The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K$_1$ and$K_2$ with thickness $\frac {3} {8}$d and $\frac{d}{2}$ , respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates $\\$If $K_1$ = 1.25 $K_2$, the value of $K_1$ is: