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Earlier discussion
1 older comment kept for context.
Victor Titus
\(S_{n} = \frac{n}{2}(2a + (n - 1)d)\) ( for an arithmetic progression)
\(S_{3} = 18 = \frac{3}{2}(2(4) + (3 - 1) d) \)
\(18 = \frac{3}{2} (8 + 2d)\)
\(18 = 12 + 3d \implies 3d = 6\)
\(d = 2\)
\(\therefore T_{1} = 4 \implies T_{2} = 4 + 2 = 6; T_{3} = 6 + 2 = 8\)
Their product = \(4 \times 6 \times 8 = 192\)