Answers:
![t_{10} = -22 \ \text{ and } S_{10} = -85](https://tex.z-dn.net/?f=t_%7B10%7D%20%3D%20-22%20%5C%20%5Ctext%7B%20and%20%7D%20S_%7B10%7D%20%3D%20-85)
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Explanation:
![t_1 = \text{first term} = 5\\t_2 = \text{first term}-3 = t_1 - 3 = 5-3 = 2](https://tex.z-dn.net/?f=t_1%20%3D%20%5Ctext%7Bfirst%20term%7D%20%3D%205%5C%5Ct_2%20%3D%20%5Ctext%7Bfirst%20term%7D-3%20%3D%20t_1%20-%203%20%3D%205-3%20%3D%202)
Note we subtract 3 off the previous term (t1) to get the next term (t2). Each new successive term is found this way
![t_3 = t_2 - 3 = 2-3 = -1\\t_4 = t_3 - 3 = -1-3 = -4](https://tex.z-dn.net/?f=t_3%20%3D%20t_2%20-%203%20%3D%202-3%20%3D%20-1%5C%5Ct_4%20%3D%20t_3%20-%203%20%3D%20-1-3%20%3D%20-4)
and so on. This process may take a while to reach ![t_{10}](https://tex.z-dn.net/?f=t_%7B10%7D)
There's a shortcut. The nth term of any arithmetic sequence is
![t_n = t_1+d(n-1)](https://tex.z-dn.net/?f=t_n%20%3D%20t_1%2Bd%28n-1%29)
We plug in
and simplify
![t_n = t_1+d(n-1)\\t_n = 5+(-3)(n-1)\\t_n = 5-3n+3\\t_n = -3n+8](https://tex.z-dn.net/?f=t_n%20%3D%20t_1%2Bd%28n-1%29%5C%5Ct_n%20%3D%205%2B%28-3%29%28n-1%29%5C%5Ct_n%20%3D%205-3n%2B3%5C%5Ct_n%20%3D%20-3n%2B8)
Then we can plug in various positive whole numbers for n to find the corresponding
value. For example, plug in n = 2
![t_n = -3n+8\\t_2 = -3*2+8\\t_2 = -6+8\\t_2 = 2](https://tex.z-dn.net/?f=t_n%20%3D%20-3n%2B8%5C%5Ct_2%20%3D%20-3%2A2%2B8%5C%5Ct_2%20%3D%20-6%2B8%5C%5Ct_2%20%3D%202)
which matches with the second term we found earlier. And,
![tn = -3n+8\\t_{10} = -3*10+8\\t_{10} = -30+8\\t_{10} = \boldsymbol{-22} \ \textbf{ is the tenth term}](https://tex.z-dn.net/?f=tn%20%3D%20-3n%2B8%5C%5Ct_%7B10%7D%20%3D%20-3%2A10%2B8%5C%5Ct_%7B10%7D%20%3D%20-30%2B8%5C%5Ct_%7B10%7D%20%3D%20%5Cboldsymbol%7B-22%7D%20%5C%20%5Ctextbf%7B%20is%20the%20tenth%20term%7D)
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The notation
refers to the sum of the first ten terms ![t_1, t_2, \ldots, t_9, t_{10}](https://tex.z-dn.net/?f=t_1%2C%20t_2%2C%20%5Cldots%2C%20t_9%2C%20t_%7B10%7D)
We could use either the long way or the shortcut above to find all
through
. Then add those values up. Or we can take this shortcut below.
![Sn = \text{sum of the first n terms of an arithmetic sequence}\\S_n = (n/2)*(t_1+t_n)\\S_{10} = (10/2)*(t_1+t_{10})\\S_{10} = (10/2)*(5-22)\\S_{10} = 5*(-17)\\\boldsymbol{S_{10} = -85}](https://tex.z-dn.net/?f=Sn%20%3D%20%5Ctext%7Bsum%20of%20the%20first%20n%20terms%20of%20an%20arithmetic%20sequence%7D%5C%5CS_n%20%3D%20%28n%2F2%29%2A%28t_1%2Bt_n%29%5C%5CS_%7B10%7D%20%3D%20%2810%2F2%29%2A%28t_1%2Bt_%7B10%7D%29%5C%5CS_%7B10%7D%20%3D%20%2810%2F2%29%2A%285-22%29%5C%5CS_%7B10%7D%20%3D%205%2A%28-17%29%5C%5C%5Cboldsymbol%7BS_%7B10%7D%20%3D%20-85%7D)
The sum of the first ten terms is -85
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As a check for
, here are the first ten terms:
- t1 = 5
- t2 = 2
- t3 = -1
- t4 = -4
- t5 = -7
- t6 = -10
- t7 = -13
- t8 = -16
- t9 = -19
- t10 = -22
Then adding said terms gets us...
5 + 2 + (-1) + (-4) + (-7) + (-10) + (-13) + (-16) + (-19) + (-22) = -85
This confirms that
is correct.
Answer:
AcORdinG To MY CaLCulAtIoNS ItS 21
Step-by-step explanation:
Logic :)
19? Lol........................
Since theres 4 numbers and 60 spins if you divide it you get the average number that all the numbers would land on. 2 would be landed on 15 times if it went exactly to the calculations. not 100% sure though
![\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}](https://tex.z-dn.net/?f=%5Cmathfrak%7B%5Chuge%7B%5Corange%7B%5Cunderline%7B%5Cunderline%7BAnSwEr%3A-%7D%7D%7D%7D%7D)
Actually Welcome to the Concept of the Equilateral Triangles.
Since, we know that, all the sides of a triangle are equal, hence it is a Equilateral triangle.
Equilateral triangle has one all angles equal to 60°.
so, major angle P = 60°
hence, 2x=60°
===> x = 30°