Mathematics - Sequence & Series Question with Solution | TestHub

API $=12,15,18,\dots ($ common difference ${\mathrm{d}}_1=3)$

AP II $=17,21,25\dots$ (common difference ${\mathrm{d}}_2=4$ )

First term of the series of common numbers $=21$

Here $\mathrm{a}=21$,

common difference of the series of common numbers $=$ L.C.M of ${\mathrm{d}}_1$ and ${\mathrm{d}}_2=12$ ∴ Required sum of first hundred terms

$$=\frac{100}{2}[2\times 21+(100-1)12]=100[21+594]=61500$$