If $a_{1}, a_{2}, a_{3}$ are three positive consecutive terms of a GP with common ratio $k$ . Then all values of $k$ for which the inequality $a_{3} > 4 a_{2} - 3 a_{1}$ is satisfied

(1, 3)

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(- \infty, 1) \cup (3, \infty)

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(- \infty, \infty)

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(0, \infty)

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Solution

The correct option is B $(- \infty, 1) \cup (3, \infty)$
$a_{3} = a k^{2}, a_{2} = a k, a_{1} = a$
$a_{3} > 4 a_{2} - 3 a_{1}$
$k^{2} > 4 k - 3$
$k^{2} - 4 k + 3 > 0$
$(k - 1) (k - 3) > 0$
either $k < 1 o r k > 3$
Option B

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If a1,a2,a3 are three positive consecutive terms of a GP with common ratio k. Then all values of k for which the inequality a3>4a2−3a1 is satisfied

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If $a_{1}, a_{2}, a_{3}$ are three positive consecutive terms of a GP with common ratio $k$ . Then all values of $k$ for which the inequality $a_{3} > 4 a_{2} - 3 a_{1}$ is satisfied

The correct option is B $(- \infty, 1) \cup (3, \infty)$
$a_{3} = a k^{2}, a_{2} = a k, a_{1} = a$
$a_{3} > 4 a_{2} - 3 a_{1}$
$k^{2} > 4 k - 3$
$k^{2} - 4 k + 3 > 0$
$(k - 1) (k - 3) > 0$
either $k < 1 o r k > 3$
Option B