Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
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.
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Evaluate
The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
Find the 20th and nthterms of the G.P.
Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Describe the sample space for the indicated experiment: A coin is tossed three times.
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Can it be solved by n(n+2)/2
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Not best