What order can X, 0, and 1/X be three consecutive terms of an arithmetic sequence?
Please clearly show all your steps.
Thanks!
Solution
As pointed out previously, 0 can not be the middle term; therefore, it must be either the first or third; it really doesn\'t matter which, as you just reverse the order, but I will go ahead and show both ways.
If it is 0, then x, then 1/x, let x = y; then, since the first term was 0, the third term must be 2y.
x = y
1/x = 2y; also; 1/x = 1/y;
2y = 1/y
2y 2 = 1
In all cases, where I have ± for x and 1/x, they have the same sign, whether it is plus or minus.
y = ±2/2
x =y = ±2/2
1/x = 2y = ±2
Consider the next possibility; the second term is 1/x and the third term is x
Then, let 1/x = y
x = 2y
Then, 1/x = y = 1/2y
y=1/2y
2y 2 = 1
y = ±2/2
1/x = y = ±2/2
x = 2y = ±2
Next, consider the other possibility, 0 is the third term.
Then, we have two possibilities, x is the second term or x is the first term.
To keep things in the same order as before, first, let x be the second term.
Then, let x = y.
As 1/x is the first term, 1/x = 2y
Then, 2y = 1/y
2y 2 = 1
y = ±2/2
x =y = ±2/2
1/x = 2y = ±2
Next, let 1/x be the second term and x be the first term.
Then, let 1/x = y
x = 2y
Then, 1/x = y = 1/2y
y=1/2y
2y 2 = 1
y = ±2/2
1/x = y = ±2/2
x = 2y = ±2
There are 4 possible plausible orders, the 4 without 0 in the middle:
0 first, x second, 1/x third
0 first, 1/x second, x third
0 third, x second, 1/x first
0 third, 1/x second, x first
In all cases, the value next to 0 is ±2/2, and the other value is ±2 (as stated before, these have the same sign).
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What order can X- 0- and 1-X be three consecutive terms of an arithmet.docx
1. What order can X, 0, and 1/X be three consecutive terms of an arithmetic sequence?
Please clearly show all your steps.
Thanks!
Solution
As pointed out previously, 0 can not be the middle term; therefore, it must be either the first or
third; it really doesn't matter which, as you just reverse the order, but I will go ahead and show
both ways.
If it is 0, then x, then 1/x, let x = y; then, since the first term was 0, the third term must be 2y.
x = y
1/x = 2y; also; 1/x = 1/y;
2y = 1/y
2y 2
= 1
In all cases, where I have ± for x and 1/x, they have the same sign, whether it is plus or minus.
y = ±2/2
x =y = ±2/2
1/x = 2y = ±2
Consider the next possibility; the second term is 1/x and the third term is x
Then, let 1/x = y
x = 2y
Then, 1/x = y = 1/2y
y=1/2y
2. 2y 2
= 1
y = ±2/2
1/x = y = ±2/2
x = 2y = ±2
Next, consider the other possibility, 0 is the third term.
Then, we have two possibilities, x is the second term or x is the first term.
To keep things in the same order as before, first, let x be the second term.
Then, let x = y.
As 1/x is the first term, 1/x = 2y
Then, 2y = 1/y
2y 2
= 1
y = ±2/2
x =y = ±2/2
1/x = 2y = ±2
Next, let 1/x be the second term and x be the first term.
Then, let 1/x = y
x = 2y
Then, 1/x = y = 1/2y
y=1/2y
2y 2
= 1
y = ±2/2
1/x = y = ±2/2
x = 2y = ±2
3. There are 4 possible plausible orders, the 4 without 0 in the middle:
0 first, x second, 1/x third
0 first, 1/x second, x third
0 third, x second, 1/x first
0 third, 1/x second, x first
In all cases, the value next to 0 is ±2/2, and the other value is ±2 (as stated before, these have
the same sign).
