Video Transcript
Find the order and the value of the last positive term in the arithmetic sequence 37, 35, 33, and so on.
We know that any arithmetic sequence has first term ๐; it also has a common or first difference ๐. In this question, the first term in the sequence is 37. So, ๐ equals 37. To get from 37 to 35 and 35 to 33, we need to subtract two. This means that our value of ๐ is negative two. The general ๐th term of an arithmetic sequence ๐ sub ๐ is equal to ๐ plus ๐ minus one multiplied by ๐. In this question, ๐ sub ๐ is equal to 37 plus ๐ minus one multiplied by negative two.
Distributing the parentheses or expanding the brackets, we can multiply negative two by ๐ and negative two by negative one. Multiplying two negative numbers gives a positive answer. So, ๐ sub ๐ is equal to 37 minus two ๐ plus two. This, in turn, simplifies to 39 minus two ๐ or negative two ๐ plus 39. We want to find the last positive term. This means that the general term must be greater than zero.
In order to solve the inequality 39 minus two ๐ is greater than zero, we begin by adding two ๐ to both sides. This gives us 39 is greater than two ๐ or two ๐ is less than 39. Dividing both sides by two gives us ๐ is less than 39 over two. This is equal to 19 and a half or 19.5. ๐ must be less than 19.5. As ๐ must be a whole number or integer, the order will be the greatest whole number that is less than 19.5. The order is equal to 19. The last positive term in the arithmetic sequence is the 19th term.
We can now calculate its value by substituting ๐ equals 19 into the expression 39 minus two ๐. Multiplying negative two by 19 gives us negative 38. 39 minus 38 is equal to one. Therefore, the 19th term in the sequence is one. We could have answered this part of the question by inspection. We notice that our sequence is the set of odd numbers from 37 downwards. This means that the last positive term must be one, as this is the smallest positive odd number.
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