If m is the A . M .   of two distinct real numbers I and n I ,   n > 1 and G 1 ,   G 2…

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

  1. 4l⁡2m2 n2
  2. 4 l⁡2mn
  3. 4 l⁡m2 n
  4. 4l⁡mn2

Solution

Given, m is the A.M. of l & n. 

⇒m=l+n2  ...1

G1, G2, G3 are G.M. between l & n.

⇒l, G1, G2, G3, n are in G.P.

⇒n=lr4  ⇒r4=nl  ...2

⇒G1=l⁡r, G2=l⁡r2, G3=l⁡r3

⇒G14+2G24+G34=l⁡4r4+2×l⁡4r8+l⁡4×r12

=l⁡4r41+2r4+r8

=l4×nl⁡1+r42=l⁡3n×1+nl⁡2

=l⁡3×n×l⁡+n2l⁡2

=l⁡n×2m2   (From (1))

=4l⁡m2n

Asked in: JEE Main 2015 (04 Apr)

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