Question 1091490
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If R,S,T are in AP, then S-R = T-S, or 

(1)  2S = R+T

If U,V,W are in GP, then {{{V/U}}}{{{""=""}}}{{{W/V}}}, or 

(2)  UW = V<sup>2</sup>
 
If X,Y,Z are in HP, then their reciprocals are in AP, so

{{{1/Y-1/X}}}{{{""=""}}}{{{1/Z-1/Y}}}, 

Multiply through by LCD = XYZ

XZ-YZ = XY-XZ

(3)  2XZ = XY+YZ

Since a,b,c are in AP, then we substitute R=a, S=b, T=c in (1)

(4)  2b = a+c               <--we know this

Since  a, mb, c are in GP, then we substitute U=a, V=mb, W=c in (2)

(5)  ac = m<sup>2</sup>b<sup>2</sup>            <--we know this

We are to prove that a, m<sup>2</sup>b, c are in HP.  So to find 
out what we need to prove, we substitute X=a, Y=m<sup>2</sup>b, Z=c in (3)

(6)  2ac = am<sup>2</sup>b + m<sup>2</sup>bc    <--we are to prove this

So we will start with 2ac, and show that it is equal to am<sup>2</sup>b + m<sup>2</sup>bc

By (5),    

2ac = 2m<sup>2</sup>b<sup>2</sup> 

2ac = m<sup>2</sup>&#8729;b&#8729;2b, use (4) to substitute a+c for 2b 

2ac = m<sup>2</sup>&#8729;b&#8729;(a+c), distribute:

1ac = m<sup>2</sup>&#8729;b&#8729;a + m<sup>2</sup>&#8729;b&#8729;c

That's the same as

2ac = am<sup>2</sup>b + m<sup>2</sup>bc

which is (6), which is what we were to prove.

Edwin</pre></font></b>