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If $(a^{2}+b^{2}),(b^{2}+c^{2})$ and $(a^{2}+c^{2})$ are in geometric progression, which of the following holds true?

  1. $b^{2}-c^{2}= \dfrac{a^{4}-c^{4}}{b^{2}+a^{2}} \\$
  2. $b^{2}-a^{2}= \dfrac{a^{4}-c^{4}}{b^{2}+c^{2}} \\$
  3. $b^{2}-c^{2}= \dfrac{b^{4}-a^{4}}{b^{2}+a^{2}} \\$
  4. $b^{2}-a^{2}= \dfrac{b^{4}-c^{4}}{b^{2}+a^{2}}$
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Given that $:(a^{2}+b^{2}),(b^{2}+c^{2})$ and $(a^{2}+c^{2})$ are in GP.

If $x,y,z$ are in GP, then $\dfrac{y}{x} = \dfrac{z}{y}\implies y^{2} = xz$

$\therefore (a^{2}+b^{2}),(b^{2}+c^{2}),(a^{2}+c^{2})$ are in GP

$\implies \dfrac{(b^{2}+c^{2})}{(a^{2}+b^{2})} = \dfrac{(a^{2}+c^{2})}{(b^{2}+c^{2})}$

$\implies (b^{2}+c^{2})(b^{2}+c^{2}) = (a^{2}+c^{2}) (a^{2}+b^{2})$

$\implies (b^{2}+c^{2})^{2} = (a^{2}+c^{2}) (a^{2}+b^{2})$

$\implies b^{4}+c^{4} + 2b^{2}c^{2} = a^{4} + a^{2}b^{2} +  c^{2}a^{2} + c^{2}b^{2}$

$\implies b^{4}+c^{4} + b^{2}c^{2} = a^{4} + a^{2}b^{2} +  a^{2}c^{2}$

$\implies b^{4} + b^{2}c^{2} – a^{2}b^{2} – a^{2}c^{2} = a^{4} – c^{4}$

$\implies b^{2}(b^{2} + c^{2}) – a^{2}(b^{2} + c^{2}) = a^{4} – c^{4}$

$\implies (b^{2} + c^{2})(b^{2}-a^{2}) = a^{4}  – c^{4}$

$\implies b^{2}-a^{2} = \dfrac{a^{4}  – c^{4}}{b^{2} + c^{2}}$

So, the correct answer is $(B).$

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