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Solution :

Given, a=AM between b and c <br> `implies " " a=(b+c)/(2) implies 2a=b+c " " "………(i)"` <br> Again, `b,G_(1),G_(2),c` are in GP. <br> ` therefore " " (G_(1))/(b)=(G_(2))/(G_(1))=(c)/(G_(2)) implies b=(G_(1)^(2))/(G_(2)),c=(G_(2)^(2))/(G_(1))` <br> and `G_(1)G_(2)=bc " " ".......(ii)"` <br> From Eqs. (i) and (ii), <br> `2a= (G_(1)^(2))/(G_(1))+(G_(2)^(2))/(G_(2))=(G_(1)^(3)+ G_(2)^(3))/(G_(1)G_(2))=(G_(1)^(3)+G_(2)^(3))/(bc)[therefore G_(1)G_(2)=bc]` <br> `implies G_(1)^(3)+G_(2)^(3)=2abc`