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Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.
Here is a Solution ...... The positive integers, which are divisible by 5, are 5, 10, 15, ..., 1000 Out of these 10,20,30,.... 1000 are divisible by 2 Thus, we have to find the sum of the positive integers 5, 15, 25, ...., 995 If n is the number of terms in it the sequence then 995 = 5 + 10(n - 1) => 1000 = 10n Therefore, n = 100. Thus the sum of the series = (n/2)(a + l) = (100/2) (5 + 995) = 50000.
Here is a Solution ......
The positive integers, which are divisible by 5, are 5, 10, 15, ..., 1000
Out of these 10,20,30,.... 1000 are divisible by 2
Thus, we have to find the sum of the positive integers 5, 15, 25, ...., 995
If n is the number of terms in it the sequence then
995 = 5 + 10(n - 1) => 1000 = 10n
Therefore, n = 100.
Thus the sum of the series = (n/2)(a + l) = (100/2) (5 + 995) = 50000.
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