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

1. ## 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.

2. 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.?

3. p.a means per annum
so in other words it means yearly

4. 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?

5. see thats were im confused aswell

6. 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.
$20, 20r+20, r(20r+20)+20,\dots$
$20, 20r+20, 20r^{2}+20r+20,\dots, 20r^{35}+20r^{34}+20r^{33}+\dots +20r^{2}+20r+1$

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

$T_{36}$
$=20r^{35}+20r^{34}+20r^{33}+\dots +20r^{2}+20r+1$
$=20(r^{35}+r^{34}+r^{33}+\dots +r^{2}+r+1)$
$=20\left[\frac{r^{36}-1}{r-1}\right]$
$=813.16$

7. 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.

8. thanx seng

9. Hello, KavX!

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.

10. never got that formula , havnt even seen it before, anyways thanx for the heads up, i'll keep that formula in mind now.

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

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