# Thread: Word Problem Reated to Compound Interest & Geometric Sequence

1. ## Word Problem Reated to Compound Interest & Geometric Sequence

The question that I am struggling with is as follows:

On the day of her birth, 1st January 1998, Mary's grandparents invested $x in a savings account. They continued to deposit$x on the first day of each month thereafter. The account paid a fixed rate of 0.4% interest per month. The interest was calculated on the last day of each month and added to the account. Let $A_n be the amount in Mary's account on the last day of the nth month, immediately after the interest had been added. As soon as Mary was 18 she decided to invest$15000 of thismoney in an account of the same type earning 0.4% interest per month. She withdraws $1000 every year on her birthday to buy herself a present. Determine how long it will take until there is no money in the account. There are 5 questions (Question A to E) related to thequestion described in the first paragraph above. I was able to answer and understand all the questions apart from the last one, which is described in the second paragraph above. The mark scheme says: r = 1.004^12 (= 1.049...) 15000r^n - [1000(r^n-1)]/(r-1) = 0 n = 27.8... So n = 28 My understanding of this working above is that "r"is the common ratio of the geometric sequence for 1 year, "15000r^n"gives us the amount of interest plus principal you can gain over "n" years, and "[1000(r^n-1)]/(r-1)" gives us the total amount that you withdraw over the same period. Correct? However, I do not understand why the total amount forwithdrawal is "[1000(r^n-1)]/(r-1)" because the question says "She withdraws$1000 every year on her birthday...".So I do not understand why it is not simply"1000n" because it is a fixed amount of $1000 annually. Obviously "[1000(r^n-1)]/(r-1)" is the sum of geometric sequence with the first term of 1000 and common ratio of r, which is 1.004^12 (= 1.049...), however, no part of the question seems to even suggest that the amount of withdrawal has to increase like this each year. I would much appreciate if someone can help me to understand this working on this mark scheme. 2. Originally Posted by Masaru As soon as Mary was 18 she decided to invest$15000 of thismoney in an account of the same type earning 0.4% interest per month. She withdraws $1000 every year on her birthday to buy herself a present. Determine how long it will take until there is no money in the account Hard to tell what your question is...is it simply the above? In other words:$1000 is withdrawn annually from an account with initial value $15000. The interest rate is .4% paid monthly. After how many years will the account be depleted? Is that what you're asking? 3. Originally Posted by Masaru However, I do not understand why the total amount forwithdrawal is "[1000(r^n-1)]/(r-1)" because the question says "She withdraws$1000 every year on her birthday...".So I do not understand why it is not simply"1000n" because it is a fixed amount of $1000 annually. Obviously "[1000(r^n-1)]/(r-1)" is the sum of geometric sequence with the first term of 1000 and common ratio of r, which is 1.004^12 (= 1.049...), however, no part of the question seems to even suggest that the amount of withdrawal has to increase like this each year. The amount of withdrawal does NOT increase. The "account" will be exactly like this: Code: YEAR PAYMENT INTEREST BALANCE 0 15000.00 1 -1000.00 736.05 14736.05 : 15000 * .04907 = 736.05 2 -1000.00 723.10 14459.15 3 -1000.00 709.51 14168.66 ... 25 -1000.00 166.65 2562.87 26 -1000.00 125.76 1688.63 27 -1000.00 82.86 771.49 28 -809.35 37.86 .00 A = 15000 P = 1000 i = .04907 n = ? Formula is: P = A*i / [1 - 1/(1 + i)^n] Simplifies to: (1 + i)^n = P / (P - A*i) So: n = LOG[P / (P - A*i)] / LOG[1 + i] = 27.8056... 4. ## Thank you for your help, Denis Thank you for your help, Denis. I have found your explanation hard to follow and I do not understand especially the formula that you put: P = A*i / [1 - 1/(1 + i)^n] However, someone else has given me a clue and pointed out the fact that when you withdraw the 1000, you not only remove the money, you forfeit the interest you would have earned had you left it on deposit. So if r is the annual interest rate to figure out the amount in the account, the balance is: Year 1: 15000r^1 - 1000 Year 2: (15000r^1 - 1000)r - 1000 = 15000r^2 - (1000r + 1000) Year 3: [15000r^2 - (1000r + 1000)]r - 1000 = 15000r^3 - (1000r^2 + 1000r + 1000) Year 4: [15000r^3 - (1000r^2 + 1000r + 1000)]r - 1000 = 15000r^4 - (1000r^3 + 1000r^2 +1000r + 1000) As you can see from above, there is a pattern in the sequence so you can generalise it as follows: Year n: 15000 r^n - (sum of geometric series of the first term of 1000 and the common ration of r) Originally Posted by Denis The amount of withdrawal does NOT increase. The "account" will be exactly like this: Code: YEAR PAYMENT INTEREST BALANCE 0 15000.00 1 -1000.00 736.05 14736.05 : 15000 * .04907 = 736.05 2 -1000.00 723.10 14459.15 3 -1000.00 709.51 14168.66 ... 25 -1000.00 166.65 2562.87 26 -1000.00 125.76 1688.63 27 -1000.00 82.86 771.49 28 -809.35 37.86 .00 A = 15000 P = 1000 i = .04907 n = ? Formula is: P = A*i / [1 - 1/(1 + i)^n] Simplifies to: (1 + i)^n = P / (P - A*i) So: n = LOG[P / (P - A*i)] / LOG[1 + i] = 27.8056... 5. Originally Posted by Masaru As you can see from above, there is a pattern in the sequence so you can generalise it as follows: Year n: 15000 r^n - (sum of geometric series of the first term of 1000 and the common ration of r) Yes, BUT your problem as presented is to calculate n; that is what the equation I gave you does. What you're showing here is the value of the account at end of year n, after the$1000 withdrawal.
If we want value 25 years later:
A = 15000
P = 1000
r = .04907...
n = 25
F = ? (future value)

