Jeremy opens a chocolate shop in the city. he pays $1,500 a month for rent and maintenance of the shop. the price of raw materials and manufacturing the chocolates is $6,000 a month. he sells the chocolates individually and in boxes of a dozen.
jeremy understands that his business needs a little time to become a success and decides that he wants to build a customer base initially. he is happy to break even for the first year.
after a year, jeremy prices his chocolates at $3 apiece. he offers a discount of 10% on boxes of chocolate to promote the sale of boxes. how many boxes of chocolate would he have to sell to recover the cost of running the business this year per month? (assume that he sells no individual chocolates.)
a.188 boxes
b.208 boxes
c.226 boxes
d.232 boxes

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