George have just started the job, plan to invest a part of his paycheck( a fixed monthly investment) into his retirement fund. After the retirement, george would like to be able to draw an income of $\$25000$ per month for $20$ years. If the account pays $0.5\% $ monthly interest, by the time george retire how much money should be able available in his fund?
$\underline{Attempt}$
Obtained a difference equation $x_{n+1}=(-25000+x_{n})+(-25000+x_{n})\times \frac{0.5}{100}$ where $x_{0}$=Available money after the retirement$=A$ and we have $x_{240}=0$
By solving difference equation I got $A \approx \$3506967 $
Is this correct? Can anyone verify my answer?
Given the difference equation you've set up, you seem to have obtained the correct answer to that. I would however question whether the difference equation is set up correctly, and I think it will be partly up to interpretation of the question.
The equation you have specified $$x_{n+1} = (x_n - 25000) + (x_n -25000) \times 0.005$$ says that $n+1$ months after retirement, $\$25000$ are withdrawn, and only then the interest gets calculated and paid out based on the money left after the withdrawal.
However, as far as I know interest is usually calculated on a daily basis, even if it may be paid out on a monthly or yearly basis. Therefore it would make more sense to calculate the interest based on the money left in the bank account right before the withdrawal, so we'd get the modified difference equation $$x_{n+1} = (x_n - 25000) + x_n \times 0.005 = 1.005 x_n - 25000$$
The standard solution to this type of difference equation then gives that $$ \begin{align} x_{240} &= 0 = 1.005^{240}x_0 - 25000 \dfrac{(1.005^{240} - 1)}{1.005-1} \implies\\ x_0 &= \dfrac{25000}{0.005}(1 - 1.005^{-240}) \approx \$3489519 \end{align} $$