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10. (a) (Spt) Find the inverse of the following funetion: y = 31- | (Spt) Find the sum of the first 40 terms Ofthe arithmetic sequence: 3.-7.-MPopc I of |...

Question

10. (a) (Spt) Find the inverse of the following funetion: y = 31- | (Spt) Find the sum of the first 40 terms Ofthe arithmetic sequence: 3.-7.-MPopc I of |

10. (a) (Spt) Find the inverse of the following funetion: y = 31- | (Spt) Find the sum of the first 40 terms Ofthe arithmetic sequence: 3.-7.-M Popc I of |



Answers

Find $a_{1}$ and $d$ for each arithmetic sequence. $$S_{31}=5580, a_{31}=360$$

We know that we can find the some off in terms off automatic progression, using the expression sn equal toe and by two in due to a one plus and minus money into the where n is the number off terms. Evan is the first time, and this is the common difference off the automatic sequence. So we have given some values related toe automatic sequence, and we are asked to find the some off 1st 30 terms. So we know that Evan is given us 40 be equal to minus three. And in this 30 so we can write the some off 1st. 30 terms as S 30 equal to any 30 30 by two into two times. Evan Evan is 40 two in the 40 plus in this 30. So 30 minus one into these minus three onda. We can simplify this to obtain this up. 15 in two. A deep minus three in do 29 on this case minus 105 This is the sum off first. 30 terms off the given automatic sequence

Okay, so it's fines are 29. You know that we're given our first term and our common difference so we can find our general formula. And then using our general formula, we confined our, um, 29th term. So our general formula is equal to our first term. Plus and minus one times are common difference. For using this, we confined our 20 night firm. That's 10 plus 29 minutes. One plays, three. So what does this give me? I'll put this into my calculator, and this gives me in 94.

This question gives me a sequence minus for 16 and that it goes dot, dot dot and then 41. It says 41 is 1/10 term. So it is 10. And they asked me to find that some. So I'm gonna notice. First of all, this is arithmetic. I'm just adding five each time. So that means my common difference is five in my first term is minus four, and that we could do, uh, some of the 1st 10 terms is we're gonna use this formula and over to under a one plus a m. Okay, so in this case and his 10 so we can substitute 10 in there. So I get 10 over to onto a oneness minus four, and a hen is 41. So that gives me an hour to its five and 41. Minus four is 37 and that gives me a total of 185

So, in order to solve the system of equations we're going to break it up into Major sees will be creating a matrix of the coefficients on the left hand side. So to one A in those times, our X matrix, which is just are variable sex And why is equal to R B matrix, which has the negative three and negative A From the right hand side note to self this we're gonna want to basically a acts. So we multiply both sides times a inverse, a inverse times a will just give us the identity matrix. So the identity matrix times X will just be X So essentially we now just have X is equal to an verse a inverse times be so our X matrix was just are variables X Why that's equal to a inverse which resell for in a previous problem it was one negative one over a negative one and positive too over. Okay. And then RB matrix is just the negative three in the negative A. So now we can solve for x and y by doing matrix multiplication start in the first row one times negative three is negative three plus negative 1/8 times Negative A is positive. One negative three plus one is positive or negative, too. Now to get the second element, be of negative one times negative. Three positive three plus two over a times negative a. So that will give us negative, too. Three plus negative too, is a positive one, and so are resulting matrix is negative 21 which tells us that eggs is equal, the negative to and why is equal to one.


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