Find 3 consecutive integers such that the sum of the three is equal to 2 times the sum of the first two integers. ?
回答 (6)
2020-09-21 5:00 pm
Let (n - 1), n and (n + 1) be the three consecutive integers.
(n - 1) + n + (n + 1) = 2 × [(n - 1) + n]
3n = 4n - 2
n = 2
The 3 consecutive integers are 1, 2 and 3.
(n - 1) + n + (n + 1) = 2 × [(n - 1) + n]
3n = 4n - 2
n = 2
The 3 consecutive integers are 1, 2 and 3.
2020-09-23 8:44 am
Find 3 consecutive integers such that the sum of the three is equal to
2 times the sum of the first two integers.3x = 2(2x - 1)
x = 2
The 3 consecutive integers are 1, 2 and 3.
2 times the sum of the first two integers.3x = 2(2x - 1)
x = 2
The 3 consecutive integers are 1, 2 and 3.
2020-09-22 4:19 am
You must first set up your formula. Let x= lowest number
X + (x+1) + (x+2)= 2(x + [x+1])
3x + 3= 2(2x + 1)
3x+3=4x+2
1=x
X + (x+1) + (x+2)= 2(x + [x+1])
3x + 3= 2(2x + 1)
3x+3=4x+2
1=x
2020-09-21 7:28 pm
Call the numbers n - 1, n, and n + 1
Then the problem says n-1 + n + n+1 = 3n = 2(n-1 + n)
3n = 4n - 2
n = 2 - the middle number, so the
numbers are 1, 2, and 3
Then the problem says n-1 + n + n+1 = 3n = 2(n-1 + n)
3n = 4n - 2
n = 2 - the middle number, so the
numbers are 1, 2, and 3
2020-09-21 7:25 pm
n + (n + 1) + ( n+2) = 2(n + (n + 1))
Hence
3n + 3 = 4n + 2
4n - 3n = 3 - 2
n = 1
n + 1 = 2
n + 2 = 3
Hence
3n + 3 = 4n + 2
4n - 3n = 3 - 2
n = 1
n + 1 = 2
n + 2 = 3
2020-09-21 6:41 pm
Letting the integers be n, n + 1 and n + 2 we have,
sum of three => n + n + 1 + n + 2 = 3n + 3
sum of first two => n + n + 1 = 2n + 1
Hence, 2(2n + 1) = 3n + 3
i.e. 4n + 2 = 3n + 3
so, n = 1
Therefore, the numbers are 1, 2 and 3
checking gives, 2(1 + 2) = (1 + 2 + 3) = 6
:)>
sum of three => n + n + 1 + n + 2 = 3n + 3
sum of first two => n + n + 1 = 2n + 1
Hence, 2(2n + 1) = 3n + 3
i.e. 4n + 2 = 3n + 3
so, n = 1
Therefore, the numbers are 1, 2 and 3
checking gives, 2(1 + 2) = (1 + 2 + 3) = 6
:)>
收錄日期: 2021-04-12 12:52:02
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