Geometric progression

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In a GP, first term is1. If $4 T_{2} + 5 T_{3}$ is minimum, then its common ratio is

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Solution

Given, a=1 and $4 T_{2} + 5 T_{3}$ is minimum. Let r be the common ratio of the GP, then
$∴ 4 T_{2} + 5 T_{3} = 4 (ar) + 5 ({ar}^{2}) [∵ a = 1]$
$\Rightarrow 4 r + 5 r^{2} = f (r) [say] . . . . (i)$
$\Rightarrow r (4 + 5 r) = f (r)$
$f (r) = 0$
$r (4 + 5 r) = 0 \Rightarrow r = 0, - 4 / 5$
We know that, $if a > 0,$ quadratic expression
${ax}^{2} + bx + c$ has least value at $x = - \frac{b}{2 a}$
From the graph it is clear that, minima occurs of point $(\frac{- 2}{5}, \frac{- 4}{5})$ .
$∴ r = \frac{- 2}{5}$
.

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Geometric progression

Remember concepts with our Masterclasses.

In a GP, first term is1. If 4T2+5T3 is minimum, then its common ratio is

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In a GP, first term is1. If $4 T_{2} + 5 T_{3}$ is minimum, then its common ratio is