The sum of first 20 arithmetic series 
Given:
Arithmetic series for 3rd term is 55
Arithmetic series for 7th term is -98
To find:
The sum of first 20 Arithmetic series
<u>Step by Step Explanation:
</u>
Solution:
Formula for calculating arithmetic series
Arithmetic series=a+(n-1) d
Arithmetic series for 3rd term 
Arithmetic series for 19th term is
Subtracting equation 2 from 1
![\left[a_{19}+18 d=-98\right]+\left[a_{1}+2 d=55\right]](https://tex.z-dn.net/?f=%5Cleft%5Ba_%7B19%7D%2B18%20d%3D-98%5Cright%5D%2B%5Cleft%5Ba_%7B1%7D%2B2%20d%3D55%5Cright%5D)
16d=-98-55
16d=-153
Also we know
First 20 terms of an AP
![a_{20}=[1106 / 16]-[2907 / 16]](https://tex.z-dn.net/?f=a_%7B20%7D%3D%5B1106%20%2F%2016%5D-%5B2907%20%2F%2016%5D)
Sum of 20 Arithmetic series is
Substitute the known values in the above equation we get
![S_{20}=\left[\frac{20\left(\left(\frac{558}{8}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right]](https://tex.z-dn.net/?f=S_%7B20%7D%3D%5Cleft%5B%5Cfrac%7B20%5Cleft%28%5Cleft%28%5Cfrac%7B558%7D%7B8%7D%5Cright%29%2B%5Cleft%28%5Cfrac%7B-1801%7D%7B16%7D%5Cright%29%5Cright%29%7D%7B2%7D%5Cright%5D)
![S_{20}=\left[\frac{\left.20\left(\frac{1106}{16}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right]](https://tex.z-dn.net/?f=S_%7B20%7D%3D%5Cleft%5B%5Cfrac%7B%5Cleft.20%5Cleft%28%5Cfrac%7B1106%7D%7B16%7D%5Cright%29%2B%5Cleft%28%5Cfrac%7B-1801%7D%7B16%7D%5Cright%29%5Cright%29%7D%7B2%7D%5Cright%5D)
![S_{20}=5\left[\frac{-695}{16}\right]](https://tex.z-dn.net/?f=S_%7B20%7D%3D5%5Cleft%5B%5Cfrac%7B-695%7D%7B16%7D%5Cright%5D)
Result:
Thus the sum of first 20 terms in an arithmetic series is
The percent of markup is 20%
Step-by-step explanation:
Selling price of the drink = $1.50
Cost of the drink last year = $1.25
Percent of markup = (selling price - cost) (100)/cost
Percent of markup = (1.50 - 1.25)(100) /1.25
= 0.25(100) /1.25
= 25/1.25
= 20%
The percent of markup is 20%
Answer:
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Answer:
See below
Step-by-step explanation:
Perpendicular lines have slopes which are negative reciprocals. When the slopes are multiplied like 3 and -1/3 is -1.
This means the line y = -6x + 7 which has slope -6 will have a perpendicular line that has slope 1/6.
Any equation which has slope 1/6 is a solution.
Example 1/6x - 1 or 1/6x + 7 or etc.
there are 12 inches in 1 foot, so 6 inches is really just half a foot, thus 3'6" is really just 3.5' or 3½ feet.
now, let's convert those mixed fractions to improper fractions and then subtract, bearing in mind our LCD will be 8.
![\bf \stackrel{mixed}{4\frac{5}{8}}\implies \cfrac{4\cdot 8+5}{8}\implies \stackrel{improper}{\cfrac{45}{8}}~\hfill \stackrel{mixed}{3\frac{1}{2}}\implies \cfrac{3\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{7}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{45}{8}-\cfrac{7}{2}\implies \stackrel{\textit{using the LCD of 8}}{\cfrac{(1)45~~-~~(4)7}{8}}\implies \cfrac{45-28}{8}\implies \cfrac{17}{8}\implies 2\frac{1}{8}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B5%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%208%2B5%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B45%7D%7B8%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B3%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B7%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B45%7D%7B8%7D-%5Ccfrac%7B7%7D%7B2%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20LCD%20of%208%7D%7D%7B%5Ccfrac%7B%281%2945~~-~~%284%297%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B45-28%7D%7B8%7D%5Cimplies%20%5Ccfrac%7B17%7D%7B8%7D%5Cimplies%202%5Cfrac%7B1%7D%7B8%7D)
