In a certain sculpture, coils of wire are arranged in rows.

Question

In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If the total of 120 coils of wire in the scuplture, how many coils are in the final row?

22
21
20
19
18

Final answer to follow. I know how to answer using brute force, but looking for the use of a formula to answer. Please show explanation.

Thanks. :-D

voltron · Answer

The formula for any row is 2(i-1) + n for i = 1 to 10 which is
2(0) + n + 2(1) + n ..... 2(9) + n

so to simplify, you get
120 = 10n + the sum of the even consecutuve integers from 2 to 18

Just use the formula for the sum of even consecutive numbers where H is the high number and L is the low number.
(H+L)/2 X (H-L)/2 + 1
(18+2)/2 X (18-2)/2 + 1
 10 X 9
 90

the formula is 120 = 10n + 90, solve for n and then plug in for i = 10.

jimmyjamesdonkey · Answer

Answer is 21. Formula seems complex. Any way to simplify?

voltron · Answer

The formula might look complex, but that's just because its hard to explain thru text.

Just think about the problem and its easy to come up with:
row 01: n + 0
row 02: n + 2
row 03: n + 4
row 04: n + 6
row 05: n + 8
....
row 09: n + 16
row 10: n + 18
-------------------------------------------------
total :10n+(the sum of 0,2,4....16,18)

then come up with this formula since the total equals 120 (based on what's given in the question)

120 = 10n + the sum of the consecutive even integers from 2 to 18 (i.e. 2,4,6,8...18)

if you get to that point and know the formula for consecutive even integers, the problem can be solved in around a minute. With "brute force," it would take much longer.

msoftceo · Answer

There's an intuitive way to solve this problem.

We've got 120 coils distributed among 10 rows. So, we know that each row has an average of 12 coils. Thus, the middle row should have 12. 

Integers 1 through 10 have no real middle, so the 5th row has 11 coils and the 6th row has 13 coils. (Making the 5.5th row, the middle, have 12) Count up from the 6th row to get 21 coils in the 10th row.

vay · Answer

Answer to this is 21

--
If first row contains a coils then based on the question 
2nd row contains - a+2, 
3rd row contains - (a+2)+2 = a+4 = a + 2(2)
4th row contaoins - {(a+2)+2}+2 = a+ 6 = a+3(2)
...

this is in AP.


a, a+d, + a=2d, a+3d....

Sn = n/2{2a+(n-1)d}

120 = 10/2{2a+(10-1)2}
=> 240 = 20a + 180
=> 20a = 60
=> a = 3

Now 10th term = a + (n-1)d
=> 3 + (10-1)2 

=> 3+18 = 21

In a certain sculpture, coils of wire are arranged in rows.

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