SOLUTION: An amphitheater has a number of rows with 30 seats in the first row, 32 in the second, 34 in the third and so on. The question is: How many rows should be there to get 3950 seat

Question 1145656: An amphitheater has a number of rows with 30 seats in the first row, 32 in the second, 34 in the third and so on.
The question is: How many rows should be there to get 3950 seats?
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I know that the formula to get the number of seats in a row is Sn(n/2)(a1+1n). But I'm struggling to figure out a formula to figure out how many rows there are given the number of seats.
Thanks
Found 3 solutions by ikleyn, greenestamps, solver91311:
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

It is arithmetic progression with the first term = 30, the common difference d= 2 and the sum of the first n terms of 3950. The formula for the sum of the first n terms is = Substitute here = 3950, = 30 and d= 2. You will get 3950 = = 30n + n*(n-1) = n^2 +29n n^2 + 29n - 3950 = 0 n = = = . Only positive value is meaningful n = = = 50. ANSWER. 50 rows.

Solved.

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For introductory lessons on arithmetic progressions see
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
in this site.

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".

Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.

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Comparing with the solution by @greenestamps, notice that I solved the problem algebraically
and presented THE METHOD to you, while he simply "guessed" the answer.

The difference should be absolutely clear to you . . .

Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!

I'm always curious why nearly all references show the sum of the terms of an arithmetic sequence as

That formula, directly translated into words, says that the sum is half the number of terms, multiplied by the sum of the first and last terms.

To me, a MUCH more meaningful form for the formula is

In that form, the formula says the sum is the number of terms, multiplied by the average of the first and last terms. Since the average of the first and last terms in an arithmetic sequence is the average of ALL the terms, this formula simply states the obvious: a sum is equal to the number of terms, multiplied by the average of the terms.

You can, of course, plug the given information into the formula an solve a quadratic equation to find the answer to the problem. However, if an algebraic solution is not required, you can find the answer with some rather easy arithmetic.

The sum of the terms of the sequence is 3950; and we know that is the product of a whole number (the number of terms) and another number which is either a whole number or halfway between two whole numbers.

In this example, with these numbers, by far the most likely case is

3950 = 79*50

So it looks as if it is very likely that there are 50 rows of seats.

Checking that, we see that the 50th row would have 30+49(2) = 30+98 = 128 seats; and that would make the average of the first and last terms (30+128)/2 = 158/2 = 79 -- as required.

So indeed the number of rows of seats is 50.
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

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