Concept:Use the arithmetic progression (A.P.) formulas for the nth term and the sum of n terms.Explanation:For an A.P., the nth term is Tn=a+(n−1)d.Given T3=6, so a+2d=6.Given T5=12, so a+4d=12.Subtract the first equation from the second: (a+4d)−(a+2d)=12−6.This gives 2d=6, so d=3.Substitute d=3 into a+2d=6: a+6=6, hence a=0.The sum of the first n terms is Sn=2n(2a+(n−1)d).For n=12: S12=212(2(0)+(12−1)(3)).Simplify: S12=6(0+33)=6×33=198.Answer:S12=198, so the correct option is C. 198.