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May 27th, 2013, 01:17 PM   #1
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arithmetic progression

I have attached the questions...[attachment=0:2ker6nlf]576.png[/attachment:2ker6nlf]please someone help me out.
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 May 27th, 2013, 02:12 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 The eighth hole would have a diameter of 18.8mm, so the ninth hole would have a diameter of 21.2mm, which is greater than the width of the plate. This is possible only if the holes are not all circular. Without knowing the shape of the holes, one can't find their area, so the question about area can't be answered.
 May 27th, 2013, 03:49 PM #3 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Re: arithmetic progression It says "diameter" of the holes, so they're probably circular. I think this is actually a trick question. Like skipjack said, the size of each hole after the 8th is larger than the width of the plate, so it is impossible to have an even larger hole "drilled" into it.
 May 27th, 2013, 07:52 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,594 Thanks: 1038 Re: arithmetic progression WHAT is the topic of what' this is about, William: arithmetic series? Tangent circles? Circle areas? Fitting circles in a rectangle? If it's arithmetic series, then sum of the 8 diameters = (number of terms) * (1st term + last term)/2 =8(2 + 18.8)/2 = 83.2 So if the holes (circles) are made consecutively, then evidently impossible. Did your teacher make that up? If so, then he/she should be shot at sunrise :wink:
 May 28th, 2013, 01:01 AM #5 Newbie   Joined: May 2013 Posts: 14 Thanks: 0 Re: arithmetic progression ^^ It's arithmetic series. Why a sum is greater than the plate itself? Could be the holes on plates are in series? *just guessing*
May 28th, 2013, 06:16 AM   #6
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Re: arithmetic progression

Hello, will.i.am1!

Where did this problem come from?
As skipjack pointed out, the problem is quite impossible.

Quote:
 A rectangular steel plate, 50 mm by 20 mm, is to have holes of various sizes drilled into it. The first hole will have a diameter of 2mm. Each subsequent hole will increase the diameter by 2.4mm on the previous hole. (a) What will be the diameter of the 8th hole?

$\text{This is an arithmetic progression.}$

$\text{The }n^{th}\text{ term is: }\:a_n \:=\:a\,+\,(n-1)d$

$\text{W\!e have: \: first term }a= 2.0,\text{ common difference }d = 2.4$

$\text{Therefore: }\:a_8 \;=\;2.0\,+\,7(2.4) \;=\;18.8\text{ mm}$

Quote:
 (b) What will be the area of the plate undrilled after the 12th hole?[color=beige] . [/color][color=red]??[/color]

$\text{The }12^{th}\text{ hole has a diameter of: }\:a_{12} \:=\:2\,+\,11(2.4) \:=\:28.4\text{ mm}$[color=beige] . [/color][color=blue]What?[/color]

Perhaps they meant the 8th hole . . .

Then we have:
[color=beige]. . . . . . . . . . . [/color]$\begin{array}{ccc}A_1=&\pi(1^2) \\ A_2=&\pi(2.2)^2 \\ A_3=&\pi(3.4)^2 \\ \vdots=&\vdots \\ A_8=&\pi(9.4)^2 \\ \hline \\=&276.8\pi \end{array}=$

$\text{The total area of the holes is about }869.6\text{ mm}^2$

$\text{The area of the undrilled plate is: }\50\,\times\,20)\,-\,869.6 \;=\;130.4\text{ mm}^2" />

 May 28th, 2013, 08:15 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 If the holes are circular and don't overlap, there isn't enough room for the first eight holes (as there's only just about enough room for the sixth, seventh and eighth). There is room for the first seven holes.
May 28th, 2013, 11:24 AM   #8
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Re: arithmetic progression

[attachment=0:303xb3ai]circ.png[/attachment:303xb3ai]
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 May 28th, 2013, 05:58 PM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,594 Thanks: 1038 Re: arithmetic progression Huh?! Why not then make one 18.8 hole and be finished May I suggest you (somehow) post your question properly. Is this from your teacher?
 May 29th, 2013, 01:06 AM #10 Newbie   Joined: May 2013 Posts: 14 Thanks: 0 Re: arithmetic progression What you meant by post your question properly? Yes, this question is given me by teacher.

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