5

A loop that evaluates its test expression after each repetition is a(n) _____ loop....

Question

A loop that evaluates its test expression after each repetition is a(n) _____ loop.

A loop that evaluates its test expression after each repetition is a(n) _____ loop.



Answers

An arithmetic sequence is a sequence in which the_____
between successive terms is constant.

Because if we assume that A F N is less than two, then a of N plus one is equal to the square root of two plus a on which is less than the square root of two plus two, which is equal to two. So therefore, event plus one is less than two. So my induction we conclude that the end is less than two for all end, and it's founded above, too, for our view of the tickets. And you're under the same assumption and that say that 8 a.m. n plus one is equal to the square root of two plus a of or if and plus one is graded on her equal to the square root of a lot and plus a on which is greater than equals in spirit to a mmm substituting am. And with two we end up with a and this one is greater than equal thio on a of end times if end or in uh, so by induction, that event is increase it using the serum bounded mon atomic sequence. Could sequences converge here? The sequence ale then is bounded about hate crazy. Then the sequences converge, so the limit does exists and the limit of a over. And it's an approaches. Infinity, easy for the l doesn't exist. So therefore we can say that the limit off as an approaches infinity off a of N is equal to two or negative one. Anything between two related one. So we're going to say that this is equal to two.

All righty. So for this question, we were given the 1st 3 terms in our genetic sequence. And we're supposed to use this information in order to write a recursive formula. So our recursive formula is going to look like this. It's going to be a to the end of term is equal to a to the N minus one. Pull us our common difference. So let's go ahead and identify this information here. So are common difference is going to be the change from each, uh, term to the next. So from 40 to 60 we're going up 20 and then from 60 to 80 were also going up 20. So it's safe to say that our common difference here is going to be 20 and then we also have to identify our first term in a recursive formula. So our first term is gonna be the 1st 1 listed up here, which is gonna be our 40. So although that might not be part of our equation, we also have to list that separate because you need that information in order to apply it for your recursive formula. Let's go ahead and plug in this information into our formulas. We've got a to the end is equal to a to the n minus one, which is going to be just our previous value, plus our 20 over here. And then we also have to identify that our first term is equal to 40 and then there we have

The general form for the recursive formula of an arithmetic sequence is a n equals a N minus one plus D, where a N is the end of term and aim and minus one is the term before that, and D is the common difference. So all we need is D the common difference to find d. We just have to subtract one term minus the previous, like 60 minus 40 to get 20. And this will be the same if we also subtract any pair of two consecutive terms like a D minus 60. So our common difference of D is 20 and we could substitute that into the formula. Getting that a end. The end term equals a and minus one plus 20. Now, we don't need for a recursive formula the initial term, but it helps because it makes things much easier to calculate. So we'll also right that a one equals 40. The first term in the sequence

Okay, Give me a N is equipped to 70 in 26. 35. We ask to find the crystal formula. So a one is 17 a and and when it's one plus our common defense. So that's 26 2017. But here's the thing. And this is where I am greater than equal to.


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