# Series and Sequence/Application (sum of a geometric series) - Supperannuation

• Mar 23rd 2008, 03:05 PM
KavX
Series and Sequence/Application (sum of a geometric series) - Supperannuation
hi

on the last question from my maths text book and i've gotten all previous questions right but this last one.

25. Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a., paid monthly? (p.a. means per annum so yearly) answer is :$813.16

btw this chapter is based on using the sum of a geometric series formula.
so ive been using this formula

Sn = a[(r^n) - 1]
aaaaaaar - 1

a: being the first amount in the geometric series
r: being the ratio or interest rate
n: being the no. of years or the time period.

unless theres another formula or way of figuring this out because i keep stuffing up something in my working out.

thx all help appreciated.
• Mar 23rd 2008, 03:59 PM
SengNee
Quote:

Originally Posted by KavX
hi

on the last question from my maths text book and i've gotten all previous questions right but this last one.

25. Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a., paid monthly? answer is :$813.16

btw this chapter is based on using the sum of a geometric series formula.
so ive been using this formula

Sn = a[(r^n) - 1]
aaaaaaar - 1

a: being the first amount in the geometric series
r: being the ratio or interest rate
n: being the no. of years or the time period.

unless theres another formula or way of figuring this out because i keep stuffing up something in my working out.

thx all help appreciated.

What is p.a.?
• Mar 23rd 2008, 04:23 PM
KavX
p.a means per annum
so in other words it means yearly
• Mar 23rd 2008, 04:40 PM
SengNee
Quote:

Originally Posted by KavX
p.a means per annum
so in other words it means yearly

Then, "the investment earns 8.2% p.a., paid monthly" means?
Each month invesment is 8.2%/12?
• Mar 23rd 2008, 05:00 PM
KavX
see thats were im confused aswell
• Mar 23rd 2008, 05:31 PM
SengNee
Quote:

Originally Posted by KavX
hi

on the last question from my maths text book and i've gotten all previous questions right but this last one.

25. Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a., paid monthly? (p.a. means per annum so yearly) answer is :$813.16

btw this chapter is based on using the sum of a geometric series formula.
so ive been using this formula

Sn = a[(r^n) - 1]
aaaaaaar - 1

a: being the first amount in the geometric series
r: being the ratio or interest rate
n: being the no. of years or the time period.

unless theres another formula or way of figuring this out because i keep stuffing up something in my working out.

thx all help appreciated.

$\displaystyle 20, 20r+20, r(20r+20)+20,\dots$
$\displaystyle 20, 20r+20, 20r^{2}+20r+20,\dots, 20r^{35}+20r^{34}+20r^{33}+\dots +20r^{2}+20r+1$

$\displaystyle r=1+\frac{0.082}{12}=1.0068333$

$\displaystyle T_{36}$
$\displaystyle =20r^{35}+20r^{34}+20r^{33}+\dots +20r^{2}+20r+1$
$\displaystyle =20(r^{35}+r^{34}+r^{33}+\dots +r^{2}+r+1)$
$\displaystyle =20\left[\frac{r^{36}-1}{r-1}\right]$
$\displaystyle =813.16$
• Mar 23rd 2008, 08:18 PM
mathceleb
kavx,

Here is a lesson you can learn:

Annuity Immediate Accumulated Value

8.2/12 for interest rate, 20 for payment, 36 for number of periods. This is the annuity immediate accumulated value formula.
• Mar 29th 2008, 03:06 AM
KavX
thanx seng
• Mar 29th 2008, 04:54 PM
Soroban
Hello, KavX!

Quote:

25. Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a., paid monthly? Answer is:$813.16

This is an Annuity, which has its own formula: . $\displaystyle A \;=\;D\,\frac{(1+i)^n-1}{i}$

. . where: .$\displaystyle \begin{Bmatrix}A &=& \text{Final amount} \\ D &=& \text{periodic deposite} \\i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}$

