The sum of 3 terms in G.P. is 21 and sum of their squares is 189. Find

1.	The sum of 3 terms in G.P. is 21 and sum of their squares is 189. Find thenumbers
Answer» Answer: 3,6,9 Step-by-step EXPLANATION: First, based on the given INFORMATION, we have: $a+ax+ax^{2} =a(1+x+x^{2} )$ = 21 (1) $a^{2}+a^{2}x^{2}+a^{2}x^{4}=a^{2}(1+x^{2}+x^{4})$ = 189 (2) SECOND, square the (1) equation, we have: $a^{2}(x^{4}+2x^{3}+2x^{2}+x^{2}+2x+1)=441$ ⇔ $a^{2}(1+x^{2}+x^{4})+a^{2}(2x^{3}+2x^{2}+2x)=441$ ⇔ $a^{2}(1+x^{2}+x^{4})+2a^{2}x(x^{2}+x+1)=441$ (3) Third, substitute (1),(2) equation into (3) equation, we have: $189+2ax.21=441$ ⇒ $ax=6$ ⇒ $a=\frac{6}{x}$ Forth, substitute $a=\frac{6}{x}$ into (1) equation, we have: $\frac{6}{x}(1+x+x^{2})=21$ ⇔ $6-15x+6x^{2}=21$ ⇒ $x=2,x=\frac{1}{2}$ Finally, use the solved x to find a by $ax=6$ THEREFORE, $a=3,a=12$ Thus, 3 numbers are 3,6,12