If $\alpha ,\beta$ are the roots of $x^2-3x+a=0$ and $\gamma ,\delta$ are the roots of $x^2-12x+b=0$ and $\alpha ,\beta ,\gamma ,\delta$ in that order from a geometric progression in increasing order with common ratio r > 1, then a + b =

$\alpha +\beta =3,\alpha \beta =a;\gamma +\delta =12,\gamma \delta =b$

Given $\alpha ,\beta ,\gamma ,\delta$ are in G.P

Let  $\alpha =\alpha ,\beta =\alpha r,\gamma =\alpha r^2,\delta =\alpha r^3$

$\alpha +\beta =3\Rightarrow \alpha (1+r)=\mathrm{3........}\left(1\right)$

$\gamma +\delta =12\Rightarrow \alpha r^2(1+r)=\mathrm{12......}\left(2\right)$

$\frac{\left(1\right)}{\left(2\right)}\Rightarrow r^2=4\Rightarrow r=2$

Equation $\left(1\right) \Rightarrow \alpha =1$

$\therefore a+b=\alpha \beta +\gamma \delta$

$=\alpha .\alpha r+\alpha r^2.\alpha r^3$

$=2+2^5=34$