Question

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, δt=16+t for 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.

Answer 1

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:

$K = 625 \times (1+ 0.06 K)$

$K = 625 +37.5 K$

At year end 2; George  amount of 400 will grow at a force interest, then the value of  $K = 400 \times e^{\int\limits^2_k {\dfrac{1}{6+t}} \, dx }$

$K =400 \times \dfrac{6+K}{6+2}$

$K =400 \times \dfrac{6+K}{8}$

$K =50 \times ({6+K})$

$K =300+50K$

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