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Sequences and Series

Section: Mathematics

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Question : 23 of 25

Marks: +1, -0

If the roots of the equation

x^3-13 x^2+K x-27=0

are in geometric progression then

K=

[AP EAPCET 19 May 2024 Shift 2]

$-30$
$30$
$39$
$-39$

Solution:

We have, the roots of

x^3-13 x^2+k x-27=0

are in GP be.

Let the roots of equation be

\;\frac{a}{r}, a, a r

\;\;\frac{a}{r} \times a \times a r = 27 \Rightarrow a^3 = 27 \Rightarrow a = 3

\;\Rightarrow \;\; \;\frac{a}{r}+a+a r = 13

\;\Rightarrow \;\; \;\frac{3}{r}+3+3 r = 13

\;\Rightarrow \;\; 3 \left(r+\frac{1}{r}\right) = 10

\;\Rightarrow \;\; 3 r^2-10 r+3 = 0

\;\Rightarrow \;\; 3 r^2-9 r-r+3 = 0

\;\Rightarrow \;\; (r-3)(3 r-1) = 0

\;\Rightarrow \;\; r=3, \;\frac{1}{3}

∴ The roots of the equation are 1,3 and 9 .

\;K = 1 \times 3 + 3 \times 9 + 9 \times 1 = 3+27+9

\;=39

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