/sci/ - Science & Math » Thread #12865914

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Anonymous Mon 22 Mar 03:04:01 2021 No.12865914 View Reply Original Report

Quoted By: >>12866383 >>12866485 >>12867427

What rate of interest was earned for $2,000 annual contribution to accumulate to be $26,362 after 10 years?

4,7,5,8,6 ?

Anonymous

Anonymous Mon 22 Mar 2021 05:46:10 No.12866383 Report

Quoted By:

>>12865914
If the interest doesn't hit until 10 years it'll be about 4 years

Anonymous

Anonymous Mon 22 Mar 2021 06:14:39 No.12866462 Report

Quoted By: >>12867298

2000*(1+x/100)^10 = 26,362

You do the rest, retard.

Anonymous

Anonymous Mon 22 Mar 2021 06:23:58 No.12866485 Report

Quoted By:

>>12865914
Compounded?

NEARAutomata

NEARAutomata Mon 22 Mar 2021 11:47:22 No.12867210 Report

Quoted By:

this is peak comfort.

Anonymous

Anonymous Mon 22 Mar 2021 12:31:04 No.12867298 Report

Quoted By:

>>12866462
Wrong, because its $2000 every year.

Anonymous

Anonymous Mon 22 Mar 2021 13:17:24 No.12867427 Report

Quoted By: >>12867666

>>12865914

For continuously compounded; using the forumula:
$A=Pe^{rt}$
you can derive a sum that gives the accumulated value after N years
$A_N = P * \sum_{n=1}^{N}e^{rn}$
After some massaging you get
$ln(A_N / P) = r * \sum_{n=1}^{N} n = r * \frac{N(N+1)}{2}$
So
$r = \frac{2ln(A_N/P)}{N(N+1)}$
Plugging in $A_N = 26362$ , $P = 2000$ , $N = 10$ you find $r \approx 0.0469$ or 4.69%

If on the other hand you're compounding quarterly the formula is much simpler:
$r = (\frac{A_N}{P})^{ \frac{1}{4N} } - 1$
which gives
$r \approx 0.0666$ or 6.66%

Anonymous

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Anonymous Mon 22 Mar 2021 13:46:37 No.12867547 Report

Quoted By:

i dont know anything about this topic btw

Anonymous

Anonymous Mon 22 Mar 2021 14:19:40 No.12867666 Report

Quoted By:

>>12867427
>.69%

Nice.

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