Finding value of a loan with increasing payments

Question

Here is the question:
A loan at a rate $i^{(12)} = 12\%$ (nominal monthly rate) is repaid with $120$ monthly repayments starting one month after the loan. The amount of the first payment is $\$ 600$ and each subsequent payment is $5$ times larger than the previous payment. Find the original amount of the loan.

Here is my attempt.
Let $L$ be the original loan amount and $v = \frac{1}{\left( 1 + \frac{0.12}{12}\right)^{12}}= \frac{1}{1.01^{12}}$ and $X_t$ is the payment at each time $t$. Therefore, $X_t = 5^{t-1} 600 = 5^{t}.120$

Then we have the relationship
$$L = X_1 v + X_2 v^2 + \ldots X_{120} v^{120}$$ $$L = 120 \sum_{t=1}^{120} (5v)^t = 120 \left( \frac{5v(1- (5v)^{12})}{1-5v}\right)$$
which evaluates to be $120 \times 5.8 \times 10^{77}$ ... which doesn't seem right at all.

Answer 1

$FV_1 = 600(1.12)^{119}$

$FV_2 = 600.5.(1.12)^{118}$

$FV_3 = 600.5^{2}.(1.12)^{117}$

...

$FV_{120} = 600.5^{119}.(1.12)^0$

$FV = 600\left[5^0.(1.12)^{119}+5^1.(1.12)^{118}\cdots + 5^{119}.(1.12)^{0}\right]$

$FV = 600\left[\frac{5^{120}-1.12^{120}}{5-1.12}\right]$ $FV = 600\left[\frac{5^{120}-1.12^{120}}{3.88}\right]$

$PV = \frac{600}{3.88.(1.12)^{120}}\left[5^{120}-1.12^{120}\right]$

$PV = 154.64 \left[(\frac{5}{1.12})^{120} - 1\right]$

$PV = 154.64\left[4.464^{120}-1\right]$

$PV \approx 154.64\times4.464^{120}$

PV IS THE ORIGINAL LOAN AMOUNT.

Edit:

If the question says 5% larger than the previous payment, follow the same pricinple and arrive at an answer

$FV = 600\left[\frac{1.12^{120}-1.05^{120}}{.07}\right]$ $PV = 8571.43\left[1-0.9375^{120}\right]$

$PV = 8571.43*0.999566935 = 8567.72$

Answer 2

Let be $G=5$ the growth factor, $P=600$ the first payment, $i^{(12)}=12\%$, $n=120$ and $v=\frac{1}{1+i^{(12)}}$. We have that the loan $L$ is $$ \begin{align} L&=Pv+PGv^2+PG^2v^3+\cdots+PG^{n-1}v^n\\ &=\frac{P}{G}\left[Gv+(Gv)^2+(Gv)^3+\cdots+(Gv)^n \right]\\ &=\frac{P}{G}\times\sum_{k=1}^n(Gv)^k \\ &=\frac{P}{G}\times Gv\frac{1-(Gv)^n}{1-Gv} \end{align} $$

If you mean $5\%$ as growth rate, i.e. $g=5\%$, you can put in the previous formula $G=1+g$ and find $$ L=P\times \frac{1-\left(\frac{1+g}{1+i^{(12)}}\right)^n}{i^{(12)}-g} $$