Question

A woman worked for 30 years before retiring. At the end of the first year of employment she deposited 5000 into an account for her retirement. At the end of each subsequent year of employment, she deposited 3% more than the prior year. The woman made a total of 30 deposits. She will withdraw 50,000 at the beginning of the first year of retirement and will make annual withdrawals at the beginning of each subsequent year for a total of 30 withdrawals. Each of these subsequent withdrawals will be 3% more than the prior year. The final withdrawal depletes the account. The account earns a constant annual effective interest rate. Calculate the account balance after the final deposit and before the first withdrawal.

Answer 1

$797,837

Step-by-step explanation:

the first withdrawal is $50,000

the second is $51,500

and so on...

the formula that used to solve the interest rate earned by the annuity is:

$50,000 x {[(1 + i)³⁰ - (1 + 3%)³⁰] / [(1 + i)³⁰ x (i - 3%)]} x (1 + i) = $5,000 x {[(1 + i)³⁰ - (1 + 3%)³⁰] / (i - 3%)}

we start to simplify the equation by cancelling {[(1 + i)³⁰ - (1 + 3%)³⁰] / (i - 3%)}

[$50,000 x (1 + i)] / (1 + i)³⁰ = $5,000

now we cancel $5,000 on each side:

[10 x (1 + i)] / (1 + i)³⁰ = 1

now lets take away (1 + i):

10 / (1 + i)²⁹ = 1

things get a little bit more simple now:

10 = (1 + i)²⁹

²⁹√10 = ²⁹√(1 + i)²⁹

1.082636734 = 1 + i

i = 1.082636734 - 1 = 0.082636734 = 8.2636734%

now we replace i in any equation:

= $50,000 x {[(1 + 0.082636734)³⁰ - 1.03³⁰] / [(1 + 0.082636734)³⁰ x (0.082636734 - 0.03)]} x (1 + 0.082636734)

= $50,000 x {[10.82636738 - 2.427262471] / [10.82636738 x 0.052636734]} x (1 + 0.082636734)

= $50,000 x {8.399104909 / 0.56986462} x (1.082636734)

= $50,000 x 14.73877236 x 1.082636734

= $797,837