SOLUTION: find what x is so that 3x-1, 5x-4 and 2x+8 are consecutive terms of an arithmetic sequence.

Algebra ->  Test -> SOLUTION: find what x is so that 3x-1, 5x-4 and 2x+8 are consecutive terms of an arithmetic sequence.      Log On


   

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) About Me  (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: %28a%2Bd%29-a=a-a%2Bd=d

If you subtract the 2nd term from the 3rd term, you get: %28a%2B2d%29-%28a%2Bd%29=%28a-a%29%2B%282d-d%29=d

If you subtract the 3rd term from the 4th term, you get: %28a%2B3d%29-%28a%2B2d%29=%28a-a%29%2B%283d-2d%29=d


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.


-------------------------------------------------


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:

%285x-4%29-%283x-1%29=%282x%2B8%29-%285x-4%29




%285x-4%29-%283x-1%29=%282x%2B8%29-%285x-4%29 Start with the given equation.


5x-4-3x%2B1=2x%2B8-5x%2B4 Distribute.


2x-3=2x%2B8-5x%2B4 Combine like terms on the left side.


2x-3=-3x%2B12 Combine like terms on the right side.


2x=-3x%2B12%2B3 Add 3 to both sides.


2x%2B3x=12%2B3 Add 3x to both sides.


5x=12%2B3 Combine like terms on the left side.


5x=15 Combine like terms on the right side.


x=%2815%29%2F%285%29 Divide both sides by 5 to isolate x.


x=3 Reduce.


----------------------------------------------------------------------

Answer:

So the answer is x=3