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There are no 4 consecutive integers whose sum is 60. 1+2+3+4=10 2+3+4+5=14 3+4+5+6=18 etc Each step you are adding a number and subtracting a number 4 less than it (eg +5-1, +6-2). So you are adding 4 each time. But 60 does not appear here but 58 and 62 do (sums starting with 13 and 14). Also 60/4=15. so the average of 4 consecutive numbers has to be 15. This average is nowhere odd and so can not be made this way.
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What Three consecutive integers have the sum of 60?
19, 20, and 21
What are four consecutive even integers whose sum is 244?
Call the lowest of the even integers n. Then from the problem statement, n + n + 2 + n + 4 = n + 6 =244. Collecting like terms results in 4n + 12 = 244, or 4n = 244 - 12 = 232, or n = 232/4 or 58. The consecutive even integers are therefore, 58, 60, 62, and 64.
60 falls between what two consecutive integers?
59 and 61
What is the sum of two consecutive even integers that equals to 118?
To work this sort of problem out we need to divide the number given by the number of integers being sought (in that case two) 118 ÷ 2 = 59. So the two numbers will average 59 and they must be consecutive and even, that means that they must be 1 above and 1 below 59..... 58 and 60. 58 + 60 = 118
How do you find the sum of the first thirty odd integers?
The arithmetic sequence of odd integers is 1, 3, 5, 7, 9, ... where the common difference is 2. The sum of the first thirty odd integers can be found by using the sum formula such as: Sn = n/2[2a1 + (n - 1)d], where a1 = 1, n = 30, and d = 2. So, S30 = (30/2)[(2)(1) + (30 - 1)(2)] = (15)[2 + (29)(2)] = (15)(60) = 900
