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

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

Sol. The natural numbers from 1 to 140 that are divisible by 4 are as follows:
4, 8, 12, 16, ..., 140

These numbers form an A.P. (Arithmetic Progression) where:
First term \(a = 4\)
Common difference \(d = 4\)

Let 140 be the \(n^{th}\) term of the A.P.
$$t_n = 140$$

Using the formula: \(t_n = a + (n - 1)d\)

\(\therefore 140 = 4 + (n - 1)4\)
\(\therefore 140 = 4 + 4n - 4\)
\(\therefore 140 = 4n\)
\(\therefore n = \frac{140}{4}\)
\(\therefore n = 35\)

\(\therefore\) 140 is the 35th term of the A.P.

Now, we have to find the sum of 35 terms, i.e., \(S_{35}\).

Using the formula: \(S_n = \frac{n}{2}[2a + (n - 1)d]\)

\(\therefore S_{35} = \frac{35}{2}[2(4) + (35 - 1)4]\)
\(\therefore S_{35} = \frac{35}{2}[8 + 34(4)]\)
\(\therefore S_{35} = \frac{35}{2}[8 + 136]\)
\(\therefore S_{35} = \frac{35}{2}[144]\)
\(\therefore S_{35} = 35 \times 72\)
\(\therefore S_{35} = 2520\)

\(\therefore\) Sum of natural numbers from 1 to 140 that are divisible by 4 is 2520.