Asked by Dannie
If the number m, 4m-8, m^2-6 are the first three terms of an arithmetic sequnce, what is m?
I don't know what equation to start with. I tried getting the m from 4m-8 which is equal to 2 but the answer is different on m^2-6. Most likely my method is wrong. What should I do in this question? Thanks!
I don't know what equation to start with. I tried getting the m from 4m-8 which is equal to 2 but the answer is different on m^2-6. Most likely my method is wrong. What should I do in this question? Thanks!
Answers
Answered by
Reiny
Use your definition of an arithmetic sequence.
The difference between any pair of consecutive terms must be the same, if subtracted in the same direction, that is ....
(4m-8) - (m) = (m^2 - 6) - (4m-8)
4m - 8 - m = m^2 - 6 - 4m + 8
0 = m^2 - 7m + 10
(m - 2)(m - 5) = 0
m = 2 or m = 5
check:
if m = 2, the terms are 2, 0, -2 , which form an AS
if m=5, the terms are 5, 12, 19 , which form an AS
your title says "Advanced Algebra" and this is a rather basic problem
I worry .
The difference between any pair of consecutive terms must be the same, if subtracted in the same direction, that is ....
(4m-8) - (m) = (m^2 - 6) - (4m-8)
4m - 8 - m = m^2 - 6 - 4m + 8
0 = m^2 - 7m + 10
(m - 2)(m - 5) = 0
m = 2 or m = 5
check:
if m = 2, the terms are 2, 0, -2 , which form an AS
if m=5, the terms are 5, 12, 19 , which form an AS
your title says "Advanced Algebra" and this is a rather basic problem
I worry .
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