Q53 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

If the roots of the equation $x^3-13 x^2+\mathrm{K} x-27=0$ are in geometric progression then $\mathrm{K}=$

  • A
    $-30$
  • B
    $30$
  • Correct answer
    C
    $39$
  • D
    $-39$

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 $r=\frac{1}{3}, \mathrm{a}=3 \Rightarrow 27+\frac{9}{3}+\frac{27}{3}=K \Rightarrow K=39$.

Hence, the correct answer is option: 'C'.

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