How to Solve an Arithmetic Sequence problem with Variable Terms

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This Article will show How to solve an Arithmetic Sequence Problem whose terms are variable terms. We will use an Example Problem to demonstrate how this is done.

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  • Given the following Arithmetic Sequence Problem. For some real number t, the first three terms of an arithmetic sequence are 2t, 5t-1, and 6t+2.
    What is the numerical value of the fourth term?
    We will explain in the following Steps how we solve this problem.

  • What Defines an Arithmetic Sequence is the common Difference between each term of the Arithmetic Sequence, that is the Difference between the Second term and the First term, should be the same or equal to the Difference between the Third term and the Second term, should be equal to the Difference between the Fourth term and the Third term, and so on.

  • In the problem given in Step #1, 2t, is the First term of the Arithmetic Sequence, 5t-1, is the Second term of the Sequence, and 6t+2, is the Third term of the Arithmetic Sequence. So since we are working with an Arithmetic Sequence, then (5t-1) - 2t should equal (6t+2) - (5t-1). that is we have an equation:
    (5t-1)-2t = (6t+2)-(5t-1),which is equivalent to 5t-2t-1 = 6t-5t+2+1.
    which is equivalent to, 3t-1 = t+3 that is 3t-t = 3+1. so 2t = 4 and t=2.

  • Since t=2, we should find the Fourth term of the Arithmetic Sequence in terms of t, then substitute t=2, for the t in that Fourth term.
    The Common Difference in our Arithmetic Sequence Problem, 2t, 5t-1, 6t+2,..., is 5t-1-2t = 3t-1. We now add 3t-1 to the Third term, 6t+2, and we get our Fourth term, 6t+2 + 3t-1 = 9t+1. by substituting t=2 in 9t+1, we get, 9(2)+1, which equals 18+1 = 19.

    So the numerical value of the Fourth term is ... 19.

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