Find the sum of integers from 1 to 100 which are divisible by 2 or 5.
Answers
Answer
The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.
This forms an A.P. with both the first term and common difference equal to 2.
⇒100 = 2 + (n –1) 2
⇒ n = 50

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.
This forms an A.P. with both the first term and common difference equal to 5.
∴100 = 5 + (n –1) 5
⇒ 5n = 100
⇒ n = 20

The integers, which are divisible by both 2 and 5, are 10, 20, … 100.
This also forms an A.P. with both the first term and common difference equal to 10.
∴100 = 10 + (n –1) (10)
⇒ 100 = 10n
⇒ n = 10

∴Required sum = 2550 + 1050 – 550 = 3050
Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050
Answer:
Answer
The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.
This forms an A.P. with both the first term and common difference equal to 2.
⇒100 = 2 + (n –1) 2
⇒ n = 50
therefore space 2 plus 4 plus 6 plus... plus 100 space equals 50 over 2 space open square brackets 2 open parentheses 2 close parentheses space plus space open parentheses 50 minus 1 close parentheses open parentheses 2 close parentheses close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 50 over 2 space open square brackets 4 space plus space 98 close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses 25 close parentheses open parentheses 102 close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2550
The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.
This forms an A.P. with both the first term and common difference equal to 5.
∴100 = 5 + (n –1) 5
⇒ 5n = 100
⇒ n = 20
therefore space 5 space plus space 10 space plus space... space plus space 100 space equals space 20 over 2 space open square brackets 2 open parentheses 5 close parentheses space plus space open parentheses 20 space minus 1 close parentheses 5 close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 10 space open square brackets 10 space plus space open parentheses 19 close parentheses 5 close square brackets
equals space 10 space open square brackets 10 space plus space 95 close square brackets space equals space 10 space cross times space 105
equals 1050
The integers, which are divisible by both 2 and 5, are 10, 20, … 100.
This also forms an A.P. with both the first term and common difference equal to 10.
∴100 = 10 + (n –1) (10)
⇒ 100 = 10n
⇒ n = 10