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Question 1108288: Hi could you plz help me with this question:
For the arithmetic sequence with t7 = 0.6 and t12 = -0.4, find t20.
Thanks
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! T7 = .6
T12 = -.4
T7 is the 7th term in the sequence.
T12 is the 12th term in the sequence.
T12 - T7 = -.4 - .6 = -1
12 - 7 = 5
the difference between each term in the sequence is therefore -1/5 = -.2
T20 = x
T12 = -.4
we assign the variable x to T20 because we don't know what the value of T2 is just yet.
T20 is the 20th term in the sequence.
T12 is the 12th term in the sequence
T20 - T12 = x - (-.4) = x + .4
20 - 12 = 8
the common difference between each term is -.2
the difference between T20 and T12 is x + .4
the difference between term number 20 and term number 12 is 8.
therefore -.2 = (x + .4) / 8
multiply both sides of this equation by 8 to get:
-1.6 = x + .4
subtract .4 from both sides of this equation to get:
-2 = x
the value of the 20th term is therefore -2.
since this is an arithmetic sequence, then there is a common difference between each term.
we found that common difference to be equal to -.2
once we found that, we could find the value of any term in the sequence from any term that we know the value of.
we know the value of the 12th term is -.4
we know that the common difference between terms is -.2
we know that there are 8 terms between the 12th term and the 20th term.
the difference between the 12th term and the 20th term is therefore 8 * -.2 = -1.6
we add -1.6 to -.4 and we get -2.
the solution to your problem is that the value of T20 is equal to -2.
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