First of all, sorry for my pour English.
Consider the situation: I'd like to save an amount $P$ of money every month for $240$ months consecutively ($m_1,\ldots,m_{240}$).
The gain with money is $1$ percent per month, that is, $i=0.01$.
Starting from month $m_{241}$ I'd like to take back $C=1000$ from the total, every month until the end of my life, that is, forever.
I know that I must have collected on $m_{240}$ the total of $$C/i=1000/0.01=10^5.$$
Question: what is the value of $P$?
Solution 1: bringing back each $P$ from the last month $M_{240}$ to today $m_0$, the total should be $$P\frac{1-(1+i)^{-240}}{i}=P\frac{1-1.01^{-240}}{0.01}.$$
Saving this total and considering the gain and sending it back to $m_{240}$ we obtain $$P\left(\frac{1-1.01^{-240}}{0.01}\right) 1.01^{240}=P\left(\frac{1.01^{240}-1}{0.01}\right)=494.63P.$$
Solving for $P$ we find $$494.63P=10^5 \Leftrightarrow P= 202.171.$$
Solution 2: Starting from $m_1$ and saving $P$ every month, with the gain of $i$ we have the sequence of money $$\begin{array}{cccccc} m_0 & m_1 & m_2 & \ldots & m_{240} \\ \hline 0 & P & \begin{array}{c}(1+i)P\\ + \\ P\end{array} & & \begin{array}{c}(1+i)^{239}P\\ +\\ \vdots \\+ \\ (1+i)P\\ + \\P \end{array} \end{array}$$ so that $$m_{240}=P\sum_{k=0}^{239}(1+i)^k=989.255P.$$
Solving for $P$, we find $989.255P=10^5 \Leftrightarrow P=101.086$.
Wow, the second solution is one half of the first one!
What is correct? If it is the first one, why should I bring the money back to $m_0$?