So it went from the 16th term to the 17th term, and went from 21 to -1... what would the "common difference"
d be then?
le'ts see
![\bf a_{16}=21\quad and\quad a_{17}=-1 \\\\\\ a_{16}+d=-1\implies 21+d=-1\implies d=-1-21\implies \boxed{d=-22}](https://tex.z-dn.net/?f=%5Cbf%20a_%7B16%7D%3D21%5Cquad%20and%5Cquad%20a_%7B17%7D%3D-1%0A%5C%5C%5C%5C%5C%5C%0Aa_%7B16%7D%2Bd%3D-1%5Cimplies%2021%2Bd%3D-1%5Cimplies%20d%3D-1-21%5Cimplies%20%5Cboxed%7Bd%3D-22%7D)
alrite.. so d = -22.. hmm what would the first term be then?
![\bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ n=16\\ d=-22\\ a_{16}=21 \end{cases} \\\\\\ a_{16}=a_1+(16-1)\boxed{-22}\implies 21=a_1+(16-1)(-22) \\\\\\ 21=a_1+(15)(-22)\implies 21=a_1-330\implies \boxed{351=a_1}\\\\ -------------------------------\\\\ a_n=a_1+(n-1)d\implies \boxed{a_n=351+(n-1)(-22)}](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%5C%5C%5C%5C%0Aa_n%3Da_1%2B%28n-1%29d%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0An%3Dn%5E%7Bth%7D%5C%20term%5C%5C%0Aa_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%0Ad%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%0A----------%5C%5C%0An%3D16%5C%5C%0Ad%3D-22%5C%5C%0Aa_%7B16%7D%3D21%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0Aa_%7B16%7D%3Da_1%2B%2816-1%29%5Cboxed%7B-22%7D%5Cimplies%2021%3Da_1%2B%2816-1%29%28-22%29%0A%5C%5C%5C%5C%5C%5C%0A21%3Da_1%2B%2815%29%28-22%29%5Cimplies%2021%3Da_1-330%5Cimplies%20%5Cboxed%7B351%3Da_1%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Aa_n%3Da_1%2B%28n-1%29d%5Cimplies%20%5Cboxed%7Ba_n%3D351%2B%28n-1%29%28-22%29%7D)
which of course, you can rewrite as
which are all the same.
Amount of money paid by Jeremy as rent and maintenance of shop per month = $1500
Cost of raw materials and manufacturing per month = $6000
Total cost that Jeremy has to spend per month = (1500 + 6000) dollars
= 7500 dollars
Number of individual chocolates sold = 2400
Number of chocolates sold in boxes = 50 boxes
= (12 * 50) chocolates
= 600 chocolates
Then
Total number of chocolates sold by Jeremy = 2400 + 600
= 3000
Now
Price of each chocolate = 7500/3000 dollars
= 75/30 dollars
= 5/2 dollars
= 2.5 dollars
Price of 600 chocolates = 600 * (5/2) dollars
= 300 * 5 dollars
= 1500 dollars
Price of 12 chocolates = (1500/600) * 12
= 15 * 2
= 30 dollars
Then
We can say that each box of chocolate should be sold at $30. All the loose chocolates should be sold at $2.5 each.
Answer:
tenth
Step-by-step explanation: