How much more is the sum of the natural numbers from 21 to 40 than the sum of the natural numbers from 1 to 20?

Natural numbers from 21 to 40 = 21, 22, 23, … , 40

Sum of these numbers = 21 + 22 + 23 + … + 40

Here, first term = a + b = 21

Common difference = a = 22 − 21 = 1

⇒ 1 + b = 21

⇒ b = 20

Therefore,

⇒ 40 = n + 20

⇒ n = 40 − 20 = 20

We know that the sum of a specified number of the consecutive terms of an arithmetic sequence is half the product of the number of the terms with the sum of the first and the last term.

∴ Sum of its first 20 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 20 terms =

Natural numbers from 1 to 20 = 1, 2, 3, … , 20

Sum of these numbers = 1 + 2 + 3 + … + 20

Here, first term = a + b = 1

Common difference = a = 2 − 1 = 1

⇒ 1 + b = 1

⇒ b = 0

Therefore,

⇒ n = 20

∴ Sum of its first 20 terms = … (1)

Here,

Putting the values in equation (1):

Sum of first 20 terms =

∴ Required difference = 610 − 210 = 400

Thus, the sum of natural numbers from 21 to 40 is 400 more than the sum of natural numbers from 1 to 20.