We have: .$\displaystyle D = 20,\;i = \frac{0.082}{12},\;n = 36$

Therefore: .$\displaystyle A \;=\;20\,\frac{\left(1 + \frac{0.082}{12}\right)^{36}}{\frac{0.082}{12}} \;=\;813.1609107 \;\approx\;\boxed{\$813.16}$~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ The derivation of the Annuity Formula is quite involved, . . but it does involve a geometric series. The formula should have been given to you. Usually, we are not expected to derive it ourselves. • Mar 30th 2008, 04:57 PM KavX never got that formula , havnt even seen it before, anyways thanx for the heads up, i'll keep that formula in mind now. • Mar 31st 2008, 05:00 PM Soroban In case anyone is interested, here's the derivation of that formula. We deposit$\displaystyle D$dollars each period. The account pays$\displaystyle i$percent interest every period. We make these deposits for$\displaystyle n$periods. Reminder: If there are$\displaystyle X$dollars in the account, . . . . . . . . after one period, the account has grown to$\displaystyle X(1+i)$dollars. We deposit$\displaystyle D$dollars. At the end of period 1, our account has$\displaystyle D(1 +i)$dollars. We deposit another$\displaystyle D$dollars, . . and the balance is:$\displaystyle D(1+i) + D\:=\:D[(1+i) + 1]$dollars. At the end of period 2, our account has$\displaystyle D[(1 +i) + 1](1+i)$dollars. We deposit another$\displaystyle D$dollars, . . and the balance is:$\displaystyle D[(1+i) + 1](1+i) + D \:=\:D\left[(1+1)^2 + (1+i) + 1\right]$dollars. At the end of period 3, our account has$\displaystyle D\left[(1 +i)^2 + (1+i) + 1\right](1+i)$dollars. We deposit another$\displaystyle D$dollars, . . and the balance is:$\displaystyle D\left[(1+i)^3 +(1+1)^2 + (1+i) + 1\right]$dollars. . . . and so on . . . At the end of period$\displaystyle n$, our final balance is: . .$\displaystyle D\underbrace{\left[(1+i)^n + (1+i)^{n-1} + (1+i)^{n-2} + \hdots + (1+i) + 1\right]}_{\text{geometric series}} \;=\;A$The geometric series has the sum: .$\displaystyle \frac{(1+i)^n-1}{(1+i)-1} \;=\;\frac{(1+i)^n-1}{i}$Therefore: .$\displaystyle \boxed{A \;=\;D\,\frac{(1+i)^n-1}{i}} \quad\hdots\quad \text{ta-}DAA!$• Apr 1st 2008, 12:06 PM Soroban In case anyone is interested, here's the derivation of the Amortization Formula. We take out a loan of$\displaystyle P$dollars. We will pay it back in$\displaystyle n$periodic installments of$\displaystyle M$dollars each. They are charging us$\displaystyle i$percent interest per period. Reminder: If we owe$\displaystyle X$dollars, . . . . . . . . after one period, we will owe$\displaystyle X(1+i)$dollars. At time 0, we owe them$\displaystyle P$dollars At the end of period 1, we owe them:$\displaystyle P(1 +i)$dollars. We pay$\displaystyle M$dollars, and our balance is:$\displaystyle P(1+i) - M$dollars. At the end of period 2, we owe them:$\displaystyle [P(1 +i) + M]\,(1+i)$dollars. We pay another$\displaystyle M$dollars, and our balance is: . .$\displaystyle [P[(1+i) - M]\,(1+i) - M\;=\;P(1+i)^2 - M(1+i) - M$dollars. At the end of period 3, we owe them:$\displaystyle [P(1+i)^2 - M(1+i) - M]\,(1+i)$dollars. We pay another$\displaystyle M$dollars, and our balance is: . .$\displaystyle [P(1+i)^2 - M(1+i) - M]\,(1+i) - M \;=\;P(1+i)^3 - M(1+i)^2 - M(1+i) - M$dollars. . . . and so on . . . At the end of period$\displaystyle n$, the final balance is zero. .(We're paid up!) . .$\displaystyle P(1+i)^n - M(1+i)^{n-1} - M(1+i)^{n-2} - \hdots - M(1+i) - M\;=\;0\displaystyle \text{We have: }\;P(1 + i)^n \;=\;M\underbrace{\left[(1+i)^{n-1} + (1+i)^{n-2} + \hdots + (1+i) + 1\right]}_{\text{geometric series}}\;\;{\color{blue}[1]} $. . The geometric series has the sum: .$\displaystyle \frac{(1+i)^n-1}{(1+i)-1} \;=\;\frac{(1+i)^n-1}{i}$Substitute into [1]: .$\displaystyle P(1+i)^n \;=\;M\cdot\frac{(1+i)^n-1}{i} $Therefore: .$\displaystyle \boxed{M \;=\;P\cdot\frac{i(1+i)^n}{(1+i)^n-1}}\quad\hdots\quad \text{There!} \$