In a sequence of 4 numbers, the difference between each number is seven. The sum is six. What are the numbers in the sequence?
回答 (3)
2016-09-06 8:08 pm
x + (x+7) + (x+14) + (x+21) = 6
4x + 42 = 6
4x = −36
x = −9
The 4 numbers are: −9, −2, 5, 12
2016-09-06 8:03 pm
a₄ = a₁ + (4-1)7 = a₁ + 21
∑ = 4(a₁+a₄)/2 = 6
(a₁ + (a₁+21)) = 3
2a₁ + 21 = 3
a₁ = -9
a₂ = -9+7 = -2
a₃ = -2+7 = 5
a₄ = 5+7 = 12
∑ = 4(a₁+a₄)/2 = 6
(a₁ + (a₁+21)) = 3
2a₁ + 21 = 3
a₁ = -9
a₂ = -9+7 = -2
a₃ = -2+7 = 5
a₄ = 5+7 = 12
2016-12-04 11:25 pm
x-y=7
x+y=6 ,,,add
2x=13,,x=13/2=
y=13/2-7=(13-14)/2=-1/2
-1/2,13/2,
x+y=6 ,,,add
2x=13,,x=13/2=
y=13/2-7=(13-14)/2=-1/2
-1/2,13/2,
收錄日期: 2021-04-21 21:39:25
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https://hk.answers.yahoo.com/question/index?qid=20160906115955AAccn9e