<![CDATA[Math Is Fun Forum / Arithmetic Means]]> 2013-10-26T07:29:27Z FluxBB http://www.mathisfunforum.com/viewtopic.php?id=20114 <![CDATA[Re: Arithmetic Means]]> That's what I was assuming.

b + e = constant => a+ c is constant.

All you need to fill in is the arithmetic mean.  Do you know what that is?

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2013-10-26T07:29:27Z http://www.mathisfunforum.com/viewtopic.php?pid=288295#p288295
<![CDATA[Re: Arithmetic Means]]> sir here beginning and last numbers are constant,fixed.
plse explane in brief ,i m in trouble

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http://www.mathisfunforum.com/profile.php?id=69939 2013-10-25T16:50:10Z http://www.mathisfunforum.com/viewtopic.php?pid=288268#p288268
<![CDATA[Re: Arithmetic Means]]> hi mukesh

I am not very clear about what is variable and what fixed here.  Let's see if I can sort this out.

Let beginning = b
end = e

first quantity = a
second quantity = c

equidistant => a - b  = e - c

=> a + c = e + b

So, if e and b are fixed it should be easy for you to finish this off. Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2013-10-25T10:32:13Z http://www.mathisfunforum.com/viewtopic.php?pid=288237#p288237
<![CDATA[Arithmetic Means]]> show that the sum of the arithmetic means between two given quantities ,equidistant from the beginning and the end is constant.

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http://www.mathisfunforum.com/profile.php?id=69939 2013-10-25T09:32:18Z http://www.mathisfunforum.com/viewtopic.php?pid=288234#p288234