**Given:**
Mass of block A, m_A = 2 kg
Mass of block B, m_B = 4 kg
Mass of block C, m_C = 2.75 kg

**Assumptions:**
1. The pulleys are massless and frictionless.
2. The threads are inextensible.
3. The system is in equilibrium before the blocks are released.

**Analysis:**
To find the acceleration of block A, we need to consider the forces acting on it.

**1. Tension in the left thread:**
The tension in the left thread is equal to the force exerted by block B on the pulley. Since the pulley is massless and frictionless, the tension in the left thread is equal to the weight of block B, which is given by:
T_left = m_B * g

**2. Tension in the right thread:**
The tension in the right thread is equal to the force exerted by block C on the pulley. Similarly, the tension in the right thread is equal to the weight of block C, which is given by:
T_right = m_C * g

**3. Acceleration of block B:**
The net force on block B is the difference between the tension in the left thread and the weight of block B. Since the system is in equilibrium, this net force is equal to zero. Therefore:
T_left - m_B * g = 0
T_left = m_B * g

**4. Acceleration of block C:**
The net force on block C is the difference between the tension in the right thread and the weight of block C. Since the system is in equilibrium, this net force is equal to zero. Therefore:
T_right - m_C * g = 0
T_right = m_C * g

**5. Acceleration of block A:**
The net force on block A is the difference between the tension in the left thread and the tension in the right thread. Therefore:
T_left - T_right = m_A * a
m_B * g - m_C * g = m_A * a
a = (m_B - m_C) * g / m_A

**Substituting the given values:**
a = (4 kg - 2.75 kg) * 9.8 m/s^2 / 2 kg
a = 13.65 m/s^2

Therefore, the acceleration of block A is 13.65 m/s^2.

Where are the pulleys