Hi in the following answer in my Text book i am struggling to understand why p$1$ and p$2$ and so on because when I did the question I did it period per period and then deduct payment and then recalculated the simple interest based on the new value. Just want to know why my answer is wrong.

I have attached the answer from Text book and my answer

Anthony borrowed R$3000$ on $4$ February at a simple interest rate of

$15\%$ per annum. He paid R$1000$ on $21$ April of the same year, R$600$ on

$12$ May of the same year, and R$700$ on $11$ June of the same year. What

was the balance due on $15$ August of the same year if payments were

subject to the same simple interest as the original debt?

The following diagram shows the debt, the payments, the dates at which

such payments are made, and the days to settlement:

Value $ofR3000=3000(1+0,15\frac{192}{365})$

$=3236,71$

The value of the debt at the end $(15$ August$)$ is R$3236,71.$

The values of the three payments on $15$ August are, respectively

$P_1=1000(1+0,15\frac{116}{365})$

$=1047,67.$

$P_2=600(1+0,15\frac{95}{365})$

$=623,42.$

$P_3=700(1+0,15\frac{65}{365})$

$=718,70.$

Thus, the total value of the payment is

R$2389,79(1047,67+623,42+718,70).$

The outstanding debt on $15$ August is thus R$846,92(3236,71-2389,79).r$~~$18+18\mathrm{.}18\%$

$(4)4$ Feb $21$ April

$S=P(1+rt)$

$\{$:$=3000(1+10,15)(\frac{76}{365}))$

$=R3093\mathrm{.}70-R1000$

$=R2093,70$

$=35-111$

$=111-35$

$=76$ days.

$21$ April$-12$may

$=111-132$

$=21$ days

$S=P(1+rt)$

$=2093,70(1+(0,15)(\frac{21}{3-6}))$

$=211,77-600$

$=1511,77\mathrm{.}12$ may $-$ "June

$=132-162$

$=30$ days.

$11$ June $-15$ Aug

$=162-227$

$=65$ days

$S=P(1+rt),\frac{NO}{\text{D}\text{A}\text{T}\text{E}}$

$=1511,77(1+(0,15)(\frac{30}{365}))$

$=1530,41-700$

$=8830,41$

$S=P(1+rt)$

$=830,41(1+(0,15)(\frac{65}{365}))$

$=852,59$