In in the expansion of `(1+px)^(q)`, `q` belongs to `N`, the coefficients of `x` and `x^(2)` are `12` and `60` respectively then `p` and `q` are

Answer

To solve the problem, we need to find the values of \( p \) and \( q \) in the expansion of \( (1 + px)^q \) given that the coefficients of \( x \) and \( x^2 \) are 12 and 60, respectively.

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The coefficient of \( x^r \) in the expansion of \( (1 + px)^q \) is given by the formula:
\[ C(r) = \binom{q}{r} p^r \]
For \( r = 1 \) (coefficient of \( x \)):
\[ C(1) = \binom{q}{1} p = qp \]
For \( r = 2 \) (coefficient of \( x^2 \)):
\[ C(2) = \binom{q}{2} p^2 = \frac{q(q-1)}{2} p^2 \]

2. **Set Up the Equations**:
From the problem, we know:
\[ qp = 12 \quad \text{(1)} \]
\[ \frac{q(q-1)}{2} p^2 = 60 \quad \text{(2)} \]

3. **Substitute Equation (1) into Equation (2)**:
From equation (1), we can express \( p \) in terms of \( q \):
\[ p = \frac{12}{q} \]
Substitute \( p \) into equation (2):
\[ \frac{q(q-1)}{2} \left(\frac{12}{q}\right)^2 = 60 \]
Simplifying this gives:
\[ \frac{q(q-1)}{2} \cdot \frac{144}{q^2} = 60 \]
\[ \frac{72(q-1)}{q} = 60 \]

4. **Cross Multiply and Simplify**:
Cross-multiplying gives:
\[ 72(q - 1) = 60q \]
Expanding this:
\[ 72q - 72 = 60q \]
Rearranging terms:
\[ 72q - 60q = 72 \]
\[ 12q = 72 \]
\[ q = 6 \]

5. **Find \( p \)**:
Substitute \( q = 6 \) back into equation (1):
\[ 6p = 12 \implies p = \frac{12}{6} = 2 \]

### Final Answer:
Thus, the values of \( p \) and \( q \) are:
\[ p = 2, \quad q = 6 \]