Current through a resistor and diode in series
But it is known that current through the diode will not increase that much(a tiny increase will happen). 


Once the diode is turned on, a change in voltage across it will cause a large change in current, not a small one.


  Isn't this a contradiction that elements connected in series shares the same current ?


No. The load line graph shows exactly what will happen. If you increase the source voltage, the green line will shift to the right. Then you will find a new intersection between the IV curve for the source+resistor and the diode. 

It will not cause a large increase in the diode current because the resistor limits the current. This is because \$V_d = V_s - V_r\$. The diode voltage is equal to the source voltage minus the resistor voltage. So as the source voltage rises, either the diode or the resistor can take up that change. And because the diode has very low differential resistance, it's the resistor that will take up most of the added voltage. 

Edit

Another way you know this is not a contradiction is that the fact the resistor and diode currents are equal was a fundamental part of drawing the load line graph.

The two curves are

$$i=I_s \exp \left(\frac{nv}{V_T}-1\right)$$

and

$$i=\frac{V_s-v}{R}$$

The \$v\$ on the x-axis is the voltage across the diode. The \$i\$ on the y-axis is the current flowing in common through all three elements in the circuit (source, resistor, and diode).