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It is given that $a{x}^2+bx+c=0$ has 1 real root, so the discriminant is zero, or $b^2=4ac$. Because a, b, c are in arithmetic progression, $b-a=c-b$, or $b=\frac{a+c}{2}$. We need to find the unique root, or $-\frac{b}{2a}$ (discriminant is 0). From $b^2=4ac$, we can get $-\frac{b}{2a}=-\frac{2c}{b}$. Ignoring the negatives(for now), we have $\frac{2c}{b}=\frac{2c}{\frac{a+c}{2}}=\frac{4c}{a+c}=\frac{1}{\frac{1}{\frac{4c}{a+c}}}=\frac{1}{\frac{a+c}{4c}}=\frac{1}{\frac{a}{4c}+\frac{1}{4}}$. Fortunately, finding $\frac{a}{c}$ is not very hard. Plug in $b=\frac{a+c}{2}$ to $b^2=4ac$, we have $a^2+2ac+c^2=16ac$, or $a^2-14ac+c^2=0$, and dividing by $c^2$ gives ${\left(\frac{a}{c}\right)}^2-14\left(\frac{a}{c}\right)+1=0$, so $\frac{a}{c}=\frac{14\pm \sqrt{192}}{2}=7\pm 4\sqrt{3}$. But $7-4\sqrt{3}<1$, violating the assumption that $a\ge c$. Therefore, $\frac{a}{c}=7+4\sqrt{3}$. Plugging this in, we have $\frac{1}{\frac{a}{4c}+\frac{1}{4}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$. But we need the negative of this, so the answer is $\boxed{-2+\sqrt{3}}.$