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Question

If x is the arithmetic mean of y and z and the two geometric means between y and z are G1 and G2, then G31+G32=.

A
xyz
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B
1xyz
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C
2xyz
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D
2xyz
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Solution

The correct option is C 2xyz
It is given that x is the arithmetic mean of y and z.
x=y+z22x=y+z
Since, G1 and G2 are two geometric means between y and z, so y, G1, G2, z are in a GP with common ratio r=(z/y)13.G1=yrG31=y3r3=y3((zy)13)3=y2zG2=yr2G32=y3r6=y3((zy)13)6=yz2
G31+G32=y2z+yz2=yz(y+z)=yz×2x=2xyz

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