Question

Find four $+$ive numbers $G_1,G_2{,G}_3,G_4$ in $G.P$ which satisfy the following conditions:
(i) $G_2^2+G_3^2=250$
(ii) $G_1{G}_4=\alpha$ where $\alpha$ is greater root of the equation $x^{2+{log}_{{10}^x}}={\left(0.001\right)}^{-8╱3}$

Answer 1

Let the numbers be $a,ar,{ar}^2,{ar}^3$
$\therefore a^2(r^2+r^4)=250$ or $a^2{r}^2(1+r^2)=250$
and $G_1{G}_4=a^2{a}^3=\alpha$ where $\alpha$ is greater root of
$(2+{\log }_{10}x){\log }_{10}x$
$=\log {}_{10}({10}^{-3}{)}^{-8/3}={\log }_{10}\left({10}^8\right)=8$
or If $\log {}_{10}x=t$ then $t^2+2t-8=0$
$\therefore t=-4,2$
$\log {}_{10}x=-4,2$ or ${\log }_{10}x={10}^{-4},{10}^2=\frac{1}{{10}^4},100$
$\therefore \alpha =100$ being greater root
$a^2{r}^3=100$ and $a^2{r}^2(1+r^2)=250$
and Dividing, $\frac{1+r^2}{r}=\frac{5}{2}$ or $2r^2-5r+2=0$
$\therefore r=2,\frac{1}{2}$
Choosing $r=2$ we get from $\left(1\right)$, $a^2.4\times 5=250$
$\therefore a^2=25/2$ and $r^2=2$
Hence the four numbers are
$a,ar,{ar}^2,{ar}^3$ i.e., $\frac{5}{\sqrt{2}},5\sqrt{2},10\sqrt{2},20\sqrt{2}$
Choosing $r=\frac{1}{2}$, we will get the same numberss in reverse order.