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Hi, A method to calculate if you do not know the formula n(n+1) is ... First number=0.. Tenth number = 0+(10-1)*2=18, 10 is total numbers and 2 is the difference between consecutive integers.

So mean = \(\frac{0+18}{2}=9\).. Sum = 9*10=90 B
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Let's add the 10 values in PAIRS, starting with the two outermost values and working inwards. Notice that 0 + 18 = 18 And 2 + 16 = 18 And 4 + 14 = 18 In fact, when we add the values like this, every pair adds to 18

There are 10 values, so there are 5 pairs (all of which add to 18) So, the sum = 5 x 18 = 90

Re: What is the sum of first 10 non-negative even integers? A) 40 B) 60 C
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20 Jun 2017, 23:28

imo b this is a tricky question as we are asked sum of first 10 non negative even integers , so we have to include 0 . our numbers will be 0 2 4 6 8 10 12 14 16 18 sum =10/2(0+18)=90
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Re: What is the sum of first 10 non-negative even integers? A) 40 B) 60 C
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01 Jul 2017, 12:45

The sum of the first 10 even integers is given by formula \(n(n+1)\)

Since we have been asked the sum of the first non-negative even integers, it includes 0 as well. So in essence, we need to find the sum of the first 9 even integers, which is \(9(9+1)\) = 90(Option B)
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