A galvanometer of resistance G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is?

Let G be the resistance of the galvanometer and S be the shunt resistance.
Let I be the main current.
When the galvanometer is shunted, the current through the galvanometer is Ig = I * (S/(S+G))
The current through the shunt is Is = I * (G/(S+G))
To keep the main current unchanged, we need to add a resistance R in series with the galvanometer such that the current through the galvanometer remains the same.
The effective resistance of the galvanometer and the series resistance R is G+R.
The current through the galvanometer with the series resistance R is Ig' = I * (S/(S+G+R))
We want Ig = Ig'
I * (S/(S+G)) = I * (S/(S+G+R))
This implies that S+G = S+G+R, which is not possible.
However, if we want the current through the galvanometer to be the same before and after the addition of the series resistance, we should consider the condition where the potential difference across the galvanometer remains the same.
Let V be the potential difference across the galvanometer.
Before shunting: V = IgG
After shunting: V = Ig' (G+R)
We want V = V', so IgG = Ig'(G+R)
I(S/(S+G))G = I(S/(S+G+R))(G+R)
G(S+G+R) = (G+R)(S+G)
GS + G^2 + GR = GS + G^2 + RS + GR
GS + G^2 + GR = GS + G^2 + RS + RG
This equation is always true regardless of the value of R.
There is an error in the question or the given solution. Let's reconsider the problem from the perspective of keeping the main current unchanged. This implies that the total resistance of the circuit should remain unchanged. The parallel combination of G and S is GS/(G+S). If we add R in series with this parallel combination, the total resistance would be R + GS/(G+S). If this is equal to the original resistance of the circuit (assuming the original resistance is R0), we have R + GS/(G+S) = R0.
However, without knowing the rest of the circuit, it is impossible to determine R. The provided solution seems incorrect.