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**mukesh****Member**- Registered: 2010-07-18
- Posts: 31

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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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,508

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

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**mukesh****Member**- Registered: 2010-07-18
- Posts: 31

sir here beginning and last numbers are constant,fixed.

plse explane in brief ,i m in trouble

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,508

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

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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