SOLUTION: 1-An aritmetic sequence has its 5th term equal to 22 and its 15th term equal to 62.Find its 100th term.
2-Find the sum of the first 50 even positive integers.
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-> SOLUTION: 1-An aritmetic sequence has its 5th term equal to 22 and its 15th term equal to 62.Find its 100th term.
2-Find the sum of the first 50 even positive integers.
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Question 739087: 1-An aritmetic sequence has its 5th term equal to 22 and its 15th term equal to 62.Find its 100th term.
2-Find the sum of the first 50 even positive integers. Answer by reviewermath(1029) (Show Source):
You can put this solution on YOUR website! 1. The common difference (slope) is equal to .
The nth term is 4n + k, we solve for k by substituting n = 5 (and equate to 5th term).
4(5) + k = 22 so k = 2.
The nth term is 4n + 2.
The 100th term is 4(100) + 2 = .
2. The sum of the first n positive integers is equal to and the sum of the first n even positive integers is .
Therefore, the sum of the first 50 positive even integers is .
