Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**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.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,445

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

Offline

**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

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,445

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

Offline

Pages: **1**