Question

Find the sum of the first 1000 even positive integers *

Answer 1

Step-by-step explanation:

sum formula : n/2 [ 2a + (n-1) ] d

Here : a= 1

d= 2-1=1

n = 1000

put all of these into formula ,

= 1000/2 [2×1 + (1000-1) ] × 1

= 500× (2+999)

= 500× 1001

= 500500

Answer 2

Given:

First 1000 positive even integers = n

To Find:

The sum of the first 1000 even positive integers

Solution:

The number series of prime even numbers is = 2, 4, 6, 8, 10, 12, _ _ _ _ 2000.

The first term a = 2

The common difference = d = 4-2 = 2

Now, using the AP formula -

Sum = n/2 x (a + Tn)

Substituting the values -

= 1000/2 x (2 + 2000)

= (1000 x 2002)/ 2

= 2002000/2

= 1001000

Answer: The sum of the first 1000 even positive integers is 1,00,1000