Suppose money grows according to simple interest accumulation function a(t) = 1 + 0.05t. How much money would you need to invest at time 3 in order to have $3200 at time 8?
I did this:
I wanted to find out how much money was invested at time 0.
$3200 = k(1 + (0.05)(8)) which implies k = $2285.71
Then I wanted to see how much money this grew to at time 3.
x = $2285.71(1 + (0.05)(3)) which implies x = $2628.57.
So invest this amount x at time 3 in order to have $3200. Is this right based on what the question is asking? I'm not really sure. Thanks for any help in advance.
Maybe, but you could be overcomplicating it.
Does the definition of a(t) suggest that it starts ONLY at t = 0? I'm guessing that it doesn't. In my mind, this function kicks in at the moment cash is deposited and the formula must start counting all over, at that point, for the new deposit.
Remember that simple interest does NOT compound. If you calculate the growth to t = 3 and then calculate the growth to t = 8, you have capitalized the interest earned up to t = 3. That's no good.
8 - 3 = 5
3200 = k*(1+0.05*5) -- Find k.