Let the given series is $$P=\sum^{n}_{t=1}[\frac{1}{(1+x)^t}]$$
I need to solve this for $x$ when $n=12$ and $P=10.83$. When I substitute the values and open the summation then I get this form: $$10.83=\frac{1}{1+x}+\frac{1}{(1+x)^2}+\frac{1}{(1+x)^3}+\frac{1}{(1+x)^4}+...+\frac{1}{(1+x)^{12}}$$ Here I need the help that how to get x.
On
Without numerical methods, you can have very simple approximations of the equation $$P=\sum^{n}_{t=1}\frac{1}{(1+x)^t}=\frac{1-(x+1)^{-n}}{x}$$ when the value of $P$ is just smaller than $t$.
You know that the solution is very small.
Use the binomial expansion or Taylor series to write $$\frac{1-(x+1)^{-n}}{x}=-\sum_{k=0}^\infty \binom{-n}{k+1}\,x^k$$ and use power series reversion.
This would give $$x=y+\frac{n+2}{3} y^2\left(1+\frac{5 n+7}{12} y +\frac{(17 n^2+44 n+29}{90} y^2+O(y^3) \right)$$
where $$y=\frac{2 (n-P)}{n (n+1)}$$
For your specific case where $n=12$, $P=\frac {1083}{100}$, $y=\frac {3}{200}$, the above truncated series would give, as an approximation $$x=\frac{25833321}{1600000000}=\color{red}{0.01614}58$$
To simplify the algebra, I would recommend computing $$(1+x)P - P = \left[1-\frac{1}{(1+x)^{12}}\right].$$ $$\implies P = \frac{1}{x}\left[1-\frac{1}{(1+x)^{12}}\right].$$
Now, if we make the substitution $x = y-1$, then $$P = \frac{1}{y-1}\left[1-\frac{1}{y^{12}}\right]$$ This is much easier to solve for $y$, and the result is the polynomial $$Py^{13}-(1+P)y^{12}+1 = 0.$$
Newton's method would work great for this based on the graph.
Finally, solve for $x$.