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Question Number 194878 by oMwarimu last updated on 18/Jul/23

	Answered by Rasheed.Sindhi last updated on 18/Jul/23

	$a + d = {a r}^{3} \land a + 9 d = {a r}^{6}$ $a (r^{3} - 1) = d \land a (r^{6} - 1) = 9 d$ $a = \frac{d}{r^{3} - 1} \land a = \frac{9 d}{r^{6} - 1}$ $\frac{d}{r^{3} - 1} = \frac{9 d}{r^{6} - 1}$ $\frac{1}{r^{3} - 1} = \frac{9}{r^{6} - 1}$ $r^{6} - 1 = 9 r^{3} - 9$ $r^{6} - 9 r^{3} + 8 = 0$ $(r^{3} - 1) (r^{3} - 8) = 0$ $r^{3} = 1, 8 \Rightarrow r = 1, 2$
	Answered by Rasheed.Sindhi last updated on 18/Jul/23
	$a+d=ar^3 ∧ a+9d=ar^6 { ((9a+9d=9ar^3 )),((a+9d=ar^6 )) :}⇒8a=9ar^3 −ar^6 ⇒8=9r^3 −r^6 ⇒r^6 −9r^3 +8=0 (r^3 −1)(r^3 −8)=0 r^3 =1,8⇒r=1,2$
	$a + d = {a r}^{3} \land a + 9 d = {a r}^{6}$ ${\begin{cases} 9 a + 9 d = 9 {a r}^{3} \\ a + 9 d = {a r}^{6} \end{cases} \Rightarrow 8 a = 9 {a r}^{3} - {a r}^{6}$ $\Rightarrow 8 = 9 r^{3} - r^{6} \Rightarrow r^{6} - 9 r^{3} + 8 = 0$ $(r^{3} - 1) (r^{3} - 8) = 0$ $r^{3} = 1, 8 \Rightarrow r = 1, 2$
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