Question

if a is the am of b and c and the two gm are g1 and g2 then prove that g1 3 g2 3 2abc

Answer 1

Heya mate,

here is your answer,
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Please find below the solution to the asked query:-

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We haveG1 and G2 as two geometric means between b and c i.e.
b,G1,G2,c are in G.P.

$c = b {(r)}^{4 - 1} \\ c = {br}^{3}$

$r \: = { (\frac{c}{b} )}^{ \frac{1}{3} }$

G1 = br = b.(c/b)^1/3
=> G1^3 = b^3 (c/b) = b^2c
G2 = br^2 = b (c/b)^2/3
=> G2^3 = b^3(c^2/b^2) = bc^2


As a is A.M. of b and c, hence

a=b+c2 ;(i)
G13+G23=b2c+bc2
=bc(b+c)
=2bc(b+c2)
=2bca    [Using (i)]
=2abc

⇒G1^3+G2^3=2abc

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# nikzz

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Answer 2

Answer:

Step-by-step explanation: