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Find the sum of all integers from 1 to 100 that are divisible by 2 and 5

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answered by (26.8k points)

Solution :

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 bracketsspace 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 bracketsspace 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 parenthesesspace 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 bracketsspace 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 bracketsspace 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 95 close square brackets space equals space 10 space cross times space 105space 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 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

therefore space 10 space plus space 20 space plus space... space plus space 100 space equals space 10 over 2 space open square brackets 2 open parentheses 10 close parentheses space plus space open parentheses 10 space minus 1 close parentheses open parentheses 10 close parentheses close square bracketsspace 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 space equals space 5 space open square brackets 20 space plus space 90 close square brackets space equals space 5 open parentheses 110 close parentheses space equals space 550

∴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.

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