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

• March 23rd 2008, 04:05 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.
• March 23rd 2008, 04:59 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.?
• March 23rd 2008, 05:23 PM
KavX
p.a means per annum
so in other words it means yearly
• March 23rd 2008, 05:40 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?
• March 23rd 2008, 06:00 PM
KavX
see thats were im confused aswell
• March 23rd 2008, 06:31 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.

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

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

$T_{36}$
$=20r^{35}+20r^{34}+20r^{33}+\dots +20r^{2}+20r+1$
$=20(r^{35}+r^{34}+r^{33}+\dots +r^{2}+r+1)$
$=20\left[\frac{r^{36}-1}{r-1}\right]$
$=813.16$
• March 23rd 2008, 09:18 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.
• March 29th 2008, 04:06 AM
KavX
thanx seng
• March 29th 2008, 05:54 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: . $A \;=\;D\,\frac{(1+i)^n-1}{i}$

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

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

Therefore: . $A \;=\;20\,\frac{\left(1 + \frac{0.082}{12}\right)^{36}}{\frac{0.082}{12}} \;=\;813.1609107 \;\approx\;\boxed{\813.16}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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.

• March 30th 2008, 05:57 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.
• March 31st 2008, 06:00 PM
Soroban
In case anyone is interested, here's the derivation of that formula.

We deposit $D$ dollars each period.
The account pays $i$ percent interest every period.
We make these deposits for $n$ periods.

Reminder: If there are $X$ dollars in the account,
. . . . . . . . after one period, the account has grown to $X(1+i)$ dollars.

We deposit $D$ dollars.

At the end of period 1, our account has $D(1 +i)$ dollars.
We deposit another $D$ dollars,
. . and the balance is: $D(1+i) + D\:=\:D[(1+i) + 1]$ dollars.

At the end of period 2, our account has $D[(1 +i) + 1](1+i)$ dollars.
We deposit another $D$ dollars,
. . and the balance is: $D[(1+i) + 1](1+i) + D \:=\:D\left[(1+1)^2 + (1+i) + 1\right]$ dollars.

At the end of period 3, our account has $D\left[(1 +i)^2 + (1+i) + 1\right](1+i)$ dollars.
We deposit another $D$ dollars,
. . and the balance is: $D\left[(1+i)^3 +(1+1)^2 + (1+i) + 1\right]$ dollars.

. . . and so on . . .

At the end of period $n$, our final balance is:

. . $D\underbrace{\left[(1+i)^n + (1+i)^{n-1} + (1+i)^{n-2} + \hdots + (1+i) + 1\right]}_{\text{geometric series}} \;=\;A$

The geometric series has the sum: . $\frac{(1+i)^n-1}{(1+i)-1} \;=\;\frac{(1+i)^n-1}{i}$

Therefore: . $\boxed{A \;=\;D\,\frac{(1+i)^n-1}{i}} \quad\hdots\quad \text{ta-}DAA!$

• April 1st 2008, 01:06 PM
Soroban
In case anyone is interested, here's the derivation of the Amortization Formula.

We take out a loan of $P$ dollars.
We will pay it back in $n$ periodic installments of $M$ dollars each.
They are charging us $i$ percent interest per period.

Reminder: If we owe $X$ dollars,
. . . . . . . . after one period, we will owe $X(1+i)$ dollars.

At time 0, we owe them $P$ dollars

At the end of period 1, we owe them: $P(1 +i)$ dollars.
We pay $M$ dollars, and our balance is: $P(1+i) - M$ dollars.

At the end of period 2, we owe them: $[P(1 +i) + M]\,(1+i)$ dollars.
We pay another $M$ dollars, and our balance is:
. . $[P[(1+i) - M]\,(1+i) - M\;=\;P(1+i)^2 - M(1+i) - M$ dollars.

At the end of period 3, we owe them: $[P(1+i)^2 - M(1+i) - M]\,(1+i)$ dollars.
We pay another $M$ dollars, and our balance is:
. . $[P(1+i)^2 - M(1+i) - M]\,(1+i) - M \;=\;P(1+i)^3 - M(1+i)^2 - M(1+i) - M$ dollars.

. . . and so on . . .

At the end of period $n$, the final balance is zero. .(We're paid up!)

. . $P(1+i)^n - M(1+i)^{n-1} - M(1+i)^{n-2} - \hdots - M(1+i) - M\;=\;0$

$\text{We have: }\;P(1 + i)^n \;=\;M\underbrace{\left[(1+i)^{n-1} + (1+i)^{n-2} + \hdots + (1+i) + 1\right]}_{\text{geometric series}}\;\;{\color{blue}[1]}$

. . The geometric series has the sum: . $\frac{(1+i)^n-1}{(1+i)-1} \;=\;\frac{(1+i)^n-1}{i}$

Substitute into [1]: . $P(1+i)^n \;=\;M\cdot\frac{(1+i)^n-1}{i}$

Therefore: . $\boxed{M \;=\;P\cdot\frac{i(1+i)^n}{(1+i)^n-1}}\quad\hdots\quad \text{There!}$