The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.

Answer

To find the arithmetic progression (AP) based on the given conditions, we can follow these steps:

### Step 1: Understand the given information
We know:
- The 19th term of the AP is equal to 3 times the 6th term.
- The 9th term of the AP is 19.

### Step 2: Write the formulas for the terms
The nth term of an AP can be expressed as:
\[ A_n = A + (n-1)D \]
where \( A \) is the first term and \( D \) is the common difference.

### Step 3: Set up the equations
From the information provided:
1. For the 19th term:
\[ A_{19} = A + (19-1)D = A + 18D \]
2. For the 6th term:
\[ A_{6} = A + (6-1)D = A + 5D \]
3. According to the problem, we have:
\[ A + 18D = 3(A + 5D) \]

### Step 4: Simplify the equation
Expanding the equation:
\[ A + 18D = 3A + 15D \]
Rearranging gives:
\[ A + 18D - 3A - 15D = 0 \]
This simplifies to:
\[ -2A + 3D = 0 \]
or
\[ 2A = 3D \]
Thus, we can express \( D \) in terms of \( A \):
\[ D = \frac{2}{3}A \]

### Step 5: Use the second condition
We also know the 9th term:
\[ A_{9} = A + (9-1)D = A + 8D \]
Setting this equal to 19:
\[ A + 8D = 19 \]

### Step 6: Substitute D in the equation
Substituting \( D = \frac{2}{3}A \) into the equation:
\[ A + 8\left(\frac{2}{3}A\right) = 19 \]
This simplifies to:
\[ A + \frac{16}{3}A = 19 \]
Combining the terms gives:
\[ \frac{3A + 16A}{3} = 19 \]
or
\[ \frac{19A}{3} = 19 \]

### Step 7: Solve for A
Multiplying both sides by 3:
\[ 19A = 57 \]
Thus,
\[ A = 3 \]

### Step 8: Find D
Now substituting \( A \) back to find \( D \):
\[ D = \frac{2}{3} \times 3 = 2 \]

### Step 9: Write the AP
Now, we can write the first few terms of the AP:
- First term \( A = 3 \)
- Second term \( A + D = 3 + 2 = 5 \)
- Third term \( A + 2D = 3 + 4 = 7 \)
- Fourth term \( A + 3D = 3 + 6 = 9 \)

Thus, the AP is:
\[ 3, 5, 7, 9, \ldots \]

### Final Answer
The arithmetic progression is:
\[ 3, 5, 7, 9, \ldots \]