If the sum of the first 40 terms of the series,  $3+4+8+9+13+14+18+\mathrm{19...........}$is (102)m, then m is equal to:

$S=3+4+8+9+13+14+18+\mathrm{19........40}$ terms

$\Rightarrow S=7+17+27+37+47+\mathrm{.....20}$ terms

$\therefore S_{40}=\frac{20}{2}\left[2\times 7+\left(19\right)10\right]=10\left[14+190\right]=10\left[2040\right]=\left(102\right)\left(20\right)$

$\Rightarrow m=20$