If $a_{1}$ , $a_{2}$ , $a_{3}$ ( $a_{1} > 0)$ are three successive terms of a G.P. with common ratio r, then find the value of r for which $a_{3} > 4 a_{2} - 3 a_{1}$ .

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Solution

$a_{3} > 4 a_{2} - 3 a_{1}$

$\Rightarrow a_{1} r^{2} > 4 a_{1} r - 3 a_{1}$

$\Rightarrow r^{2} > 4 r - 3$

$\Rightarrow r^{2} - 4 r + 3 > 0$

$\Rightarrow (r - 1) (r - 3) > 0$

$\Rightarrow r < 1 o r r > 3$

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If $a_{1}$ , $a_{2}$ , $a_{3}$ ( $a_{1} > 0)$ are three successive terms of a G.P. with common ratio r, the value of r for which $a_{3} > 4 a_{2} - 3 a_{1}$ is

If $a_{1}$ , $a_{2}$ , $a_{3}$ ( $a_{1} > 0)$ are three successive terms of a G.P. with common ratio r, the value of r for which $a_{3} > 4 a_{2} - 3 a_{1}$ holds is given by