The above synchronous sequential circuit built using JK flip-flops is initialized with $Q_2Q_1Q_0 = 000$. The state sequence for this circuit for the next 3 clock cycles is

From the circuit diagram, the inputs to the JK flip-flops are connected as follows:

• Flip-flop 2 ($Q_2$):

  $$J_2 = \overline{Q}_0, \quad K_2 = \overline{Q}_0$$

  Characteristic equation: $Q_2^+ = J_2 \overline{Q}_2 + \overline{K}_2 Q_2 = \overline{Q}_0 \overline{Q}_2 + Q_0 Q_2$

• Flip-flop 1 ($Q_1$):

  $$J_1 = Q_2, \quad K_1 = \overline{Q}_2$$

  Characteristic equation: $Q_1^+ = J_1 \overline{Q}_1 + \overline{K}_1 Q_1 = Q_2 \overline{Q}_1 + Q_2 Q_1 = Q_2$

• Flip-flop 0 ($Q_0$):

  $$J_0 = Q_1, \quad K_0 = \overline{Q}_1$$

  Characteristic equation: $Q_0^+ = J_0 \overline{Q}_0 + \overline{K}_0 Q_0 = Q_1 \overline{Q}_0 + Q_1 Q_0 = Q_1$

Now, tracking the state transitions starting from $Q_2 Q_1 Q_0 = 000$:

1. Clock Cycle 1:

   • Current state: $Q_2 Q_1 Q_0 = 000$

   • Inputs: $J_2=1, K_2=1 \implies Q_2^+ = 1$

   • Inputs: $J_1=0, K_1=1 \implies Q_1^+ = 0$

   • Inputs: $J_0=0, K_0=1 \implies Q_0^+ = 0$

   • Next State 1: $100$

2. Clock Cycle 2:

   • Current state: $Q_2 Q_1 Q_0 = 100$

   • Inputs: $J_2=1, K_2=1 \implies Q_2^+ = 1$

   • Inputs: $J_1=1, K_1=0 \implies Q_1^+ = 1$

   • Inputs: $J_0=0, K_0=1 \implies Q_0^+ = 0$

   • Next State 2: $110$

3. Clock Cycle 3:

   • Current state: $Q_2 Q_1 Q_0 = 110$

   • Inputs: $J_2=1, K_2=1 \implies Q_2^+ = 1$

   • Inputs: $J_1=1, K_1=0 \implies Q_1^+ = 1$

   • Inputs: $J_0=1, K_0=0 \implies Q_0^+ = 1$

   • Next State 3: $111$

Thus, the state sequence for the next 3 clock cycles is 100, 110, 111, which corresponds to option (C).