If the sum of the first 40 terms of the series, $3+4+8+9+13+14+18+19+...$is  $102m$, then  $m$ is equal to 

The given series is $3+4+8+9+13+14+18+19+...(40 terms)$

This can be written as 

$(3+4)+(8+9)+(13+14)+(18+19)+...\left(20 brackets\right)$

This is equal to $7+17+27+37+...\left(20 terms\right)$

Observing the above series, it is sum of 20 terms of Arithmetic progression with the first term $a=7$ and the common difference $d=10$ 

We know that the sum of $n$ terms of an Arithmetic progression having first term $a$ and the common difference $d$ is $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

Therefore, the required sum is 

$\begin{array}{rcl}{S}_{20}& =& \frac{20}{2}\left(2\left(7\right)+\left(20-1\right)10\right)\\ & =& 10\left(14+19·10\right)\\ & =& 10\left(14+190\right)\\ & =& 10\left(204\right)\\ & =& 2040\end{array}$

Hence, the value of the given series is $2040$ is equal to $102m$

Therefore, $m=\boxed{\mathbf{20}}$