Question

Find the sum of all natural numbers between 500 and 800 which leaves a reminder 2 when divided by5

Answer 1

592, 597, 692, 697, 792, 797,

Answer 2

Answer:

98910

Step-by-step explanation:

ap= 102 ,107 ,112 . . . 997

a= 102

d= 5

an= a+(n-1) d

997           = 102 +(n-1) 5

997-102    = (n-1) 5

895           = (n-1) 5

895/5       = (n-1)

179            = n-1

n  =  180

Sn = n/2 ( a+An )

    =180/2 (102+997)

    = 90 * 1099

    = 98910

=)