Given that the sum to infinity of a G.P. is denoted by S.
Show that if a > 0, then S > 4ar except for one value of r.
Hello, moonchildvulcan!
$\displaystyle \text{Given that the sum to infinity of a G.P. is denoted by }S.$
$\displaystyle \text{Show that if }a > 0,\,\text{ then }S > 4ar\,\text{ except for one value of }r.$
The sum of an infinite G.P. is:.$\displaystyle \frac{a}{1-r},\,\text{ for }|r| < 1.$
If $\displaystyle S$ is equal to $\displaystyle 4ar$, we have: .$\displaystyle \frac{a}{1-r} \:=\:4ar$
. . $\displaystyle \frac{1}{1-r} \:=\:4r \quad\Rightarrow\quad 1 \:=\:4r-4r^2 \quad\Rightarrow\quad 4r^2 - 4r + 1 \:=\:0$
. . $\displaystyle (2r-1)^2\:=\:0 \quad\Rightarrow\quad 2r-1 \:=\:0 \quad\Rightarrow\quad \boxed{r \:=\:\tfrac{1}{2}}$
I started by assuming the conclusion to see where it lead.
$\displaystyle \frac{a}{1-r}>4ar$
$\displaystyle \Rightarrow a>4ar-4ar^2$
$\displaystyle \Rightarrow a(4r^2-4r+1)>0$
$\displaystyle \Rightarrow a(2r-1)^2>0$
but this works in reverse too so starting with $\displaystyle a(2r-1)^2 > 0$ if $\displaystyle r \ne \frac{1}{2}$ you get the result you want.