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}$