If the roots of the equation $x^3-13 x^2+\mathrm{K} x-27=0$ are in geometric progression then $\mathrm{K}=$
If the roots of the equation $x^3-13 x^2+\mathrm{K} x-27=0$ are in geometric progression then $\mathrm{K}=$
$-30$
$30$
$39$
$-39$
Solution
Given the equation, $x^3-13 x^2+K x-27=0$
Let roots are $\frac{a}{r}, a, a r$
Now, $\frac{a}{r} \cdot a \cdot a r=\frac{-(-27)}{1} \Rightarrow a^3=27 \Rightarrow a=3$
and, $\frac{3}{r}+3+3 r=\frac{-(-13)}{1} \Rightarrow \frac{3}{r}+3 r=10 \Rightarrow r=3, \frac{1}{3}$
Since, $\frac{a}{r} \cdot a+a \cdot a r+\frac{a}{r} \cdot a r=K$
If $r=3, a=3 \Rightarrow \frac{9}{3}+27+\frac{27}{3}=K \Rightarrow K=39$
If $r=\frac{1}{3}, \mathrm{a}=3 \Rightarrow 27+\frac{9}{3}+\frac{27}{3}=K \Rightarrow K=39$.