An AP has first term $\frac{3}{4}$ and common difference $\frac{5}{4}$. Find the sum of its first 16 terms.

$S=\frac{16}{2}$[$2(\frac{3}{4})+15(\frac{5}{4})$] = 8[$\frac{3}{2}+\frac{75}{4}$] $=8(\frac{81}{4})=162$. The value 172 uses 16d instead of 15d in the bracket. 324 drops the factor $\frac{1}{2}.\frac{39}{2}$ is only $a_{16}$, the last term, not the sum of all sixteen.