Find the sum of those integers from 1 to 500 which are multiples of 2 or 5
Answers
Case 1: (multiples of 2)
Given:
a = 2, d = 4 - 2 = 2, l = 500, n = 500/2 = 250
Sum = (n/2)*(a + l)
= (250/2)*(2 + 500)
= (250/2) * 50
= 125 * 502
= 62750
Case 2 : (multiples of 5)
Given :
a = 5, d = 10 - 5 = 5, l = 500, n = 500/5 = 100
Sum = (n/2)*(a + l)
= (100/2)*(5 + 500)
= 50 * 505
= 25250
Both has multiples of ten. So,
Given :
a = 10,d = 20 - 10 = 10, l = 500, n = 500/10 = 50
Sum = (n/2)*(a + l)
= (50/2)*(10 + 500)
= 25 * 510
= 12750
So ,the sum of those integers from 1 to 500 which are multiples of 2 or 5
= 62750 + 25250 - 12750
= 75250
THEREFORE THE SUM IS 75250.
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Case 1: (multiples of 2)
Given:
a = 2, d = 4 - 2 = 2, l = 500, n = 500/2 = 250
Sum = (n/2)x(a + l)
= (250/2)x(2 + 500)
= (250/2) x 50
= 125 x 502
= 62750
Case 2 : (multiples of 5)
Given :
a = 5, d = 10 - 5 = 5, l = 500, n = 500/5 = 100
Sum = (n/2)x(a + l)
= (100/2)x(5 + 500)
= 50 x 505
= 25250
Both have multiples of ten. So,
Given :
a = 10,d = 20 - 10 = 10, l = 500, n = 500/10 = 50
Sum = (n/2)x(a + l)
= (50/2)x(10 + 500)
= 25 x 510
= 12750
∴the sum of those integers from 1 to 500 which are multiples of 2 or 5
= 62750 + 25250 - 12750
= 75250