1. ## Geometric series

A mortgage is taken out for 80,000$. It is to be paid by annual installments of 5000$ with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4% is then charged on any outstanding debt. Find the total time taken to pay off the mortgage.

The answer given is 25 years.

2. Hello, Seirin!

A mortgage is taken out for $80,000. It is to be paid by annual installments of$5000 with the first payment
being made at the end of the first year that the mortgage was taken out.
Interest of 4% is then charged on any outstanding debt.
Find the total time taken to pay off the mortgage.

The answer given is 25 years. . . . . I don't agree
This is an Amortization . . .

The formula is: . $A \;=\;P\frac{i(1+i)^n}{(1+i)^n-1}$

. . where: . $\begin{Bmatrix}A &=& \text{periodic payment} \\
P &=& \text{principal amount of loan} \\ i &=& \text{interest rate per period} \\ n &=& \text{number of periods} \end{Bmatrix}$

We have: . $A = 5000,\;P = 80,000,\;i = 0.04$

Then: . $5000 \;=\;\frac{80,000(0.04)(1.04^n)}{(1.04)^n-1} \quad\Rightarrow\quad 5000 \;=\;\frac{3200(1.04)^n}{(1.04)^n-1} \quad\Rightarrow\quad 25 \;=\;\frac{16(1.04)^n}{(1.04)^n - 1}$

. . $25(1.04)^n - 25 \;=\;16(1.04)^n \quad\Rightarrow\quad 9(1.04)^n \;=\;25 \quad\Rightarrow\quad (1.04)^n \;=\;\frac{25}{9}$

Take logs: . $\ln(1.04)^n \;=\;\ln\left(\frac{25}{9}\right) \quad\Rightarrow\quad n\!\cdot\!\ln(1.04) \:=\:\ln\left(\frac{25}{9}\right)$

Therefore: . $n \;=\;\frac{\ln(\frac{25}{9})}{\ln(1.04)} \;=\;26.04876774 \;\approx\;\boxed{26\text{ years}}$

3. Thanks Soroban

But is there a way to solve this using geometric series formula? Coz this is a question from the geometric series chapter.