Answer:
26
Step-by-step explanation:
Given that:
At time, t=0, Billy puts 625 into an account paying 6% simple interest
At the end of year 2, George puts 400 into an account paying interest at a force of interest, 1/(6+t), for all t ≥ 2.
If both accounts continue to earn interest indefinitely at the levels given above, the amounts in both accounts will be equal at the end of year n. Calculate n.
In order to calculate n;
Let K constant to be the value of time for both accounts
At time, t=0, the value of time K when Billy puts 625 into an account paying 6% simple interest is:
At year end 2; George amount of 400 will grow at a force interest, then the value of
Therefore:
If K = K
Then:
625 + 37.5 = 300 +50 K
625-300 = 50 K - 37.5 K
325 = 12.5K
K = 325/12.5
K = 26
the amounts in both accounts at the end of year n = K = 26