Starting from @python_enthusiast's answer, the equation to be solved for $i$ is
$$\text{PV}=\frac{\text{FV}}{(1+i)^n}+\frac{\text{PMT}}{i}\Bigg[1-\frac 1{(1+i)^n} \Bigg]$$ which does not show any explicit solution but which does not make any problem using a numerical method.
However, we can make good approximations. For conveniency, I shall use $a=\text{PV}$, $b=\text{FV}$ and $c=\text{PMT}$ and I shall take advantage of the fact that $i \ll 1$.
Using Taylor expansion for the rhs and series reversion, we have
$$i=t+\frac {(3 b+2 c)+3 n (b+c)+c n^2 } { 3(2 b+ c)+3 c n} t^2+O(t^3)$$
where $t=2\frac{(b-a)+c n}{ (2 b+c)n+c n^2 }$
Let me try for $a=2000$, $b=1000$, $c=100$ and $n=15$. This would give
$$i=\frac{823}{39366}=0.0209064$$ while the "exact" solution is $i=0.0213729$.
If this is not sufficiently good, add the next term which is
$$\frac {(n+1)\left(\left(24 b^2+36 b c+14 c^2\right)+n \left(48 b^2+78 b c+31 c^2\right)+n^2 \left(30
b c+22 c^2\right)+5 c^2 n^3 \right)}{36[(2 b+c)n+c n^2]^2}\,t^3$$ This would give
$$i=\frac{1221967}{57395628}=0.0212902$$
Edit
Another formulation
$$i=\frac{-6 ((a-b) (2 b+c))+6 c (-a+3 b+c)n+6 c^2 n^2 } {2 (a-b) (3 b+2 c)+ \left(6 b (a+b)+6 a c-c^2\right)n+2 c (a+2 b)n^2+c^2
n^3 }$$ gives, for the worked example
$$i=\frac{27}{1270}=0.0212598$$