# If $\;a_{1},a_{2},a_{3}\;(a_{1}\;\geq\;0)$ are in GP with common ratio r . the value of r for which inequality $\;a_{3}\;\geq\;4a_{2}-3a_{1}$ holds is given by ,

$(a)\;r=6\qquad(b)\;r=1\qquad(c)\;r=2\qquad(d)\;r=3$

Answer : (a) r=6
Explanation : $a_{3}\;>\;4a_{2}-3a_{1}$
$a_{1}\;r^2\;>\;4a_{1}r-3a_{1}$
$r^2\;>\;4r-3$
$(r-3)(r-1)\;>\;0$
$r\;<\;1\quad\;or\quad\;r\;>\;3$
r=6 satisfies above condition .
answered Jan 22, 2014 by