Question

# Find out the sum of all-natural numbers between 1 and 145 which are visible by 4.

Hint: We can form a series of natural numbers divisible by 4 using arithmetic progression within the desired range. Then we can find the sum of that series.

Given the problem, we need to find out the sum of all-natural numbers between 1 and 145 which are divisible by 4.
We know that the smallest natural number divisible by 4 is 4 itself.
The next natural numbers divisible by 4 are 8,12, 16, till 144, as 144 is the largest natural number less than 145 divisible by 4.
It is observed that the numbers in this series have a common difference of 4.
Hence the above series forms an arithmetic progression,
4,8,12,16…144 (1)
In the above A.P series,
First term $a = 4$
And last term $l = 144$
Also, we know that the general nth term in an A.P series is given by
${a_n} = a + \left( {n - 1} \right)d$, where d is the common difference.
The nth term in A.P series (1) is 144.
Using the above formula, we get
$144 = 4 + \left( {n - 1} \right)4 \\ \Rightarrow \dfrac{{140}}{4} = \left( {n - 1} \right) \\ \Rightarrow n = 36 \\$
Hence the no. of terms in series (1) is 36.
Now we can find the sum of this series.
Also, we know that sum of an arithmetic progression is given by
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) \\ \Rightarrow {S_{36}} = \dfrac{{36}}{2}\left( {8 + \left( {36 - 1} \right)4} \right) = 18\left( {148} \right) = 2664 \\$
Therefore, the sum of terms of A.P series (1) is equal to 2664.
Hence, the sum of all-natural numbers between 1 and 145 which are visible by 4 is equal to 2664.

Note: The multiples of a number differ by the same number. Arithmetic progression formulas should be kept in mind while solving problems like above. Same procedure could be used for finding the sum of natural numbers divisible by any other number.