Question

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Answer 1

$Let \: ( {a}^{2} + {b}^{2} ) \: (ab + bc) \: and \: ( {b}^{2} + {c}^{2} ) \: are \: in \: GP. \\ \\ Then \: by \: using \: the \: condition\: of \: GP \\ \\ {(ab + bc)}^{2} = ( {a}^{2} + {b}^{2} )( {b}^{2} + {c}^{2} ) \\ \\ {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + 2a {b}^{2} c = {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {b}^{4} + {b}^{2} {c}^{2} \\ \\ {b}^{4} + {a}^{2} {b}^{2} - 2a {b}^{2} c = 0 \\ \\ {( {b}^{2} - ac)}^{2} = 0 \\ \\ {b}^{2} - ac = 0 \\ \\ {b}^{2} = ac$
It is the condition of GP.

So, a,b,c are in GP.

Answer 2

Answer:

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