If m is the A.M. of two distinct real numbers I and n (l,n>1) and G1,G2 and G3  are three geometric means between I and n, then G14+2G24+G34 equals :

If m is the A.M. of two distinct real numbers I and n (l,n>1) and G1,G2 and G3  are three geometric means between I and n, then G14+2G24+G34 equals :

  1. A

    4 lmn2

  2. B

    4 l2m2n2

  3. C

    4 l2mn

  4. D

    4 lm2n

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    Solution:

    We have, m=l+n2

    Let r be the common ratio of the  G.P. l,G1,G2,G3,n Then,

    n=lr4r=nl1/4 G1=lr=l3/4n1/4,G2=lr2=l2/4n2/4,G3=lr3=l1/4n3/4 G14+2G24+G34=l3n+2l2n2+ln3 =ln(l+n)2=ln(2m)2=4lm2n.

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