Question

If a, x, b are in

a.P ; a, y, b are in g.P ; a, z, b are in h.P such that x=9z and a>0,b>0 then:

Answer 1

i think its the answer

Answer 2

$i ) a,x,b\:are \: in \: A.P \: (given)$

$Arithmetic \: Mean (x) = \frac{a+b}{2}\\\implies 2x =( a + b ) \: --(1)$

$ii ) a,y \: and \: b \: are \:in \:G.P . (given)$

$Geometric \:mean (y) = \sqrt{ab} \\\implies y^{2} = ab \: ---(2)$

$iii ) a,z \: and \: b \: are \:in \:H.P . (given)$

$Harmonic \: mean (z) = \frac{2ab}{a+b} \\= \frac{ 2 \times y^{2}}{ 2x} \: [From \:(1) \:and \:(2) ]$

$\implies z = \frac{y^{2}}{x} \\\implies zx = y^{2} \:--(3)$

$But , \: x = 9z \: (given)\: ---(4)$

$z \times 9z = y^{2}$

$\implies 9z^{2} = y^{2}$

$\implies (3z)^{2} = y^{2}$

$\implies 3z = y$

Therefore.,

$\red{If \: x = 9z }\green { \: then \: y = 3z }$

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