Question

Find the sum of all natural number between 1 and145 which
are divisible by 4​

Answer 1

Answer :

The sum of all natural numbers divisible by 4 between 1 and 145 is 2664.

Step-by-step explanation :

From 1 to 145,

the first natural number divisible by 4 = 4

the last natural number divisible by 4 = 144

The sequence is as follows :

4 , 8 , 12 , 16 , ... , 144

Since each term is obtained by adding 4 to the previous term, the sequence is in Arithmetic Progression. (A.P)

first term, a = 4

common difference, d = 4

Find which term is 144

nth term of an A.P. is given by,

aₙ = a + (n - 1)d

144 = 4 + (n - 1)(4)

144 - 4 = (n - 1)(4)

140 = 4n - 4

4n = 140 + 4

4n = 144

n = 144/4

n = 36

∴ 36th term is 144.

It means, there are 36 terms divisible by 4 between 1 to 145.

We have to find the sum of the 36 terms.

In an A.P., the sum of n terms is given by,

$\bf S_n=\dfrac{n}{2}[2a+(n-1)d]$

$\sf \rightarrow S_{36}=\dfrac{36}{2}[2(4)+(36-1)(4)] \\\\ \sf \rightarrow S_{36}=18[8+35(4)] \\\\ \rightarrow \sf S_{36}=18[8+140] \\\\ \sf \rightarrow S_{36}=18[148] \\\\ \rightarrow \sf S_{36}=2664$

Therefore, sum of all natural numbers divisible by 4 between 1 and 145 is 2664.