Find the 10th term of a geometric sequenc and common ratio is =0.2
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find the sum of all positive numbers from 5 to 1,555 that are divisible by 5

The numbers from 5 to 1555 that are divisible by 5 are: 5, 10, 15, 20, ... 1545, 1550, 1555

which is five times 1, 2, 3, 4, ..., 309, 310, 311.

So, the sum of all positive numbers from 5 to 1,555 that are divisible by 5 is 5 times the sum of all positive numbers from 1 to 311.

The sum of all positive numbers from 1 to n is: (1/2)n(n+1)

The sum of all positive numbers from 1 to 311 is: (1/2)*311*(311 + 1) = 311*156 = 48,516

Answer: 242, 580

by Level 11 User (81.5k points)

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