The AP 1.5, 2.9, 4.3, 5.7, ... is given. Find the sum of its first 12 terms.

$a=1.5,d=1.4,n=12.S=\frac{12}{2}$$[2(1.5)+11(1.4)]=6[3+15.4]=6(18.4)=110.4$. The trap 118.8 comes from using 12$d$ in the bracket instead of $(n-1)d=11d$. 220.8 drops the factor $\frac{1}{2}$. 16.9 is only the 12th term $a_{12}$, not the sum.