SOLUTION: The first term of an arithmetic sequence is 1. The fifth term is 6. Find the 2009th term. Please show me all your work.Thanks in advance.

Algebra ->  Sequences-and-series -> SOLUTION: The first term of an arithmetic sequence is 1. The fifth term is 6. Find the 2009th term. Please show me all your work.Thanks in advance.       Log On


   



Question 535095: The first term of an arithmetic sequence is 1. The fifth term is 6. Find the 2009th term. Please show me all your work.Thanks in advance.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
The following applies for an arithmetic sequence:
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If the first term of the sequence is A, the next term in the sequence is A + D where D is the common difference between terms. So we can say:
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Term #1 = A
Term #2 = A + D
Term #3 = Term #2 + D = (A + D) + D = A + 2D
Term #4 = Term #3 + D = (A + 2D) + D = A + 3D
Term #5 = Term #4 + D = (A + 3D) + D = A + 4D
Term #6 = Term #5 + D = (A + 4D) + D = A + 5D
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And so forth. You may be able to see the pattern. The nth term is found by taking the first term and adding to it the quantity n - 1 times D, the difference between terms. In the above example, if we were given A, the first term, we could have found the 6th term by adding to A the product of (6 - 1) times the Difference. In equation form this could be written as the nth term (call it T%5Bn%5D) is given by:
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T%5Bn%5D+=+A+%2B+%28n+-+1%29%2AD
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That's the equation that we can use to solve this problem.
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From the problem we were told that the first term in the sequence is 1. So we can substitute 1 for A in our equation. We were also told that we are trying to find the nth term when n is 2009. So we can substitute 2009 for n in the equation.
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All we are missing is D, the difference between terms. We can find D by subtracting the first term (1) from the 5th term (6) and dividing by 4 (the number of terms between the first and 5th terms and it is one less than the number of the last (5th) term. In this instance it is (6 - 1) divided by 4 = 1.25.
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As a check we can generate the series for the first 5 terms as follows:
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Term 1 = 1.00
Term 2 = 2.25
Term 3 = 3.50
Term 4 = 4.75
Term 5 = 6.00 <--- just as the problem says it should be
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Note that each term after the first was generated by adding 1.25 to the preceding term. So now we know that D is 1.25 and we can write the equation as:
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T%5B2009%5D+=+1+%2B+%282009+-+1%29%2A1.25
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Combine the numbers in the parentheses to get:
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T%5B2009%5D+=+1+%2B+%282008%29%2A1.25
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Multiply out the product term on the right side:
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T%5B2009%5D+=+1+%2B+2510
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Now just do the addition on the right side and you have the answer:
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T%5B2009%5D+=+2511
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Hope this helps you to understand how arithmetic series are generated and how to develop and use the equation for finding a term in the series.
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