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

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In a certain sculpture, coils of wire are arranged in rows. [#permalink]

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New post 02 Jan 2010, 15:44
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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 there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

A. 22
B. 21
C. 20
D. 19
E. 18
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Re: Sequencing PS Help Needed [#permalink]

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New post 02 Jan 2010, 16:01
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x13069 wrote:
Can anyone help with this one?

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 there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

1)22
2)21
3)20
4)19
5)18


We have arithmetic progression with common difference \(d=2\), number of terms \(n=10\) and the sum \(S=120\).

The sum of AP \(S=120=n\frac{2a+d(n-1)}{2}=10\frac{2a+2(10-1)}{2}\), where \(a\) is the first term. --> \(120=10\frac{2a+2(10-1)}{2}=10a+90\) --> \(a=3\).

\(a_n=a_{10}=a_1+d(n-1)=3+2(10-1)=21\)

Answer: B.
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Re: In a certain sculpture, coils of wire are arranged in rows. [#permalink]

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New post 19 Jan 2018, 08:40
x13069 wrote:
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 there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

A. 22
B. 21
C. 20
D. 19
E. 18


Method 1:

Let \(n\) be the number of rings in the final row. Each preceding row is 2 less than the current row going down to 10 rows in total. So,

\(n + n - 2 + n - 6 + n - 8 + n - 10 + n - 12 + n - 14 + n - 16 + n - 18 = 120\)
\(10n - 90 = 120\)
\(10n = 210\)
\(n = 21\)

Method 2:

We know that

Average * number of terms = Sum

putting in values

\(Average = \frac{120}{10} = 12\)

Since there are 10 terms and 2 apart, 5th and 6th term would be 11 and 13 respectively. adding \(4*2 = 8\) to \(13\) would give us the 10th term.
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Re: In a certain sculpture, coils of wire are arranged in rows. [#permalink]

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New post 20 Mar 2018, 04:26
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x13069 wrote:
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 there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

A. 22
B. 21
C. 20
D. 19
E. 18



the coils are

x + x+ 2 + x+ 4 + x+ 6 + x+ 8 + x+ 10 + x+ 12 + x+ 14 + x+ 16 + x+ 18

= 10x + 2 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 10x + 2 * (9 * 10)/2

=) 10x + 90 = 120

=) x = 3

thus, the coils in the final row are
=x + 18 = 21 = B the answer

thanks
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Re: In a certain sculpture, coils of wire are arranged in rows. [#permalink]

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New post 21 Mar 2018, 16:11
x13069 wrote:
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 there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

A. 22
B. 21
C. 20
D. 19
E. 18


We can let x = the number of coils in the first row, then x + 2 = the second row, x + 4 = the third row, and so on. Thus, x + 18 = the number of coils in the last (or tenth) row.

Since the number of coils in each row forms an arithmetic progression, the sum of all the terms equals the number of terms times the average of the terms.

We can calculate the average of the terms in this equally spaced set by averaging the first and last terms. Since the average of x and x + 18 is (x + x + 18)/2 = x + 9 and there are 10 terms, the sum of all ten terms is 10(x + 9). Setting this expression equal to 120, we have:

10(x + 9) = 120

x + 9 = 12

x = 3

So there are 3 + 18 = 21 coils in the last row.

Answer: B
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Re: In a certain sculpture, coils of wire are arranged in rows.   [#permalink] 21 Mar 2018, 16:11
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