SOLUTION: List the first three terms of an arithmatic sequence with a first term of 6 and a twentieth term of 63

Question 574659: List the first three terms of an arithmatic sequence with a first term of 6 and a twentieth term of 63
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
An arithmetic sequence has a constant difference between each of its terms. If we call this difference D, and the first term is 6, then the second term is 6 + D and the third term is the second term which is 6 + D, plus D. So the third term is 6 + 2D. Similarly the fourth term is the third term which is 6 + 2D plus D. Therefore, the fourth term is 6 + 3D.
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There is a pattern here. Notice that each term is 6 plus D times one less than the number of the term we are trying to find. So we can see that the twentieth term would be 6 + 19*D.
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And we are told that the twentieth term equals 63. Therefore, we can write the equation:
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6 + 19*D = 63
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From this equation we can find the value of D, the constant difference between terms. First subtract 6 from both sides of the equation to get:
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19*D = 57
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Solve for D by dividing both sides of this equation by 19. As a result of this division the equation becomes:
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D = 57/19 = 3
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So now we know that the difference between terms is 3. From the problem we were told that the first term is 6. Then the second term is 6 + 3 or 9. Then the third term is the second term plus 3, which means that it is 9 + 3 = 12. And the fourth term is 12 + 3 = 15. And so on ...
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The answer is that the first three terms of the sequence are 6, 9, and 12.
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Hope that this helps you to understand a little bit more about arithmetic sequences.
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