Taking the 9th root:
\[ r = \left(\frac{3}{2}\right)^{\frac{1}{9}} \]

### Step 4: Calculate \( a_7 \) and \( q_{19} \)
Now we can find \( a_7 \):
\[ a_7 = a + 6d = 2 + 6 \cdot \frac{1}{9} = 2 + \frac{6}{9} = 2 + \frac{2}{3} = \frac{6}{3} + \frac{2}{3} = \frac{8}{3} \]

Next, we calculate \( q_{19} \):
\[ q_{19} = ar^{18} = 2 \left(\left(\frac{3}{2}\right)^{\frac{1}{9}}\right)^{18} = 2 \left(\frac{3}{2}\right)^{2} = 2 \cdot \frac{9}{4} = \frac{18}{4} = \frac{9}{2} \]

### Step 5: Calculate \( a_7 \cdot q_{19} \)
Now we multiply \( a_7 \) and \( q_{19} \):
\[ a_7 \cdot q_{19} = \frac{8}{3} \cdot \frac{9}{2} = \frac{8 \cdot 9}{3 \cdot 2} = \frac{72}{6} = 12 \]
Since 12 is an integer, the statement in option A is incorrect.

### Step 6: Calculate \( a_{19} \) and \( q_7 \)
Next, we find \( a_{19} \):
\[ a_{19} = a + 18d = 2 + 18 \cdot \frac{1}{9} = 2 + 2 = 4 \]

Now, we calculate \( q_7 \):
\[ q_7 = ar^6 = 2 \left(\left(\frac{3}{2}\right)^{\frac{1}{9}}\right)^{6} = 2 \left(\frac{3}{2}\right)^{\frac{2}{3}} = 2 \cdot \frac{3^{2/3}}{2^{2/3}} = 2^{1 - \frac{2}{3}} \cdot 3^{2/3} = 2^{\frac{1}{3}} \cdot 3^{\frac{2}{3}} = \frac{2^{1/3} \cdot 9^{1/3}}{2^{2/3}} = \frac{9^{1/3}}{2^{1/3}} \]

### Step 7: Check if \( a_{19} \cdot q_7 \) is an integer
Now we check \( a_{19} \cdot q_7 \):
\[ a_{19} \cdot q_7 = 4 \cdot \frac{9^{1/3}}{2^{1/3}} = 4 \cdot \frac{3^{2/3}}{2^{1/3}} = \frac{4 \cdot 9^{1/3}}{2^{1/3}} = \frac{4 \cdot 3^{2/3}}{2^{1/3}} = \frac{12^{1/3}}{2^{1/3}} = \text{not an integer} \]

### Conclusion
After checking the options, we find that:
- Option A is incorrect (as \( a_7 \cdot q_{19} = 12 \), which is an integer).
- Option B is incorrect (as \( a_{19} \cdot q_7 \) is not an integer).
- Option C is correct (as \( a_7 \cdot q_{19} = a_{19} \cdot q_7 = 12 \)).