Please help with the steps

2821.51

73425.64

28124.24

4124.24

Step-by-step explanation:

Effective rate: .058/2 = .029

This question is kind of ambigous and I'll make the assumption that there is no payment at time 0

\begin{gathered}175000=x\frac{(1+.029)^{2*18}-1}{.029}\\x=2821.511\end{gathered}

175000=x

.029

(1+.029)

2∗18

−1

x=2821.511

interested earned:

175000-2821.51*36= 73425.64

2.)

Same assumption as question 1 (there is no payment at time 0)

effective rate: .063/12= .00525

400(\frac{(1+.00525)^{60}-1}{.00525})=28124.24400(

.00525

(1+.00525)

60

−1

)=28124.24

Interest earned: 28124-400*60=4124.24

9514 1404 393

Step-by-step explanation:

Two formulas come into play here. One is the future value of a series of payments. The other is the payment amount available from an ordinary annuity.

  FV = An/r((1 +r/n)^(nt) -1) . . . . future value of payments of A made n times per year for t years

  A = P(r/n)/(1 -(1 +r/n)^(-nt)) . . . . payment from principal P at invested at rate r for t years, compounded n times per year.

__

Problem 1

You want FV = $350,000, A = $400, r = 0.057, n = 12, and we want to find t. We can solve the FV equation for t to get ...

  log(FV·r/(An) +1)/log(1 +r/n) = nt . . . . . number of months

  log(350000·0.057/(400·12) +1)/log(1 +0.057/12) = nt ≈ 346.1

The retirement account can be funded by 346.1 monthly payments. That will take 28.8 years.

__

Problem 2

Again, solving for t, we get ...

  log(1 -Pr/(An))/(-n·log(1 +r/n)) = t

The parameters for this are P = 700,000, A = 5000, n = 12, r = 0.054, so the account can be expected to last for ...

  log(1 -700000·0.054/(5000·12))/(-12·log(1 +0.054/12)) ≈ 18.5 . . . years

Changing the withdrawal to $3000 per month makes the account last forever.  It earns $3150 in interest each month, so the account balance continues to increase.

The account will support withdrawals of $5000 per month for 18.5 years.

The account will support withdrawals of $3000 per month forever.