Question 1075313: determine the value of m so that 2m+1, m, and 5-3m form an arithmetic progression
Found 2 solutions by Boreal, Theo: Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! arithmetic progressions have equal distances between the terms.
2m+1-m=m+1
m-(5-3m) must equal m+1, and it is 4m-5
m+1=4m-5
3m=6
m=2
5, 2, -1 is the progression, with common distance 3.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! if it's an arithmetic progression, then:
An = A1 + (n-1)* d
this means that A2 = A1 + d and A3 = A1 + 2d
based on this, then:
A2 - A1 = A1 + d - A1 = d
A3 - A2 = A1 + 2d - A1 - d which results in A3 - A2 = d
so, you have:
A2 - A1 = d
A3 - A1 = d
in other words, the common difference is d and each succeeding term in the sequence is increased by the same value of d and each term subtracted from the term immediately above is is equal to d.
your terms are:
A1 = 2m + 1
A2 = m
A3 = 5 - 3m
A2 - A1 is therefore equal to m - 2m - 1 which is equal to -m - 1
A3 - A2 is therefore equal to 5 - 3m - m which is equal to 5 - 4m
since both -m - 1 and 5 - 4m are equal to d, then they are equal to each other and you get:
-m - 1 = 5 - 4m
add 4m to both sides and add 1 to both sides sand you get:
3m = 6
solve for m to get m = 2.
check your sequence is if it correct using m = 2.
our sequence will be:
A1 = 2m + 1 = 4 + 1 = 5
A2 = m = 2
A3 = 5 - 3m = 5- 6 = -1
your arithmetic sequence is 5, 2, -1.
your common difference is -3.
everything checks out ok when m = 2.
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