T3 = 6 T5 = 12 S12 = ? Tn = a + (n - 1)d ⇒ T3 = a + 2d = 6 ----- (i) ⇒ T5 = a + 4d = 12 ----- (ii) Subtract equation (ii) from (i) ⇒ -2d = -6
⇒ d−6−2 = 3
Substitute 3 for d in equation (i) ⇒ a + 2(3) = 6 ⇒ a + 6 = 6 ⇒ a = 6 - 6 = 0
Sn = n(2a+(n−1)d)2
⇒ S12 = 12(2×0+(12−1)3)2
⇒ S12 = 6(0 + 11 x 3) ⇒ S12 = 6(33) ∴ S12 = 198
