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For the reaction, the equilibrium constant values are given $$A \rightleftharpoons B$$, $$K_1 = 2$$; $$B \rightleftharpoons C$$, $$K_2 = 4$$; $$C \rightleftharpoons D$$, $$K_3 = 3$$. The equilibrium constant for the reaction $$A \rightleftharpoons D$$ is
Adding the three reactions gives $$A \rightleftharpoons D$$, and when reactions are added their equilibrium constants multiply. So $$K = K_1 K_2 K_3 = 2 \times 4 \times 3 = 24$$.
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