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 \( q \) given the coefficients of \( x \) and \( x^2 \) in the expansion of \( (1 + px)^q \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the expansion of \( (1 + px)^q \) is given by:
\[ T_{r+1} = \binom{q}{r} (1)^{q-r} (px)^r = \binom{q}{r} p^r x^r \]

2. **Find the Coefficient of \( x \)**:
The coefficient of \( x \) corresponds to \( r = 1 \):
\[ \text{Coefficient of } x = T_2 = \binom{q}{1} p^1 = qp \]
Given that this coefficient is 12, we have:
\[ qp = 12 \quad \text{(1)} \]

3. **Find the Coefficient of \( x^2 \)**:
The coefficient of \( x^2 \) corresponds to \( r = 2 \):
\[ \text{Coefficient of } x^2 = T_3 = \binom{q}{2} p^2 = \frac{q(q-1)}{2} p^2 \]
Given that this coefficient is 60, we have:
\[ \frac{q(q-1)}{2} p^2 = 60 \quad \Rightarrow \quad q(q-1)p^2 = 120 \quad \text{(2)} \]

4. **Substituting \( qp \) into Equation (2)**:
From equation (1), we can express \( p \) in terms of \( q \):
\[ p = \frac{12}{q} \]
Substitute \( p \) into equation (2):
\[ q(q-1) \left(\frac{12}{q}\right)^2 = 120 \]
Simplifying this gives:
\[ q(q-1) \frac{144}{q^2} = 120 \]
\[ 144(q-1) = 120q \]
\[ 144q - 144 = 120q \]
\[ 24q = 144 \quad \Rightarrow \quad q = 6 \]

5. **Finding \( p \)**:
Substitute \( q = 6 \) back into equation (1):
\[ 6p = 12 \quad \Rightarrow \quad p = 2 \]

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