The given sequence is
a₁, a₂, ...,
![a_{n}](https://tex.z-dn.net/?f=%20a_%7Bn%7D%20)
Because the given sequence is an arithmetic progression (AP), the equation satisfied is
![a_{n}=a_{1}+(n-1)d](https://tex.z-dn.net/?f=%20a_%7Bn%7D%3Da_%7B1%7D%2B%28n-1%29d%20)
where
d = the common difference.
The common difference may be determined as
d = a₂ - a₁
The common difference is the difference between successive terms, therefore
d = a₃ - a₂ = a₄ - a₃, and so on..
The sum of the first n terms is
![S_{n}= \frac{n}{2} (a_{1}+a_{n})](https://tex.z-dn.net/?f=S_%7Bn%7D%3D%20%5Cfrac%7Bn%7D%7B2%7D%20%28a_%7B1%7D%2Ba_%7Bn%7D%29)
Example:
For the arithmetic sequence
1,3,5, ...,
the common difference is d= 3 - 1 = 2.
The n-th term is
![a_{n}=1+2(n-1)](https://tex.z-dn.net/?f=a_%7Bn%7D%3D1%2B2%28n-1%29)
For example, the 10-term is
a₁₀ = 1 + (10-1)*2 = 19
Th sum of th first 10 terms is
S₁₀ = (10/2)*(1 + 19) = 100
A.) For the Junior Varsity Team, mean would be the appropriate measure of center since the data is <span>symmetric or well-proportioned while we should use standard deviation for getting the measure of spread since it also measures the center and how far the values are from the mean.
b.) For the Varsity Team, the median would be the appropriate measure of the center since the data is skewed left and not evenly distributed so median could be used since it does not account for outliers while we use IQR or interquartile range in measuring the spread of data since IQR does not account for the data that is skewed. </span>
Answer:
5/8
Step-by-step explanation:
if 1/8 is used for one batch . for five batch=1/8*5=5/8
That would be a right-angled triangle.