Mathematics - Sequence & Series Question with Solution | TestHub

Sum of odd numbers between 1 and 1000 , which is divisible by $3=3+9+15+21+27+\dots +999=\mathrm{S}$ (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. $\begin{array}{l}999=3+(n-1)\times 6\\ n-1=\frac{999-3}{6}=\frac{996}{6}\\ \Rightarrow n-1=166\\ \Rightarrow n=167\\ \therefore \mathrm{S}=\frac{n}{2}[2a+(n-1)d]\\ =\frac{167}{2}[2\times 3+(167-1)\times 6]\\ =\frac{167}{2}[1002]=167\times 501=83667\end{array}$