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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19, 20, and 21
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.
59 and 61
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
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
The integers are 19, 20 and 21.
58, 60, 62, 64
The numbers are 60, 62, 64, 66 and 68.
63
60 and 61
19, 20, and 21
58, 59, 60
Thanks for sharing. The integers are 58 and 60.
lol I need to know this too.
Their sum must be 60.
They are: 59+60+61 = 180
Let's denote the first even integer as x. The next consecutive even integer would be x + 2. Since they sum up to 60, we can write the equation x + (x + 2) = 60. Simplifying this equation gives us 2x + 2 = 60. Solving for x, we find x = 29. Therefore, the two consecutive even integers are 29 and 31, and their sum is 60.