In a GP, first term is1. If $4{\mathrm{T}}_2+5{\mathrm{T}}_3$ is minimum, then its common ratio is

 Given, a=1 and $4{\mathrm{T}}_2+5{\mathrm{T}}_3$  is minimum. Let r be the  common  ratio  of the  GP,  then

$\therefore 4{\mathrm{T}}_2+5{\mathrm{T}}_3=4\left(\mathrm{ar}\right)+5\left({\mathrm{ar}}^2\right)[\because \mathrm{a}=1]$

$\Rightarrow 4\mathrm{r}+5{\mathrm{r}}^2=\mathrm{f}\left(\mathrm{r}\right) \left[\mathrm{say}\right]....\left(\mathrm{i}\right)$

$\Rightarrow \mathrm{r}(4+5\mathrm{r})=\mathrm{f}\left(\mathrm{r}\right)$

     $\mathrm{f}\left(\mathrm{r}\right)=0$

$\mathrm{r}(4+5\mathrm{r})=0\Rightarrow \mathrm{r}=0,-4/5$

We know that,$\text{ if }\mathrm{a}>0,$ quadratic expression

${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$ has least value at  $\mathrm{x}=-\frac{\mathrm{b}}{2\mathrm{a}}$

From  the  graph  it is clear that, minima  occurs  of point $\left(\frac{-2}{5},\frac{-4}{5}\right)$.

$\therefore \mathrm{r}=\frac{-2}{5}$

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