Question:
if a is the A.M. of b and c and G1 and G2 are two GM's between them then prove that sum of their cubes is 2abc
Answer:
Given, a is the A.M. of b and c
=> a = (b + c)
=> 2a = b + c ............1
Again, given G1 and G2 are two GM between b and c,
=> b, G1 , G2 , c are in the GP having common ration r, then
=> r = (c/b)1/(2 + 1) = (c/b)1/3
Now,
G1 = br = b*(c/b)1/3
and G1 = br = b*(c/b)2/3
Now,
(G1 )3 + (G2 )3 = b3 *(c/b) + b3 *(c/b)2
=> (G1 )3 + (G2 )3 = b3 *(c/b)*( 1 + c/b)
=> (G1 )3 + (G2 )3 = b3 *(c/b)*( b + c)/b
=> (G1 )3 + (G2 )3 = b2 *c*( b + c)/b
=> (G1 )3 + (G2 )3 = b2 *c*( b + c)/b ..............2
From equation 1
2a = b + c
=> 2a/b = (b + c)/b
Put value of(b + c)/b in eqaution 2, we get
(G1 )3 + (G2 )3 = b2 *c*(2a/b)
=> (G1 )3 + (G2 )3 = b *c*2a
=> (G1 )3 + (G2 )3 = 2abc
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