SOLUTION: Can someone help me? An amphitheater is set up so that there are 10 seats in the first row, 13 seats in the second row, 16 in the third row, 19 in the fourth row, etc., until ther

Question 1090838: Can someone help me?
An amphitheater is set up so that there are 10 seats in the first row, 13 seats in the second row, 16 in the third row, 19 in the fourth row, etc., until there are 85 seats in the last row of the amphitheater.
a. how many rows are in the amphitheater?
b. How many total seats are in the amphitheater?
Thanks

Answer by ikleyn(52788) (Show Source): You can put this solution on YOUR website!
.

This sequence 10, 13, 16, . . . is an arithmetic progression with the first term of 10 and the common difference of 3.


The formula for the n-th term of any arithmetic progression is

 = .


So, to find the number of terms (the number of rows) you need to solve this equation

 = 85,   or

10 + 3*(n-1)*3 = 85  ====>  3*(n-1) = 85 - 10 = 75  ====>  n-1 =  = 25  ====>  n = 26.


Thus there are 26 rows in the amphitheater. 


To answer the second question, you need to find the sum of the 26 terms of this arithmetic progression.

The sum is   =  =  = 95*13 = 1235.


Answer.  There are 26 rows and 1235 seats in the amphitheater.

Solved.

There is a bunch of lessons on arithmetic progressions in this site:
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
    - Mathematical induction and arithmetic progressions
    - One characteristic property of arithmetic progressions
    - Solved problems on arithmetic progressions

Also, you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".

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