The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

Hi!

Let the three numbers in G.P. be

*a*,*ar*and*ar*^{2}.*a*+

*ar*+

*ar*

^{2}= 21 (Given)

⇒

*a*^{2}( 1+*r*+*r*^{2})^{2}= 441 ... (1)(

*a*)^{2}+ (*ar*)^{2}+ (*ar*^{2})^{2}= 189 (Given)⇒

*a*^{2}(1 +*r*^{2}+*r*^{4}) = 189 ... (2)From (1) and (2), we get

Thus, the three numbers in G.P. are 3, 6 and 12.

Cheers!

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