$\mathrm{\Delta x}\text{ at }\mathrm{O}=\mathrm{d}$ [path difference is maximum at O]

So, if $\mathrm{d}=\frac{7\mathrm{\lambda }}{2}$, O will be a minima. If $\mathrm{d}=\mathrm{\lambda },$ O will be a maxima.

If $\mathrm{d}=\frac{5\mathrm{\lambda }}{2}$, O will be a minima and hence intensity is minimum. 

If $\mathrm{d}=4.8 \mathrm{\lambda }$, then total 10 minima can be observed on the screen, 5 above O and 5 below O, which correspond to

           $\mathrm{\Delta x}=\pm \frac{\mathrm{\lambda }}{2},\pm \frac{3\mathrm{\lambda }}{2},\pm \frac{5\mathrm{\lambda }}{2},\pm \frac{7\mathrm{\lambda }}{2},\pm \frac{9\mathrm{\lambda }}{2}$