~TLDR at the bottom~
1. The problem statement, all variables and given/known data

The Big question:
 Suppose you can put $250/month into a retirement account that pays a 6% (annualized) return on investment. When you reach 65 years of age, how much will you have in your retirement account? [“Annualized” refers to “on a yearly basis” —this is why you see things like “i/12” in formulas (or “i/n” in general). 6% annually would correspond to 6%/12 = 0.5% per month.] At this point when you retire, you move your money into a low-risk account that pays 3% (annualized) return on investment. If you plan to live for 15 more years, how much can you draw out each month to live on? Variables Account 1: Code: P=250 R=.005 N=12 T=46 Variables Account 2: Code: A1=155123.83 R=.0025 N=12 T=15 2. Relevant equations Annuity Formula(Given): Code: A=P[(1+(r/n))nt)-1] ______________ (r/n) 3. The attempt at a solution For the savings account it was easy. My Attempt: Code: A=250((1+(.005/12))^((12)(46))-1)/(.005/12) A=$155123.83

After this point it shows my attempts at solving the second part of the question.
If you know how to solve it skip this next very long and very tedious section.
If you are trying to solve it and am not sure how,
the next part may inspire you or just confuse you.
_______________________________________________________________________ ___
The second part is tricky. I must find the compound interest while removing money.
My First Attempt:
Solve for P using annuity
Variables:
Code:
A=155123.83
R=.0025
T=15
N=12
Formula:
Code:
155123.83=P[(1+(.0025/12))^((12)(15))-1]
______________
(.0025/12)
P=$12912.18 This cannot be true if A is the annuity than P would be what you add per month to get A. _______________________________________________________________________ _____ My Seccond Attempt: Sum of a Geometric Sequence Variables: Code: A1=155123.83 R=1.0025 N=180 Formula: Code: ((155123.83)(1-1.0025^180)) ____________________ (1-1.0025) I felt this would give me the end amount but without the money removed each month which is unhelpful :(. _______________________________________________________________________ __ My Third Attempt: Indicated Sum Variables: Code: X=180 I=155123.83 F=(i-x)(1-1.0025^180))/(1-1.0025) Assuming: X $\Sigma$ F I Formula: Code: 180 Ʃ (i-x)(1.0025) 155123.83 Which i see no way i would be abel to solve for x this was also a dead end :( __________________________________________________________________ My Final Attempt: Recursion Variables: Code: A1=155123.83 AN=An-1-x)(1.0025) N>=180 Formula: Code: AN=An-1-x)(1.0025) A1=155123.83 A2=(155123.83-x)(*1.0025) But as you see i cannot solve for x without knowing what A180 equals. and i cannot find what A180 equals without X. _______________________________________________________________________ ____ I'm sure there is a way to do it that i am just missing. As you can see i did try.... I do not necessarily need the answer just a push in the right direction. And if you get the answer show your work so i do not lose my mind wondering. _______________________________________________________________________ ____ TLDR: Answer the problem and show work ;) TLDR EXTENDED: $250/month into a retirement account that pays a 6% (annualized) return on investment. When I am 65 the amount in the account will be A. A=$155123.83  you put the money from the first account into another account. this account pays a 3% (annualized) return on investment. At age 65 If you plan to live for 15 more years, how much can you draw out each month to live on?  PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity  I figured the way to find it would be A=P[(1+(r/n)nt)-1] ______________ (r/n) x=Amount in the account P=Amount you remove each month. R=3% (annualized) n=12 but P would initially have to be the amount deposited. than the ammount in the account subtracted from the amount needed to live on. My brain hurts  A(180) =0. Think of what you are doing. You have a sum of money deposited on your 65th birthday. Each month you earn interest, then withdraw some fixed amount, call it m. At the end of 180 months you make the last withdrawal of m, leaving 0 in the account. Write out the account balance for the first couple of months and you will see that what you have is the original principal growing minus a monthly