can anyone help me with a formula for this?
What interest rate is required to earn $250 interest on a $999 investment over 3 years compounding daily
Thank You
$\displaystyle 999=250\times i^x$ where $\displaystyle i$ is interest (plus 1) and $\displaystyle x$ is the number of time periods.
So how many days are in 3 years, let's assume there's no leap-year. The answer is 1095 days
So: $\displaystyle 999=250\times i^{1095}$
Thus: $\displaystyle \frac{999}{250}=i^{1095}$
Then: $\displaystyle \sqrt[1095]{\frac{999}{250}}=i\approx 1.00126591$
So interest is: $\displaystyle \approx .126591\%$
Hello, xxpink__saltxx!
What interest rate is required to earn $250 interest on a $999 investment
over 3 years compounding daily?
The compound interest formula is: .$\displaystyle A \;= \;P(1 + i)^n$
where: $\displaystyle A$ = final amount, $\displaystyle P$ = principal, $\displaystyle i$ = periodic interest rate, $\displaystyle n$ = number of periods.
We are given: .$\displaystyle A = 999,\;P = 250$
Since the interest is compounded daily, the interest rate is: $\displaystyle \frac{I}{365}$
. . and the number of periods is: $\displaystyle 3 \times 365 \,=\,1095$
So we have: .$\displaystyle 999 \;=\;250\left(1 + \frac{I}{365}\right)^{1095}$
. . and we must solve for $\displaystyle I$, the annual interest rate.
We have: .$\displaystyle \left(1 + \frac{I}{365}\right)^{1095} \:=\:\frac{999}{250} \:=\:3.996$
Raise both sides to the $\displaystyle \frac{1}{1095}$ power:
. . $\displaystyle \left[\left(1 + \frac{I}{365}\right)^{1095}\right]^{\frac{1}{1095}} \;=\;(3.996)^{\frac{1}{1095}} \quad\Rightarrow\quad 1 + \frac{I}{365}\:=\:1.001265909$
. . Then: .$\displaystyle \frac{I}{365}\:=\:0.001265909\quad\Rightarrow\quad I \:=\;0.462056836$
Therefore, the annual interest rate is about $\displaystyle 46.2\%.$
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Check
$\displaystyle A \;= \;\$250\left(1 + \frac{0.462}{365}\right)^{1095} \:=\:\$998.8298956 \:\approx\:\$999$ . . . Yes!