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(kN)1 2 3 410 11 12x(m)...

Question

(kN)1 2 3 410 11 12x(m)

(kN) 1 2 3 4 10 11 12 x(m)



Answers

$$\begin{array}{|r|rrrrr} \hline n & 1 & 2 & 3 & 4 & 5 \\ \hline f(n) & 3 & 1 & -1 & -3 & -5 \end{array}$$

In the given question We have a table in which values of N&A 4th an argument and And a 4th n. The values of N are given us 1, 2, 3, 4 and five. And for each value of N, we have the corresponding value of a four pin Which will be our sequence. And this problem. So 149 16 and 25. So let's grow table around this. So this is the table that is given in the christian. So the sequence is given by each door of a 4th and right. So in N equal to one, we will have 1/4 1 which will be the first time, that is one. The second term will be for the third term will be nine and so on. So the sequence that we had Taking into consideration as 14 9 1625. And to determine whether we are told to determine whether the given sequences, arithmetic or not. And to do that. What we do is to find the difference between the constitutive terms in the sequence and if the difference between them are the same, then we can call it and arithmetic sequence. So let's take the first, the 2nd and 1st term right four minus one gives us a difference of three. Next up we have 9, 9, 4, the next constitutive pair. So we have nine minus food, which gives us A difference of five and next we have 16 and 19. So we can see over here that each constitutive term is different by a different number, Right? -9 is seven. Right? So there is no common difference in being between the constituted terms of the sequence. And since there is no common difference, we can say that the given sequence is not arithmetic. So if it were arithmetic, then each of these differences. That is the difference between each consecutive term would have been the same. But over here it's not common. It's different. So since there is no common difference, we can say this is not and arithmetic sequence. So this is how we determine whether a given sequences, arithmetic or not. I hope you understood the method. Thank you. Uh.

And this question were given the age N C n is equal to, we have to find Yes. We know date. N C N is equal to in sectorial, divided by and minus and factorial into and victoria. Yeah. So you cannot. This is in fact a really went by In my tradition to zero. Factorial. In fact area infertile. Infertile can slow and we are left with one of our satisfaction which is going to one. So the answer is that and different option is option number two, which is a quarter to one. This is the answer. This question. Thank you for what can we do?

Hi we're given here for and long natural London through the park three over the part and plus one is divisible by what value we have. We just hold it now so uh is divisible by if you just check here let's say for Option three We have three department plus one. Take care 3 to the bar and plus one giving that envelope. Natural numbers go to the park freeing over the bar and plus one and usually by to to the bar and Plus one here via the school. So now just check for real activity natural numbers and it will to want to take any calls one we have to part three of the bar one that is eight plus one that's given us nine. So nine is individual by Vienna Three to the bar to 19 years old by nine. Maybe just check for let's say to the par three to the bar too Plus one that is To the bar 9-plus 1. And here we have this three to the power we have two plus 13 of the park three. So if you just calculate here to the part nine plus one that is the truth about nine plus one of about nine. We'll get to plus one times one of them will be happy. But two plus one is uh we will have free here We get here 513 irreducible like 27. So therefore we have option trees valid it is valid for all the numbers that We have been able to one take anybody to and so on. So we can say that option three first doctor, correct choice. Thank you.

Health insurance in this problem. We have to find some of this sequence and we can observe that this is an ap because the difference here is constant. The difference of 1st and 2nd time is four. The difference of 2nd and 3rd is also for so the difference of terms difference of two consecutive terms is constant and it is an ap. Now we know that if I calculate any system of this ap using the formula A Plus and -1 times b. So this is coming out to be one plus n minus one times the common difference which is four, so one plus four and minus four and this is four and minus three which is the last time. So the number of times here would be and and we have to find the sum of first in terms here. So let me use the formula for the sum of first in terms of a P which is and upon to in the bracket two times the first term, plus the number of times -1 times that common difference. So this is an upon to two Plus 4 and -4. This is an upon to four and -2, and this is an Times two and -11 and which is the third option. So third option would be the correct answer to this problem. Thank you.


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