Future value of 15000:
A * (1 + r)^n = 49682.69 [1]

Future value of the 1000 payments:
P * [(1 + r)^n - 1] / r = 47119.82 [2]

F = [1] - [2] = 2562.87 : as at end of 25th year of my printout.

Remember that your problem is SAME as a loan of $15000 repayable at$1000 annually: YOU loaned the bank $15000, the bank is repaying YOU at$1000 annually !
That's the formula I used to calculate n.

6. Originally Posted by Masaru
As soon as Mary was 18 she decided to invest $15000 of thismoney in an account of the same type earning 0.4% interest per month. She withdraws$1000 every year on her birthday to buy herself a present. Determine how long it will take until there is no money in the account.

The mark scheme says:
r = 1.004^12 (= 1.049...)
15000r^n - [1000(r^n-1)]/(r-1) = 0
n = 27.8...
So n = 28
The others who have answered don't seem to have dealt with your actual question (though they are probably right about this not being the easiest way). What you say here,
However, someone else has given me a clue and pointed out the fact that when you withdraw the 1000, you not only remove the money, you forfeit the interest you would have earned had you left it on deposit.

So if r is the annual interest rate to figure out the amount in the account, the balance is:

Year 1: 15000r^1 - 1000
Year 2: (15000r^1 - 1000)r - 1000 = 15000r^2 - (1000r + 1000)
Year 3: [15000r^2 - (1000r + 1000)]r - 1000 = 15000r^3 - (1000r^2 + 1000r + 1000)
Year 4: [15000r^3 - (1000r^2 + 1000r + 1000)]r - 1000 = 15000r^4 - (1000r^3 + 1000r^2 +1000r + 1000)

As you can see from above, there is a pattern in the sequence so you can generalise it as follows:

Year n: 15000 r^n - (sum of geometric series of the first term of 1000 and the common ration of r)
is a correct explanation of what the mark scheme said.

Are you satisfied with this as the answer to your question?

Originally Posted by Masaru
However, I do not understand why the total amount forwithdrawal is "[1000(r^n-1)]/(r-1)" because the question says "She withdraws $1000 every year on her birthday...".So I do not understand why it is not simply"1000n" because it is a fixed amount of$1000 annually. Obviously "[1000(r^n-1)]/(r-1)" is the sum of geometric sequence with the first term of 1000 and common ratio of r, which is 1.004^12 (= 1.049...), however, no part of the question seems to even suggest that the amount of withdrawal has to increase like this each year.

I would much appreciate if someone can help me to understand this working on this mark scheme.
The amount you ask about, "[1000(r^n-1)]/(r-1)", is not the total amount withdrawn; it is the amount that must be deducted from the balance to account for both the withdrawals and the resulting lost interest.

What I don't understand is how they solved their horrible equation for n with almost no explanation. Are you okay with that step?

7. Originally Posted by Masaru
However, I do not understand why the total amount for withdrawal is
"[1000(r^n-1)]/(r-1)" because the question says "She withdraws
\$1000 every year on her birthday"
Ahhh...I see...so THAT was your question...
Dr. P gave you the answer...
When you "see" it, you'll kick yourself

8. ## Thank you for your concern, Dr. Peterson.

Originally Posted by Dr.Peterson

What I don't understand is how they solved their horrible equation for n with almost no explanation. Are you okay with that step?
Yes, I should be OK because we are allowed to use graphing calculator for this question.

9. Originally Posted by Masaru
Yes, I should be OK because we are allowed to use graphing calculator for this question.
Excellent. That's what I suspected.

Of course, the other way you were shown (using a formula you evidently have not learned) makes it possible to solve without that. But most real-life problems can't be solved by direct algebraic means anyway.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•