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Permutations And Combinations

Question
CBSEENMA11015456

If m is the AMN of two distinct real numbers l and n (l,n>1) and G1, G2, and G3 are three geometric means between l and n, then straight G subscript 1 superscript 4 space plus 2 straight G subscript 2 superscript 4 space plus space straight G subscript 3 superscript 4 equals

  • 4l2 mn

  • 4lm2n

  • 4 lmn2

  • 4l2m2n2

Solution

B.

4lm2n

Given, 
m is the AM of and n
l +n = 2m
and G1, G2, G3, n are in GP
Let r be the common ratio of this GP
G1 = lr
G2 =lr2
G3= lr3
n = lr4
rightwards double arrow space straight r space equals space open parentheses straight n over l close parentheses to the power of 1 divided by 4 end exponent
Now comma space straight G subscript 1 superscript 4 space plus space 2 straight G subscript 2 superscript 4 space plus space straight G subscript 3 superscript 4 space equals left parenthesis l straight r right parenthesis to the power of 4 space plus space left parenthesis l straight r squared right parenthesis to the power of 4 space plus space left parenthesis l straight r cubed right parenthesis to the power of 4
space equals space straight l to the power of 4 space straight x space straight r to the power of 4 left parenthesis 1 plus 2 straight r to the power of 4 plus straight r to the power of 8 right parenthesis
equals space straight l to the power of 4 space straight x space straight r to the power of 4 space left parenthesis straight r to the power of 4 space plus 1 right parenthesis squared
equals space straight l to the power of 4 space straight x space straight n over straight l open parentheses fraction numerator n italic plus italic 1 over denominator l end fraction close parentheses
equals space l n italic space x italic 4 m to the power of italic 2 italic space italic equals italic space italic 4 l m to the power of italic 2 n

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