Question

if a:b=b:c then prove that

plz friends don't answer Irrelevant​

Answer 1

Answer:

I am sorry :( Idk, maybe someone has already asked this?

Step-by-step explanation: ...

Answer 2

Step-by-step explanation:

Given that:

$\tt \red{If \:   \blue b \:  is \: a \: mean \: proportional \: between \:   \blue a  \: and \: \blue c, }$

To prove:

$\tt That \: \red {\frac{ { {a}^{2} - b}^{2} + {c}^{2} }{ {a}^{ - 2} - {b}^{ - 2} {c}^{ - 2} } } = {b}^{4}$

Solution:

$\tt{Now \: b \: is \: mean \: proportional \: between \:  a, \: c \:   then \:   {b}^{2} =ac} \\ \\ </u><u>[</u><u>/</u><u>tex]</u></p><p><u>[tex] \tt{Now \: b \: is \: mean \: proportional \: between \:  a, \: c \:   then \</u><u>:</u><u>{b}^{2} =ac}$

$\tt \red{ \frac{ {a}^{2} - {b}^{2} + {c}^{2} }{ \frac{1}{ {a}^{2} } - \frac{1}{ {b}^{2} } + \frac{1}{ {c}^{2} } } } = {b}^{4}$

$\tt \: Putting \: tht \: value \: of \: \red{ b ^{2} }$

$⟹ \tt \red{\frac{ {a}^{2} - ac + {c}^{2} }{ \frac{1}{ {a}^{2} } - \frac{1}{ac} + \frac{1}{ {c}^{2} } } } = {b}^{4}$

$⟹ \tt \red{ {a}^{2} {c}^{2} ( \frac{ {a}^{2} - ac + {c}^{2} }{ {c}^{2} - ac + {a}^{2} } ) }= {b}^{4}$

$⟹ \tt \red{{a}^{2} {c}^{2}} = {b}^{4}$

$⟹ \tt \red{{b}^{4}} = {b}^{4}$

$\tt \underline {\red{LHS} =RHS}$

Hence proved.