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 the difference between the sum of the first 14 terms of {b} and the sum of the first 14 terms of {an}.

Accepted Answer

Hello everybody every night today that are going to look at this question that states evaluate S of subscript A 20 minus s of subscript B 20 where S subscript A 20 SBS subscript B 20 are the sum of the 1st 20 terms of a subscript N and B subscript N respectively. And use the graphs of the arithmetic sequences. A subscript N and B subscript N shown. Now the graphs shown have a Y axis going up in terms of five. So we have 5 10 15 and an X axis showing one and three. Both graphs show this the graph or, or the Y axis on the left graph is labeled as a subscript N and has points at one comma 42 comma seven and three comma 10. And the graph on the right has the Y axis labeled as B subscript N with points at one comma 92 comma five and three comma one. Now the X axis on both cases is labeled as N. Now the answer choice provided R A 70 B 520 C 1230 and the 840. So let's go through this question little by little and we'll get to a final answer. It's going to be a bit of work, but we will get there. So our task at hand is to get the answer for S of subscript A minus s of subscript B 20. That's our task. So let's break this down bit by bit and start off with the left hand side. So S of subscript A 20 which means we have to find the sum of the 1st 20 terms of the arithmetic sequence of a subscript N. So a subscript N, if we look at that graph, we know we have the points, one comma four, two comma seven and three comma 10. Now these points can be translated one comma four, for example, could function as that a of one, a subscript one is equal to four. So the X functions as the, as the subscript and the Y functions as the equal to. So two comma seven would be a of subscript two equals seven and three. Comma 10 is a of subscript three equals 10. So our next step here is to find a common difference between these terms. So a common difference, common difference D So here we'll have that a of subscript three minus a subscript two is equal to 10 minus seven which is equal to three. And we have that a subscript two minus a subscript one is equal to seven minus four, which is equal to three meaning that our common difference D, we have D equals three since that was three was the answer for both subtractions. So another thing that we should note is the sum, some of the first N terms of an arithmetic sequence. This has a general formula that we can use. So the sum of the first and terms of an arithmetic sequence can be, is given by S of subscript N S capital S meaning sum of subscript N is equal to open parentheses. N divided by two closed parentheses, multiplying open parentheses. A of subscript one plus a of subscript N closed parentheses. Now, something else that we should know when it comes to arithmetic sequence sequences is that the nth term of an arithmetic sequence with the first term being a one and a common difference D that nth term is given by a of subscript N is equal to a of subscript one plus open parentheses, N minus one closed parentheses, multiplying the common difference in D. So these are two rules that are very important when it comes to this and that we will, we are going to be using for the rest of the question. So I'm going to be moving my work to have a bit of more space to work with and we'll continue forward. So we had that a subscript N is equal to a subscript one plus open parentheses, N minus one plus parentheses multiplying D and we're dealing with the 1st 20 terms. So we have that a of subscript or because N is the nth term. So we have a of subscript 20 is equal to our A one which we said was four. So we have four plus open parentheses, N minus one and our N is 20. So we have 20 minus one, closed parentheses, multiplying the common difference which we said was three. So we have four plus 19, multiplied by three, which is four plus 57 which means that a subscript 20 is equal to 61. So that is a valuable answer here because now we can solve for the sum of subscript A 20. So we have subscript A 20 is equal to. Now we use the sum of the first and terms of an arithmetic sequence formula. So we have N divided by two, our N is 20. So we have 20 divided by two to close that parentheses, multiplying an open parentheses. A one which is four plus a of subscript N which we just figured out is a of subscript 20 which is 61. So we're left with 20 divided by two is 10 and four plus 61 is 65. So we have 10 multiplying 65 we have that the sum of subscript A 20 is equal to 650. And that is the first step in this equation. So now, because we have the sum of eight of subscript A 20 minus the sum of subscript B 20. Now, we have to do the same thing, but for the B subscript so I'm just going to scroll down a bit to have a bit of more room to work with. And I'll move my work over to the left. So now let's repeat our steps for the B F subscript N so B F subscript and if we looked at the graph, we saw that we had points at one comma nine two comma five and three comma one. Meaning that I B one B, one B of subscript one is equal to nine B A subscript two is equal to five NBA subscript three is equal to one. So now same thing, we want to look for a common difference between these terms. The common difference being labeled as D. So we have that B of subscript 20 minus B of subscript two is equal to one minus five, which is equal to negative four. And we have that B of subscript two minus B of subscript one is equal to five minus nine, which is equal to negative four. Therefore, our common difference D is D is equal to negative four. So now we go back to our formula where we have B of subscript N is equal to B of subscript one plus open parentheses, N minus one, closed parentheses, multiplying the common difference D. And now we can just do plugging in again, we're dealing with the 1st 20 terms of the sequence. So we have B of subscript 20 is equal to nine, which is our B one plus open parentheses 20 which is our N minus one closed parentheses, multiplying negative four, which is our D and that will leave us with nine plus 19 multiplying negative four. So we have to do the 19 multiplying the negative four first and we'll be left with nine minus 76 which means that B of subscript 20 is equal to negative 67. So with that information, we can move on to the sun of, of B of subscript B 20 and the sum of subscript B 20. Once again, we look at the formula that we had used before and we'll have that the sum of cells can be is equal to 20 which is our N divided by two closed parentheses, multiplying and open parentheses nine, which is our B one minus 67 which is our A subscript 20 which means that we're left with 10 multiplying A negative which means that subscript or sum of subscript B is equal to negative 580. And that is the answer for our second son. Now remember that we are trying to solve for the sum of subscript A 20 minus the sum of subscript B 20. Now this is equal to, we just plug in the values that we have. So this is equal to which is our sum of eight of subscript A 20 minus negative 580 which is the sum of substrate B 20. And that is equal to 650 plus which is equal to 1230. So after all of that, we have reached a final answer where when we have the sum of subscript A 20 minus the sum of subscript B 20 that is equal to answer choice. C 1230. 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 the difference between the sum of the first 14 terms of {b} and the sum of the first 14 terms of {an}.

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