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

If $\frac{1}{a + b}, \frac{1}{2 b}, \frac{1}{b + c}$ are three consecutive terms of an A.P., prove that a, b, c are the three consecutive terms of a G.P.

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If a,b,c be three consecutive terms of a G.P. with common ratio r, which satisfies the inequality c>4b−3a then prove that either r<1 or r>3.

ar2−4ar+3a>0 as b=ar, c=ar2or (r−1)(r−3)=+ive.∴r<1 or r>3

If $a, b, c$ be three consecutive terms of a $G . P .$ with common ratio $r$ , which satisfies the inequality $c > 4 b - 3 a$ then prove that either $r < 1$ or $r > 3$ .

$a r^{2} - 4 a r + 3 a > 0$ as $b = a r$ , $c = a r^{2}$
or $(r - 1) (r - 3) = + i v e$ .
$∴ r < 1$ or $r > 3$