What is the relation between Op-Amp input current and input impedance?
Input resistance (1.5T\$\Omega\$ typical) is the change in input current for change in input voltage 

\$R_{in}\$ = \$\frac {\Delta V_{in}}{\Delta I_{in}}\$ 

It is not clear whether this figure is intended to apply to differential input voltage or to common mode voltage or both. 

Input current (10pA typical) is the current flowing into or out of the input pin. 

If the input was an ideal current source, the input current would be constant so the input impedance would be infinite. 

You can model the input (at DC) as a 1.5T ohm resistor to the other input (probably, given the disposition of the input protection diodes) and two +/-10pA current sources, one connected to each input. 

Some op-amps (the ancient CA3140 is not one of them) have a rather high input resistance when the two inputs are close to each other in voltage but nonlinear networks across the inputs that turns into k-ohms if you apply more than a diode drop differentially. Not a problem in normal op-amp applications, but problematic if you're using it in applications where it might saturate (precision comparator, some precision rectifier circuits etc.).