The sum of all odd numbers between 1 and 1000 which are divisible...

Sum of odd numbers between 1 and 1000 , which is divisible by $3=3+9+15+21+27+\ldots+999=\mathrm{S}$ (let) $\therefore$ Let $\mathrm{n}$ be the number of terms in series and $\mathrm{a}$ is first term. $\therefore l=a+(n-1) d$, where $l$ is last term and $d$ is is common difference. $$\begin{array}{l} 999=3+(n-1) \times 6 \\ n-1=\frac{999-3}{6}=\frac{996}{6} \\ \Rightarrow n-1=166 \\ \Rightarrow n=167 \\ \therefore \mathrm{S}=\frac{n}{2}[2 a+(n-1) d] \\ =\frac{167}{2}[2 \times 3+(167-1) \times 6] \\ =\frac{167}{2}[1002]=167 \times 501=83667 \end{array}$$