T\(_3\) = 6 T\(_5\) = 12 S\(_{12}\) = ? T\(_n\) = a + (n - 1)d ⇒ T\(_3\) = a + 2d = 6 ----- (i) ⇒ T\(_5\) = a + 4d = 12 ----- (ii) Subtract equation (ii) from (i) ⇒ -2d = -6 ⇒ d\(\frac{-6}{-2}\) = 3 Substitute 3 for d in equation (i) ⇒ a + 2(3) = 6 ⇒ a + 6 = 6 ⇒ a = 6 - 6 = 0 S\(_n\) = \(\frac{n(2a + (n - 1)d)}{2}\) ⇒ S\(_{12}\) = \(\frac{12(2 \times 0 + (12 - 1)3)}{2}\) ⇒ S\(_{12}\) = 6(0 + 11 x 3) ⇒ S\(_{12}\) = 6(33) ∴ S\(_{12}\) = 198