If the sum of the first 4 terms of an arithmetic progression is p, the sum of the first 8 terms is q and the sum of the first 12 terms is r, express 3p+r in terms of q.

Answer

To solve the problem, we need to express \(3p + r\) in terms of \(q\) given the sums of the first few terms of an arithmetic progression (AP).

### Step-by-Step Solution:

1. **Recall the formula for the sum of the first \(n\) terms of an AP:**
\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \]
where \(a\) is the first term and \(d\) is the common difference.

2. **Calculate \(p\):**
The sum of the first 4 terms is given as \(p\). Using the formula:
\[ p = S_4 = \frac{4}{2} \left(2a + (4 - 1)d\right) = 2(2a + 3d) = 4a + 6d \]
Let's denote this as Equation (1):
\[ p = 4a + 6d \quad \text{(1)} \]

3. **Calculate \(q\):**
The sum of the first 8 terms is given as \(q\). Using the formula:
\[ q = S_8 = \frac{8}{2} \left(2a + (8 - 1)d\right) = 4(2a + 7d) = 8a + 28d \]
Let's denote this as Equation (2):
\[ q = 8a + 28d \quad \text{(2)} \]

4. **Calculate \(r\):**
The sum of the first 12 terms is given as \(r\). Using the formula:
\[ r = S_{12} = \frac{12}{2} \left(2a + (12 - 1)d\right) = 6(2a + 11d) = 12a + 66d \]
Let's denote this as Equation (3):
\[ r = 12a + 66d \quad \text{(3)} \]

5. **Express \(3p + r\):**
Now we can express \(3p + r\):
\[ 3p = 3(4a + 6d) = 12a + 18d \]
Adding \(r\):
\[ 3p + r = (12a + 18d) + (12a + 66d) = 24a + 84d \]

6. **Relate \(3p + r\) to \(q\):**
From Equation (2), we have:
\[ q = 8a + 28d \]
We can factor out a 3 from \(q\):
\[ 3q = 3(8a + 28d) = 24a + 84d \]

7. **Final Expression:**
Therefore, we can express \(3p + r\) in terms of \(q\):
\[ 3p + r = 3q \]

### Conclusion:
Thus, the final answer is:
\[ 3p + r = 3q \]