Use the graphs of the arithmetic sequences {a} and {b} to solve E...

Question

Use the graphs of the arithmetic sequences {a} and {b} to solve Exercises 51-58.

Find a14+b12.

Accepted Answer

Hello everybody. Everything I today we going to look at this question that states evaluate the following expression using the graphs of the arithmetic sequences. A subscript N and B subscript N where the expression that we have to evaluate is a of subscript 22 plus B of subscript 25. Now the graphs provided on the left hand side, we have a graph with the Y axis being labeled as a sub subscript N and the X axis being labeled as N with points at one comma 42 comma seven and three comma 10. And on the right hand side, we have another graph with the Y axis labeled B. Hello everybody. Everything right. Today, today we're going to be looking at this question that states evaluate the following expression using the graphs of the arithmetic sequences. A subscript N and B subscript N where the expression that we have to evaluate is a subscript 22 plus B subscript 22. Now the graphs provided on the left hand side, we have a graph with the Y axis being labeled as a subscript N and the X axis being labeled as N with points at one comma 42, comma seven and three comma 10. And on the right hand side, we have the graph with the Y axis being labeled B subscript N the X axis being labeled as N with points at one comma 92, comma five and three comma one. Now the answer choice is provided R A B negative 62 C D negative 20. No. The, our task at hand is to calculate a of subscript 22 plus B of subscript 25. So what we're going to do is break that down into two parts. First, we gotta find out what a of subscript 22 is. And then we have to find out what B of 25 is. So let's focus on a of subscript 22 1st. So if we look at the graph of a subscript N, we can see that it has points at one comma four, two comma seven and three comma 10. So we can turn these points into terms. So the X value becomes the subscript of our A. So if we have a point at one comma four, that means we have a of subscript, one is equal to and the Y value in our point becomes what our, a subscript equals. So if we have one comma four, we have a of subscript one equals four. If we have two comma seven, we have a of subscript two equals seven. If, if we have three, call 10, we have a S subscript three equals 10. So now with that information, we can move on to something else that we call a common difference. And this is important once we start using arithmetic sequence formulas that I will talk about soon. So a common difference which is labeled as D, it's basically that when we subtract two of the terms, is there a common answer between them? So in this case, we're going to have a of subscript three minus a of subscript two that is equal to 10 minus seven, which is equal to three. And that is our first answer for those two terms. And then if we have a of subscript two minus a of subscript one that is equal to seven minus four, which if you see the pattern, this will also be equal to three and the common difference will be three. So we have that D is equal to three. So now is when we move on to a formula dealing with arithmetic sequences, so we have that the NTH term of and arithmetic sequence with the first term being a subscript one. And the common difference being D, the formula for this states that A of subscript N is equal to a of subscript one plus open parentheses, N minus one closed parentheses, multiplying D. So this is a formula that we can use right now. So we know what we want to solve for a of sub group 22. That is what we want to solve at the end of the day. So we have that A F subscript 22 is equal to a of subscript one which we already calculated was four. So it's equal to four plus open parentheses N minus one. Our N here is 22 because we had subscript N and we're solving for a of subscript 22. Sir. N is 22 minus one, close parentheses, multiplying the D which we obtained was three. So we have that A of sub group 22 is equal to four plus open parentheses 22 minus one plus parentheses, multiplying A three. Now that is equal to four plus 22 minus one is 21 multiplied by three, which is equal to four plus 63 which gives us that A of subscript 22 is equal to 67. So that is our first answer for this question. And we just did the first part. So we wanted to break down the A F subscript 22 plus B of subscript 25. We just finished a F subscript 22. So now we have to do the same thing to soft for B of subscript 25. So when we look at the graph of B A subscript N, we see that we have the points one comma nine two, comma five and three comma one, meaning that at A B of subscript one that would be equal to nine, be a subscript two is equal to five and B of subscript three is equal to one. So now, same thing we look for the common difference or, or D and we have that B of subscript three minus B of subscript two is equal to one minus five, which is equal to negative four. And then we have that B of subscript two minus B of subscript one is equal to five minus nine, which is once again equal to negative four, meaning that our common difference D is equal to negative four. So now we continue to use that information and we go back to our formula where we have B F subscript N is equal to B of subscript one plus open parentheses, N minus one closed parentheses, multiplying the common difference D. And in this case, we're trying to solve for B A for 25. So we plug that in and we have B of subject 25 is equal to B of subscript one which is nine plus open parentheses N which is 25 minus one closed parentheses, multiplying the D which is negative four. So we have that nine plus 25 minus one is 24 multiplied by negative four, which is equal to nine minus which means that B of subscript 25 is equal to negative 87. And now we solved for the second part of our question and now we can combine them. So remember we had a of subscript 22 plus B of subscript 25 that is equal to, we have a F section 22 which is plus B of subscript 25 which is negative 87. This is the same thing as saying 67 minus which means that a of subscript 22 plus B of subscript 25 is equal to negative 20. And just like that, we've reached our final answer which matches with answer so and answer choice. D so I hope that was helpful. And until next time.

Struggling with College Algebra?

Use the graphs of the arithmetic sequences {a} and {b} to solve Exercises 51-58.
Find a14+b12.

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