The statement is true.
\(a \propto b\) \(\therefore a = k_1 b\) [where \(k_1\) is a non-zero constant] \(b \propto \frac{1}{c}\) \(\therefore b = \frac{k_2}{c}\) [where \(k_2\) is a non-zero constant] \(c \propto d\) \(\therefore c = k_3 d\) [where \(k_3\) is a non-zero constant] Now, \(a = k_1 b = \frac{k_1 k_2}{c} = \frac{k_1 k_2}{k_3 d}\) Or, \(a d = \frac{k_1 k_2}{k_3}\) = constant [since \(k_1, k_2, k_3\) are constants] \(\therefore a \propto \frac{1}{d}\)