1. ## Geometric Sequence Question...

Question:
Fatima invests $\100$ per month for a complete year, with interest added every month at the rate of $\frac{1}{2}\%$ per month at the end of the month. How much would she have had to invest at the beginning of the year to have the same total amount after the complete year?

Attempt:
$\frac{100(1-1.005^{12})}{(1-1.005)} = \1233.56$
Don't know what to do next!

2. Originally Posted by looi76
Question:
Fatima invests $\100$ per month for a complete year, with interest added every month at the rate of $\frac{1}{2}\%$ per month at the end of the month. How much would she have had to invest at the beginning of the year to have the same total amount after the complete year?

Attempt:
$\frac{100(1-1.005^{12})}{(1-1.005)} = \1233.56$
Don't know what to do next!
According to your question the interest rate is 6% p.a. She wants to get the same amount of money at the end of the year:

If $A_I$ is the initial amount then solve the equation

$\1233.56 =(1+0.06) \cdot A_I$

for $A_I$. I've got $A_I = \1163.74$

3. Originally Posted by earboth
According to your question the interest rate is 6% p.a. She wants to get the same amount of money at the end of the year:

If $A_I$ is the initial amount then solve the equation

$\1233.56 =(1+0.06) \cdot A_I$

for $A_I$. I've got $A_I = \1163.74$
from where did u get $6\%$

4. Originally Posted by looi76
Question:
Fatima invests $\100$ per month for a complete year, with interest added every month at the rate of $\frac{1}{2}\%$ per month at the end of the month. How much would she have had to invest at the beginning of the year to have the same total amount after the complete year?
[snip]
Originally Posted by looi76
from where did u get $6\%$
Given: $\frac{1}{2}\%$ per month.

Known: 12 months in the year.

Therefore: Interest rate per annum is $12 \times \frac{1}{2}\% = 6 \%$.