Question

If between two numbers one geometric mean is G and two arithmetic means are p and q, prove that :G^2 =(2p-q) (2q-p)

Answer 1

g = (ab)1/2  is quite simple.now a , p , q, b are in AP .b =  a +(4-1) d  ;   d = b -a/3thereofore ,  p =  a +(b-a/3) =  2a+b/3                     q  = p +d =  (2a+b/3) + (b -a)/3 = 2b+a/3so 2p-q  = 3a/3 =a  and p -2q  = -bso , (2p-q)(p-2q) =  -ab =  -g^2

Answer 2

Let the two nos. Be a and b.
Geometric mean (G) between a and b will be √ab
G^2 = ab ---------(1)
Given : p and q are 2 arithmetic mean between a and b
Then A.P. is a,p,q,b
--> p=(a+q)/2 and q=(p+b)/2
--> 2p = a+q and 2p = p+b ------ (2)
To prove : G^2 =(2p-q)(2q-p)
RHS = (2p-q)(2q-p)
= (a+q-q)(p+b-p) {from (2)}
= ab
= G^2 {proved above in (1)}
= LHS
HENCE PROVED
HOPE IT IS HELPFUL :)