Find the sum of first $22$ terms of an AP in which $d=7$ and the $22$nd term is $149$.

First find the first term from $a_{22}=a+21d$: $$149=a+21\cdot 7=a+147,$$ so $a=2$. Then $$S_{22}=\frac{22}{2}[2a+(22-1)d]=11[4+147]=11\cdot 151=1661.$$ But check carefully: using $S=\frac{n}{2}(a+l)$ is simpler: $$S_{22}=\frac{22}{2}(2+149)=11\cdot 151=1661.$$ Therefore the correct sum is $1661$.