Question

1) Find the sum of natural numbers between 1 and 140 which are divisible by 4

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Answer 1

Answer:

It is an Question of Arithmetic Progression

Let a = 4 (as its is first natural number divisible by 4)

Let d = 4

an = 140

an = a + (n - 1)d

140 = 4 + (n - 1)4

136 = 4n - 4

140 = 4n

n = 35

There are 35 numbers between 1 and 140 which are divisible by 4

Answer 2

Given: The series of natural numbers between 1 to 40 which are divisible by 4

To find: Sum or these natural numbers

Explanation: The natural numbers between 1 to 140 which are divisible by 4 are as follows:

4,8,12,16,20....136,140

This is an arithmetic progression in which first term (a) = 4 and and common difference (d)=4.

To find the number of these natural numbers, let the nth number be 140.

140= 4 + (n-1) 4

=> 136 = (n-1) 4

=> n-1 = 34

=> n = 35

There are 35 terms in this series where first term (a)= 4 and last term(l)=140

Formula for calculating sum of these numbers

=$\frac{n}{2} (a + l)$

=$\frac{35}{2} (4 + 140)$

=$\frac{35 \times 144}{2}$

=2520

Therefore, the sum of the natural numbers between 1 to 140 which are divisible by 4 is 2520.