If one GM 'g' and two AM's 'p' and 'q' are inserted between two numbers a and b, then prove that :-

(2p-q)(p-2q)= -g^2

Dear sridhar ,

g = (ab)^1/2  is quite simple.

now a , p , q, b are in AP .

b =  a +(4-1) d  ;   d = b -a/3

thereofore ,  p =  a +(b-a/3) =  2a+b/3

                     q  = p +d =  (2a+b/3) + (b -a)/3 = 2b+a/3

so 2p-q  = 3a/3 =a  and p -2q  = -b

so , (2p-q)(p-2q) =  -ab =  -g^2

proved . this is the straight forward  method and i think is most intuitive so i suggest follow this method.

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