Question

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

Answer 1

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