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# If a, b, c are in G.P thenA. ${{\text{a}}^{\text{2}}}{\text{,b}}{}^{\text{2}}{\text{,}}{{\text{c}}^{\text{2}}}$are in G.PB. ${{\text{a}}^{\text{2}}}{\text{(b + c),}}{{\text{c}}^{\text{2}}}{\text{(a + b),b}}{}^{\text{2}}{\text{(a + c)}}$are in G.PC. $\dfrac{{\text{a}}}{{{\text{b + c}}}}{\text{,}}\dfrac{{\text{b}}}{{{\text{a + c}}}}{\text{,}}\dfrac{{\text{c}}}{{{\text{a + b}}}}$are in G.PD. None of these

Hint: Using the given concept of G.P, ${{\text{b}}^{\text{2}}}{\text{ = ac}}$,as known that the product of first and last term is equal to square of the middle term. Hence, using this and applying trial and error method to calculate the above question.
As we know that ${{\text{b}}^{\text{2}}}{\text{ = ac}}$
${{\text{(}}{{\text{b}}^{\text{2}}}{\text{)}}^{\text{2}}}{\text{ = (ac}}{{\text{)}}^{\text{2}}} \\ {{\text{(}}{{\text{b}}^{\text{2}}}{\text{)}}^{\text{2}}}{\text{ = (}}{{\text{a}}^{\text{2}}}{\text{)(}}{{\text{c}}^{\text{2}}}{\text{)}} \\$
Hence, ${{\text{a}}^{\text{2}}}{\text{,b}}{}^{\text{2}}{\text{,}}{{\text{c}}^{\text{2}}}$are in G.P