Sum of 3 real numbers in GP is $\alpha \mathrm{S}$ and the sum of their squares is $\mathrm{S}^2$. T

Question

Sum of 3 real numbers in GP is $\alpha \mathrm{S}$ and the sum of their squares is $\mathrm{S}^2$. Then

Professor PC Thomas Classes · Answer

Let the GP be a, ${\operatorname{ar}, \operatorname{ar}^2}^2 \frac{\mathrm{a}\left(1-\mathrm{r}^3ight)}{1-\mathrm{r}}=\alpha \mathrm{S}$ $$ \begin{align*} & a_2^2 a^2 r^2, a^2 r^4:  	ag{I}\ & a^2\left[\frac{1-r^6}{1-r^2}ight]=S^2 \quad(	ext { II })  	ag{II}\ & \frac{a^2\left(1-r^3ight)^2}{(1-r)^2}=\alpha^2 \cdot \frac{a^2\left(1-r^6ight)}{1-r^2} \ & \frac{1-r^3}{1-r}=\alpha^2 \frac{\left(1+r^3ight)}{1+r}, 1+r+r^2=\alpha^2\left(1-r+r^2ight) \ & r^2\left(1-\alpha^2ight)+r\left(1+\alpha^2ight)+1-\alpha^2=0 \end{align*} $$ Since r is real, $\mathrm{D} \geq 0$. $$ \begin{aligned} & \left(1+\alpha^2ight)^2-4\left(1-\alpha^2ight)^2 \geq 0 \ & 3 \alpha^4-10 \alpha^2+3 \leq 0 \ & \left(3 \alpha^2-1ight)\left(\alpha^2-3ight) \leq 0 \ & \Rightarrow \frac{1}{3} \leq \alpha^2 \leq 3 \end{aligned} $$