Question 187236
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: {{{(a+d)-a=a-a+d=d}}}


If you subtract the 2nd term from the 3rd term, you get: {{{(a+2d)-(a+d)=(a-a)+(2d-d)=d}}}


If you subtract the 3rd term from the 4th term, you get: {{{(a+3d)-(a+2d)=(a-a)+(3d-2d)=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. 



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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:


{{{(5x-4)-(3x-1)=(2x+8)-(5x-4)}}}





{{{(5x-4)-(3x-1)=(2x+8)-(5x-4)}}} Start with the given equation.



{{{5x-4-3x+1=2x+8-5x+4}}} Distribute.



{{{2x-3=2x+8-5x+4}}} Combine like terms on the left side.



{{{2x-3=-3x+12}}} Combine like terms on the right side.



{{{2x=-3x+12+3}}} Add {{{3}}} to both sides.



{{{2x+3x=12+3}}} Add {{{3x}}} to both sides.



{{{5x=12+3}}} Combine like terms on the left side.



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



{{{x=(15)/(5)}}} Divide both sides by {{{5}}} to isolate {{{x}}}.



{{{x=3}}} Reduce.



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Answer:


So the answer is {{{x=3}}}