Voltage through resistor connected to positive terminal of op amp
since no current flows from the negative to the positive terminal, how
  does this affect the voltage of the two inputs of the op amp?


In reality, all op-amps do have a small current flowing into or out from their input terminals.  This parameter is listed in their datasheet and is called input bias current (\$I_{B}\$) .

If the 10 ohm resistor wasn't there, and V1 = 0, your output voltage wouldn't be exactly zero, and one of the main reasons for this is the effect of \$I_{B}\$.

Assuming the rest of the op amp is ideal (infinite open loop gain, zero input voltage offset, etc), we can calculate this effect:

Lets call the 500 ohm resistor \$R_{F}\$, the 50 ohm resistor \$R_{IN-}\$ and the 10 ohm resistor \$R_{IN+}\$.

\$I_{B}\$ would have to come through \$R_{F}\$, so \$V_{o}=I_{B}R_{F}\$.  The larger your feedback resistor, the bigger the deviation from the expected output voltage.

But if you add \$R_{IN-}\$, it will drop a voltage of \$V_{IN+} = -I_{B}R_{IN-}\$.

The op amp will adjust Vo so that \$V_{IN-} = V_{IN+}\$, so now \$R_{IN-}\$ can also contribute with some current.  After solving this you'll find that if you choose \$R_{IN+} = R_{IN-}||R_{F}\$, it will properly compensate the effects of the input bias currents, and your output Vo will be as expected.

As a last remark, I'd like to mention that with this new knowledge you may want to always add input bias current compensation, but this is not always a good thing, because of the effects of thermal noise of resistors.  A classic example is when amplifying a small ac signal.  It is better to live with a DC offset (that you can later remove or compensate for) than to introduce the noise of the additional resistor right at the input of the op-amp, which will be amplified by the gain of the configuration.