Progressions, anyone?

Question

Would be grateful if you can please tell me how to work these out. Thank you very much!
1) A fund is created with an initial amount of $20000. Each year, an interest of 6% is added into the amount. If a total of $1800 is used up each year, calculate how many withdrawal times(years) of $1800 can be made.
2) In an arithmetic progression, the sum of its first odd numbers is more than the sum of its first even numbers by 3n. Find the common difference of this arithmetic progression.

Answer 1

1) Let A_n be the amount at the very beginning of year n, n=0, 1,2....., that is, after the withdrawal. Then,

A_0 = C, the initial capital
A_n = (1 +i) A_(n-1) - W, for n-1,2,3...., i = the interest rate, W = the yearly withdrawal.

It follows that
A_1 = C(1 + i) - W
A_2 = C(1 + i)^2 - W( 1 + (1 +i)
A_3 = C(1+ i)^3 - W(1 + (1 +i) +(1 + i)^2).

If we proceed by induction, we conclude that,
A_n = C(1 + i)^n - W Sum (k = 0 , n-1) (1 + i)^k

Each term of the sum is a term of the GP with constant (1 + i) and initial term 1. As we know,

Sum (k = 0 , n-1) (1 + i)^k = ((1 + 1)^n - 1)/(1 + i -1) = ((1 + i)^n -1)/i. So, our expression for A_n is

A_n = C(1 + i)^n - W((1 + i)^n -1)/i. In our case, C = 20000, W = 1800 and i = 0.6, so that

A_n = 20000(1.06))^n - 1800 * ((1.06)^n -1))(0.6)

We can withdraw 1880 a year while A_n >= 0 So, we have to find the largest integer n such that A_n >=0

Using an Excel spreadsheet, we get A_17 = 1456.6 and A_18 = -255.9, so that it's possible to withdraw 1800$ from the end of the 1st year to the end of the 17th year, that is 17 withdrawals. These formulas can be computed by means of the financial formulas of Excel.

2) Let a_1, a_2....a_n.... be the terms of the AP. The sum of the first n odd numbers is
So = a_1 + a_3 +......a-(2n -1) and the sum of the first n even numbers is
Se = a_2 + a_4......+ a_2n If r is the common diffrerence, then
a_2 = a_1 + r, a_4 = a_3 + r, ....a_2n = a_(2n-1) + r, it follows that
Se = (a_1 + r) + (a_3 + r( .....+ (a_(2n -1)) + r = So + n*r (we have n termis in this sum). So

Se = So + n*r => So - Se = -n*r. Since So - Se = 3n, it follows that 3n = -nr , so that r = -3.

Answer 2

(1) Assuming you extract the money at the end of the year when the interest is awarded:

Amount: $20,000
Year 1: $1,200 interest, $1,800 withdrawal
Amount: $19,400
Year 2: $1,164 interest, $1,800 withdrawal
Amount: $18,764
Year 3: $1,125.84 interest, $1,800 withdrawal
Amount: $18,089.84
Year 4: $1,085.39 interest, $1,800 withdrawal
Amount: $17,375.23
Year 5: $1,042.51 interest, $1,800 withdrawal
Amount: $16,617.74
Year 6: $997.06 interest, $1,800 withdrawal
Amount: $15,814.80
Year 7: $948.89 interest, $1,800 withdrawal
Amount: $14,963.69
Year 8: $897.82 interest, $1,800 withdrawal
Amount: $14,061.51
Year 9: $843.69 interest, $1,800 withdrawal
Amount: $13,105.20
Year 10: $786.31 interest, $1,800 withdrawal
Amount: $12,091.51
Year 11: $725.49 interest, $1,800 withdrawal
Amount: $11,017.00
Year 12: $661.02 interest, $1,800 withdrawal
Amount: $9,878.02
Year 13: $559.68 interest, $1,800 withdrawal
Amount: $8,637.70
Year 14: $518.26 interest, $1,800 withdrawal
Amount: $7,355.96
Year 15: $441.36 interest, $1,800 withdrawal
Amount: $5,997.32
Year 16: $359.84 interest, $1,800 withdrawal
Amount: $4,557.16
Year 17: $273.43 interest, $1,800 withdrawal
Amount: $3030.59
Year 18: $181.84 interest, $1,800 withdrawal
Amount: $1,142.43

For year 19, there's not $1,800 left to withdraw.

(2) Is a little unclear: you don't say how many numbers you're subtracting. Assuming it's "three" it's still a little unclear:

If you mean the difference between the 1st+3rd+5th term of the series and the 2nd+4th+6th, then the difference would be three times the common difference in the series. (2nd-1st is the common difference, 4th-3rd is the common difference, and 6th-5th is the common difference). Since that is 3n, and since the 1/3/5 terms are greater, the common difference is negative and 1/3 of the reported value: -n.

If instead you start numbering with the 0th term (or if they mean the 1st term is an even number and the 2nd is odd and so forth), then the difference is the 2nd+4th+6th minus the 1st+3rd+5th. Since that is 3n, and since the 2/4/6 terms are greater, the common difference is positive and 1/3 of the reported value: n.

I would go with (positive) n, but it's unclear in wording.

Answer 3

For #1, the direct way to solve is as follows:
20000 = 1800*[1 - (1/1.06)^n]/.06, then
1200/1800 = 1 - (1/1.06)^n, or
(1/1.06)^n = 1/3.
n ln(1/1.06) = ln (1/3), or
n = 18.854. The withdrawal in year 19 will be less than 1800, so 18 complete withdrawals of 1800 ca nbe made.

Answer 4

1) a=20000, r=1.06
T1= ar-1800
T2= (T1)r -1800
= (ar-1800)r - 1800
= ar^2 - 1800r -1800
= ar^2 - 1800(r+1)
T3 = (T2)r - 1800
= r[ar^2 - 1800(r+1)]-1800
= ar^3 - 1800(r^2 + r + 1)

Tn = ar^n -1800 [1+r+..r^(n-1)]
Tn = 20000(1.06)^n - 1800 [1+r+..r^(n-1)] ... (1)

For [1+r+..r^(n-1)] is another geometric series, with summation till n-1, which can be written as
Sn-1 = (1)[1.06^(n-1) - 1] / [1.06-1] ... (2)

Substitute (2) in (1)

Tn = 20000(1.06)^n - 1800{ [1.06^(n-1) - 1] / [1.06-1]}
Tn = 20000(1.06)^n - 30000 [1.06^(n-1) - 1]

Withdraw till the fund become negative, Tn < 0.
1.06^n < (1.5) [1.06^(n-1) - 1]
1.5 < (1.5) [1.06^(n-1) ] - 1.06^n
1.5 < [1.06^(n-1) ] (1.5 - 1.06)
1.5 < [1.06^(n-1) ] (0.44)
[1.06^(n-1) ] > 1.5/0.44

lg for both side,
(n-1)lg 1.06> lg 3.4091
(n-1) > 0.5326/0.0253
n - 1> 21.05
n > 22.05

The fund will be withdraw completely on year 22.

* However if based on McFate list down the net fund each year, then 19 years is the correct answer. The answer also same as 19 years if using Excel to build a formula list.

Therefore I doubt there is somewhere mistake in my calculation, pls advise if found.

Answer 5

1) 18 years... in the 19th year the amount will be only $1.456

2) the first odd numbers... how many?