Question

how many years would it take you to have 2400 if you saved 100 each month at 15%?

Answer 1

I am not sure what I am doing wrong...
Future Value is 2500, interest effective 15%, and monthly

Answer 2

payment of 100
I used the formula F=A[(1+I)^-1/I]
But not right

Answer 3

\(\huge A = P (1+\dfrac{r}{n})^{tn}\)

Answer 4

for monthly interest, n=12

A= final amount = 2400
P = initial amount =100
r = rate of interest = 15% = 0.15
t= time period in years, which you need to find

Answer 5

Thank you. I have just done it with your formula but still didn't get it right. Might be me doing the calculations wrong. I have the answer and I am comparing with my one and it's not the same

Answer 6

oh, 100 is not initial amount...you're saving 100 EACH month!
ok..
first month 100
2nd month 100 +100*0.15 =x
3rd month x+ 100*0.15 =y
4th month y+100*0.15 =z
...
and so on....
and sum of all those =2400

100 + (100+100*0.15) + (100+100*0.15+100*0.15) +... (100+n*100*0.15)= 2400

can you post the correct answer ? it could help me to verify my logic...

Answer 7

yes. I am new in this and I am going crazy with that. Sorry I made a mistake also, it's 2500 no 2400 as I wrote earlier
The answers is 22 months, the way I did it, I get 11

Answer 8

\[2500=100[(1+0.15)*-1/0.15 being * what I want \to calculate\]

Answer 9

Total saved each month
$$
\large{
1st~month: a_1=100\\
2nd~month: a_2=100 + 100*1.15=100+1.15*a_1\\
3rd~month: a_3=100+(100+ 100*1.15)*1.15=100+1.15*a_2\\
4th~month: a_4=100+(100+(100+ 100*1.15)*1.15)*1.15\\
=100+1.15*a_3\\
\cdots
}
$$
Let x=1.15 and P=100
$$
a_4=P+(P+(P+P*x)*x)*x=P(1+(1+(1+x)x)x)=P(1+(1+x+x^2)x)\\
=P(1+x+x^2+x^3)
$$
So we need to solve
$$
\large{
2400=a_n=100\sum_{i=0}^{n-1}x^{i}
}
$$
Where x=1.15

Answer 10

Correction
$$
x=1+.15/12\\
$$
Then
$$
\large{
a_{22}=100\sum_{i=0}^{21}{(1+.15/12)}^{i}\\
y = 2514.31
}
$$

Answer 11

I need to check what you have done. But I am even more confused now hehe

Answer 12

Note that you can use the geometric formula for the sum above to solve for \(a_{22}\)
$$
\large{
a_{22}=\sum_{i=0}^{21}y^{i}
}
=\cfrac{y^{22}-1}{y-1}
$$

So
$$
\large{
a_{n}=100\cfrac{1.0125^n-1}{1.0125-1}
}
$$
We want this sum to be greater than 2400. What is the n that will give us this?
$$
\large{
100\cfrac{1.0125^n-1}{1.0125-1}>2400\\
1.0125^n-1>24*.0125\\
1.0125^n>24*.0125+1\\
n\ln1.0125>\ln{(240*.0125+1)}\\
n>\cfrac{\ln{(24*.0125+1)}}{\ln1.0125}
}
$$
So we want the integer, n, that makes this true.

We get n=22

Is any part of this confusing?