Question

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 = `square`

Answer 1

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

a = 4, d = 4

Now, 

\[t_n = a + \left( n - 1 \right)d\]

\[136 = 4 + \left( n - 1 \right)4\]

\[ \Rightarrow 136 = 4 + 4n - 4\]

\[ \Rightarrow 4n = 136\]

\[ \Rightarrow n = 34\]

Thus, number of terms (n) = 34.

We know that,

\[S_n = \frac{n}{2}\left( 2a + \left( n - 1 \right)d \right)\]

\[ \therefore S_{34} = \frac{34}{2}\left( 2(4) + \left( 34 - 1 \right)\left( 4 \right) \right)\]

\[ = 17\left( 8 + 132 \right)\]

\[ = 17\left( 140 \right)\]

\[ = 2380\]

Hence, the sum of numbers from 1 to 140, which are divisible by 4 = 2380