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What is the sum of the first 113 consecutive even numbers?
It should be 12882 if I calculated correctly. However I cannot show you how I got the answer here because of the limited mathematical symbols input here. The steps involve using sigma with upper and lower limits. If you want to see my work, visit quickanswerz.com and post this same question there and I can show you exactly how it's worked out!The even number starts with 2 and goes on.2, 4, 6, 8… It forms a sequence with the first term 'a' is 2 and the common difference 'd' is 2. We need to find the sum of the first 113 consecutive even numbers. Sum of n terms in a sequence is given by the formula Sn = (n/2) (a + l) (Where 'a' is the first term of the sequence and 'l' is the last term of the sequence.) First, we need to find the last term of the sequence. We know that the nth term of a sequence is given by the formula tn = a + (n-1) d So, t113 = a + (113-1) d. Substitute the value of a and d in the above equation. t113 = 2 + (113-1) 2 = 2 + (112) 2 t113 = 2 + 224. t113 = 226. So, the last term l is 226. Now, substitute all the values in the formula Sn = (n/2) (a + l). Sn = (113/2) (2 + 226) Sn = (113/2) (228) Sn = (113) (114) Sn = 12882 So, the sum of the first 113 consecutive integers is 12882. Source: www.icoachmath.com Option 2 An easier solution would be to use the formula N(N+1), where N is the number of consecutive even numbers. Eg: sum of 113 consecutive even integers would be = 113* 114 = 12882. Use this formula to calculate the sum of consecutive even numbers starting from 2. If the number starts from the middle (20,22,24,26) you need to use Option 1 to find the solution.
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