Complete the following activity to find the sum of natural numbers from 1 to 140 which are divisible by 4.

Sum of numbers from 1 to 140, which are divisible by 4 = $\boxed{}$

From 1 to 140, natural numbers divisible by 4 are 4, 8, ... 136.
a = 4
d = 4

Now,
$$\begin{gathered} t_n=a+\left(n-1\right)d \\ 136=4+\left(n-1\right)4 \\ \Rightarrow 136=4+4n-4 \\ \Rightarrow 4n=136 \\ \Rightarrow n=34 \end{gathered}$$

Thus, number of terms (n) = 34.

We know that,
$$\begin{gathered} S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right) \\ \therefore S_{34}=\frac{34}{2}\left(2\left(4\right)+\left(34-1\right)\left(4\right)\right) \\ =17\left(8+132\right) \\ =17\left(140\right) \\ =2380 \end{gathered}$$



Hence, the sum of numbers from 1 to 140, which are divisible by 4 = $\boxed{2380}$.