Let there be n resistors R1 .......... R2, with Rmax=maxR1 ............. Rn and Rmin=minR1 ............. Rn. Show that when they are connected in parallel, the resultant resistance Rp<Rmin and when they are connected in series, the resultant resistance Rs>Rmax. Interpret the result physically.

Hint: Use the concept of series and parallel combination of resistors.
Step 1:
When all resistances are connected in parallel, the resultant resistance R is given by;
1Rp=1R1+..+1Rn
On multiplying both sides by Rmin, we have;
RminRp=RminR1+RminR2+..+RminRn
Here, in RHS, there exist one term RminRmin = 1 and other terms are positive, so we have;
RminRp=RminR1+RminR2+.+RminRn>1
This shows that the resultant resistance Rp<Rmin.
Thus, in parallel combination, the equivalent resistance of resistors is less than the minimum resistance available in a combination of resistors.
Step 2:
Now, in series combination, the equivalent resistance is given by:
Rs=R1++Rn
Here, in RHS, there exist one term having resistance Rmax.
So, we have;
Rs=R1++Rmax++Rn=Rmax+(R1++)Rnor RsRmax
Thus, in a series combination, the equivalent resistance of resistors is greater than the maximum resistance available in a combination of resistors.
Physical interpretation:
In fig. (b), Rmin provides an equivalent route as in Fig. (a) for current. But in addition, there are (n-1) routes by the remaining (n-1) resistors. The current in fig. (b) is greater than the current in Fig. (a). Effective resistance in Fig. (b)< Rmin. The second circuit evidently affords a smaller resistance.
In Fig. (d), Rmax provides an equivalent route as in Fig. (c) for current. Current in Fig. (d)< current in Fig. (c). Effective resistance in Fig (d)>Rmax. The second circuit evidently affords a greater resistance.