hlp me on consecutive arithmetic sequence?

Question

find the values of m and n if 3m-n,2m+n,5m+1 are consecutive terms of an arithmetic sequence.... tnx

Answer 1

I answered this already about 10 minutes ago and you DELETED THE QUESTION.



Why on Earth did you do that? Now I have to type out my solution all over again.



Don't do that again, because (as you may have noticed) nobody else is going to give you a complete solution like I am, and I'm not going to type it out again!



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Summary:

There are infinitely many solutions, whenever 3n = 1 + 4m. This gives d = (5m+2)/3.



If you restrict yourself to integers, then a single integer T will generate all possible solutions:



(m,n,d) = (3T-1, 4T-1, 5T-1)



Not only will this generate all possible integer solutions for integers T, but it will generate other real-number solutions if you plug in other real numbers for T besides integers.



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Solution:



If these three are in arithmetic sequence, then:



d = (2m+n) - (3m-n) = 2n - m

d = (5m+1) - (2m+n) = 3m - n + 1



Setting those equal gives you:



2n - m = 3m - n + 1

3n = 1 + 4m



This is the only restriction, which means you can use any values of m and n and obtain consecutive terms of an arithmetic sequence.



For example, m=2, n=3, gives you: 3, 7, 11 (d=4).



But m=5, n=7, gives you: 8, 17, 26



If you are restricted to integers, then you know:



n = (4m+1)/3



which means m has to be a multiple of 3, minus 1. This way, n works out to be an integer too.



You can parametrize the solutions based on this:



m = 3T-1

n = (4m+1)/3 = ( 4(3T-1) +1)/3 = 4T - 1



Use an expression from above to find d:



d = 2n - m = 2(3T-1) - (4T-1) = 5T - 1



This gives you all possible solutions:



(m,n,d) = (3T-1, 4T-1, 5T-1)



Here are a few examples:



m=2, n=3 → 3, 7, 11 → d=4

m=5, n=7 → 8, 17, 26 → d=9

m=11,n=15 → 18,37,56 → d=19



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BEWARE of the answer by Michael Rasch. He has made SERIOUS algebra mistakes that make his answer wrong and very misleading. He thinks:



(2m+n)-(3m-n) = -m



But it is not. It is:



(2m+n)-(3m-n) = 2n - m



You cannot use:



[third term] - [first term] = 2d



to help you in this problem because the third term given is the same as the first d plus the second d we already have:



[2m+n] - [3m-n] = d = 2n - m

[3m-n] - [5m+1] = d = 3m - n + 1



Now what happens if you subtract:

[5m+1] - [3m-n] = 2d



You get a new equation:



2d = 2m + n + 1



But that's just the sum of the previous two equations:



d + d = (2n-m) + (3m - n + 1) = 2m + n + 1



and not only does it not provide new information. It is impossible under any circumstances for it to do so. You cannot use this false "third" equation to help you eliminate one of the variables and find a specific single solution.



This user is mislead and has made an algebra mistake that steered him off a path of circular logic that leads to no conclusions, onto a road of flawed logic that leads to false ones.

Answer 2

Since these numbers are consecutive terms of an arithmetic sequence, each differs by the same quantity from the next.

So

(5m+1)-(2m+n)=(2m+n)-(3m-n) or

4m+1 = 3n



There are infinitely many values of m&n that meet this condition. Once any pair of m,n values satisfy the above equation, the terms will fall in an arithmetic sequence.

Answer 3

a million. enable the arithmetic distinction be x. Then we would write ok + 7 + x = ok^2 and ok^2 + x = 21. We would cut back the project-loose unknown x to style the equation 2*ok^2 - ok - 28 = 0 showing the perfect result as ok = 4, ignoring the different destructive fraction. 2. Geometric progression sequence must be written as 15 / (ok-7) = (7*ok+5) / 15 keeping the geometric ratio. We would cut back it as 7*ok^2 - 40 4*ok-260 = 0 which shows ok = 10 as suitable answer.

Answer 4

If d is the difference,



(2m+n)-(3m-n)=d

(5m+1)-(2m+n)=d

(5m+1)-(3m-n)=2d



simplify each equation to get



-m=d

3m-n+1=d

2m+n+1=2d



substitute -m for d in the 2nd and 3rd equations above



3m-n+1=-m

2m+n+1=2(-m)



simplify and get variables on 1 side, constants on the other



4m-n=-1

4m+n=-1



if necessary multiply each equation by constants so the n terms are opposites (this is already the case) and add the equations



8m=-2



so, m=-1/4

Answer 5

You can't solve one eqn in 2 unknowns. So unique answer is not possible or there are several solutions.

2(2m+n) = (3m-n)+(5m+1)

3n-1 = 4m but nothing can be done further for unique answer.

Answer 6

suppose this part of the sequence is x x+ 1 x+2

then 3m -n = x 2m+n= x+1 5m+1 = x+ 2

eliminate x by subtracting 1st and 2nd, 2nd and 3rd

and solve these two simultaneous equations in m and n