annuity. Decide whether you are withdrawing at the beginning or end of the month. Set this quantity equal to 0 and solve for the monthly withdrawal. ## Finding your annuity, Finding your reverse annuity like this? Code: AN=An-1-m)(1.0025) A1=155123.83 A2=(155123.83-m)(*1.0025) A180=0 i just dont know how to solve for m in a situation like this. Or is it more like Code: 0=m((1+(.0025/12))^((12)(15))-1)/(.0025/12)  I guess what I'm saying is that once you understand the general process and how to sum a geometric series, then you can do any variation of an annuity. I think that you're trying to shove numbers through a formula until it works and your code is confusing me. let A(n)=amount in account at time n, 0  (Yes its a precalc class) i understand geometric series in terms of a formula, that's the way i was taught it. i was taught geometric series is: Code: an=a1(r)n-1 An being the result of a part of a series. A1 being the first step in a series. R being the common ratio. but from what your saying the equation to find m would be. Code: 0=((1+.0025)180)155123.83-(m/.0025)((1+.0025)180-1) solve for m? attempt to get m by itself: (m/.0025)=((1.0025)^180)155123.83-((1.0025)^179) (m/.0025)=243144.44887108957942 m=607.86112217772394855 this not being the sane answer you received what did i do wrong? another attempt at what you said. 0=(1.5674)155123.83-(m/.0025)(1.5635) 0=243141.091142-(m/.0025)(1.5635) -243141.091142=(-m/-.0025)(1.5635) -155510.7714=(-m/-.0025) 388.7769=-m m=-388.7769 :/  i just am not sure i understand lol Recognitions: Homework Help  Quote by yumito i just am not sure i understand lol I get something very different from you. For the amount in the account at age 65 (starting from age 19) I get A = 738,250.83, not your value A1 = 155,123.83. I used a monthly growth factor r = 1.005 for a total of 46*12 = 552 months, giving $A = 250 \times \sum_{n=1}^{552} r^n = 738250.83 .$ RGV  Thanks Ray. I never even checked his principal at age 65, I just went from his amount. My trusty financial calculator tells me$734,577.94 at age 65. But you have to check your algebra Yumito, your solution for m is all wrong. My answer above is correct if I assumed that your principal at age 65 was correct. After you make the correction that Ray pointed out you should find m=$5,073. Recognitions: Homework Help  Quote by alan2 Thanks Ray. I never even checked his principal at age 65, I just went from his amount. My trusty financial calculator tells me$734,577.94 at age 65. But you have to check your algebra Yumito, your solution for m is all wrong. My answer above is correct if I assumed that your principal at age 65 was correct. After you make the correction that Ray pointed out you should find m=\$5,073.
Since your trusty financial calculator and my computational package (Maple) give different answers, we must be computing different things or using different data, or something. Could you please say _explicitly_ what formula you are calculating, and what data you use?

RGV
 you guys are going about it wrong. this is not actual financial information. The formula i set as (given) is the formula i am suppose to use...

Recognitions:
Homework Help
 Quote by yumito you guys are going about it wrong. this is not actual financial information. The formula i set as (given) is the formula i am suppose to use...
Sorry, but that is nonsense. If the data you gave is accurate, and if your description of the problem is correct, then your numerical results are wrong. There must be something you are not telling us, or else you are using the formulas incorrectly.

Personally, I do not like the use of "canned" formulas; I much prefer to work the problem from first principles, stating exactly what calculations are involved. I generally do not trust "financial" packages, etc., without thoroughly testing them first before use, or reading reviews and tests of them.

RGV
 I'm sure you're correct Ray, I didn't check. You used beginning of period. I used end so he didn't have a gap in his stream. I'm embarrassed that I took his earlier work for granted, I should have noticed that it made no sense. I think what's occurring is that he's given some formulas without explanation. It makes me crazy that this is how math is taught.

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