Asked by Riana
a hockey arena has a total seating capacity of 15690.the first row of seats around the rink has 262 seats. the number of seats in each subsequent row increases by 18. how many rows of seats are in the arena?
Answers
Answered by
Stephen
Solve using aritmetic progression and quadratic equation, how?
Sum of AP = n/2(2a+(n-1)d)
Where n is the number of terms, ie in your case the number of rows of seats, 'a' is the number in the first row (a=262) and 'd' is the difference in the next row (d=18). Sum of AP is the number of seats in the arena (Sum=15690). Put this together and you get:
15690=n/2[2*262+(n-1)18]
18n^2+506n-31380=0
equation to: ax^2+bx+c,; a=18, b=506, c=-31380
Now comes the quadratic, solution to a quadratic =
[-b(+/-)Sqrt(b^2-4ac)]/2a
substitute the values of a,b anc from above into the quadratic and solve.
n=30.
Please always check your answer, in this case use sum of AP.
Sum of AP = n/2(2a+(n-1)d)
Where n is the number of terms, ie in your case the number of rows of seats, 'a' is the number in the first row (a=262) and 'd' is the difference in the next row (d=18). Sum of AP is the number of seats in the arena (Sum=15690). Put this together and you get:
15690=n/2[2*262+(n-1)18]
18n^2+506n-31380=0
equation to: ax^2+bx+c,; a=18, b=506, c=-31380
Now comes the quadratic, solution to a quadratic =
[-b(+/-)Sqrt(b^2-4ac)]/2a
substitute the values of a,b anc from above into the quadratic and solve.
n=30.
Please always check your answer, in this case use sum of AP.
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