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.