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Question

If m is the AM 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 G41+2G42+G43 equals

A
4l2mn
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B
4lm2n
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C
lmn2
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D
l2m2n2
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Solution

The correct option is B 4lm2n
Given, m is the AM of l and n
l+n=2m(i)
and G1,G2,G3 are geometric means between l and n
l,G1,G2,G3,n are in GP
Let r be the common ratio of this GP
G1=lr, G2=lr2, G3=lr3, n=lr4
r=(nl)14
Now, G41+2G42+G43=(lr)4+2(lr2)4+(lr3)4
=l4×r4(1+2r4+r8)=l4×r4(r4+1)2
=l4×nl(n+ll)2=ln×4m2=4lm2n

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