Question

Find the sum of all numbers from 1 to 140 which are divisible by 4.

Answer 1

All the numbers from 1 to 140 that are divisible by 4 are 4, 8, 12, 16,…, 140.
Now, we observe that the numbers are in an A.P.
Here,
First term, a = 4
Common difference, d = t₂ – t₁ = 8 – 4 = 4
Let the number of these odd numbers be n.
We have: t_n = 140
We know that the n^th term of an A.P. is t_n = a + (n – 1)d.
Thus, we have:
140 = 4 + (n – 1)4
$\Rightarrow$140 – 4 = 4(n – 1)
$\Rightarrow$136 = 4(n – 1)
$\Rightarrow$(n – 1) = 34
$\Rightarrow$n = 34 + 1 = 35
Thus, there are 35 numbers from 1 to 140 that are divisible by 4.
Now,
We know that the sum of n terms of an A.P is $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.
To find the sum of the numbers that are divisible by 4, we need to take a = 4, d = 4 and n = 35.
Thus, we have:

$$\begin{gathered} S_{35}=\frac{35}{2}\left[2\times 4+(35-1)4\right]=\frac{35}{2}\left[8+136\right] \\ =\frac{35}{2}\times 144 \\ =35\times 72 \\ =2520 \end{gathered}$$