SOLUTION: find what x is so that 3x-1, 5x-4 and 2x+8 are consecutive terms of an arithmetic sequence.
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Question 187236: find what x is so that 3x-1, 5x-4 and 2x+8 are consecutive terms of an arithmetic sequence.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
An arithmetic sequence is of the form: a, a+d, a+2d, a+3d, ..., a+nd
Note: I'm just adding a constant number to 'a' each time.
So if you subtract the 1st term from the second, you get:
If you subtract the 2nd term from the 3rd term, you get:
If you subtract the 3rd term from the 4th term, you get:
So if you subtract ANY term from the next term, you will ALWAYS get the result of "d"
In other words, each term is separated by the same distance.
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In our case, the first term is 3x-1, the second term is 5x-4 and the third term is 2x+8
So according to the logic above, this should be true (if the sequence is an arithmetic one):
2nd term - 1st term = 3rd term - 2nd term
which looks like:
Start with the given equation.
Distribute.
Combine like terms on the left side.
Combine like terms on the right side.
Add to both sides.
Add to both sides.
Combine like terms on the left side.
Combine like terms on the right side.
Divide both sides by to isolate .
Reduce.
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Answer:
So the answer is
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