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# Arithmetic/ Geometric Sequence?

Topic: Arithmetic/ Geometric Sequence?
June 16, 2019 / By Alicia
Question: May plans to stack rows of cups in such a way that each rose has one less cup than the row below it. She begin to lay out 41 cups that will form the bottom row of the stack. If she has 783 cups, how many rows will she be able to complete before she runs out of cups?

## Best Answers: Arithmetic/ Geometric Sequence?

Uel | 2 days ago
If the number of cups in each row is: n = 1: 41 n = 2: 40 n = 3: 39 then there is a "constant difference" of 1 between entries, which means that this sequence of numbers is an "arithmetic sequence". The sum of an arithmetic sequence can be calculated quickly using the formula Sum = (number of terms)(average of first & last terms) = (n)(first Plus last) / 2 [Note that Plus means an addition symbol. Yahoo Answers eats Plus signs when you edit an answer.... arrrg!] For this problem, our challenge is to determine "n"... and if we do not know "n", we also do not know precisely what the last term will be. However we can write an expression for it, because the formula for the Nth term of an arithmetic sequence is A_n = A_1 Plus (n-1)(D) where D is the constant difference. So A_n = 41 Plus (n - 1)(-1) = 41 - n Plus 1 = 42 - n Putting this information into the Sum formula from above produces: 783 = (n)(41 Plus 42 - n) / 2 Now we can solve for n 783 = (n)(83 - n) / 2 1566 = 83n - n^2 n^2 - 83n Plus 1566 = 0 Using the quadratic formula to solve the above produces possible solutions of 29 and 54. We need to test each in the sum formula to verify they produce sensible solutions in the context of this problem: (n)(41 Plus 42 - n) / 2 (29)(41 Plus 42 - 29) / 2 = (29)(54)/2 = (29)(27) = 783 Therefore 29 works However note that trying 54 in the formula would result in a negative "last term" (42 - 54 = -12): (54)(41 Plus 42 - 54) / 2 which implies that you would have to have negative cups in the last rows... which does not make sense in the context of this problem. So, the answer is 29 rows.
👍 192 | 👎 2
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Rimmon
The number of cups will stack up in the following manner: 41, 40, 39, 38, ...... Notice that this forms an arithmetic sequence with common difference -1. The sum can be expressed is: Sum = n/2 * (First term + nth term) where n is the total number of rows. n/2 * (41 - n + 42) = 783 n * (83 - n) = 1566 n^2 - 83n + 1566 = 0 n = (83 +/- √(83^2 - 4(1566))/2 = 29, 54 Note that n = 54 won't work since when plugged back into the nth term equation (42 - n), you'll have a negative number of cups. Therefore the total number of rows May can complete before she runs out of cups is 29 rows!
👍 80 | 👎 -5

Meryl
She has 783 cups in total and starts with: row 1 = 41 cups row 2 = 40 row 3 = 39 ..... row 29 = 13 There will be 29 rows and all 783 cups will be used because: (41+40+39+38.......+15+14+13) = 783
👍 79 | 👎 -12

Joseph
It should be 41+40+39+....+n < or = 783. Clearly there are 42-n terms so (42-n)(41+n)/2 < or = 783 1722 + n - n^2 < or = 1566 0
👍 78 | 👎 -19

Haniel
This is just arithmetic sequence. For an arithmetic sequence, if the common difference is d, number of terms is n, and initial term is a, The sum, S(n) of the terms is given by, S(n) = n{2a + (n - 1)d}/2 For our case, d = 1 and a = 41. We have to find the smallest n for which S(n) > 783 or, n(40 + n)/2 > 783 or, n^2 + 40n - 1566 > 0 Take out factors of the expression on the left: (n - 24.33)(n + 64.33) > 0 The sign of the expression on the left varies over the range of n as shown: +ve with n < -64.33 -ve with -64.33 < n < 24.33 +ve with n > 24.33 Then, smallest allowable value for n is 25, which is the solution we seek.
👍 77 | 👎 -26

Originally Answered: Which term of the arithmetic sequence 3, 8, 13,. is 73?
13-8=5 8-3=5 so the sequence is going up by 5. basically the question is asking starting at 3, how many times do you have to go up by 5 to reach 73. the answer is 15.

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