Series
and Applications
The
definitions of series, term, nth term and sum to n terms
should be understood.
Arithmetic
Progressions (AP)
The definition of an arithmetic
series/progression and its common difference should be understood.
The formulae for the nth
term and the sum to n terms should be known.
The nth term of an AP :
Partial sums of APs : 
Geometric
Progressions
The definitions of a geometric
series and its common ratio should be understood, and the formulae for the
nth term and sum to n terms should be known.
The nth term of a GP :
Sum of n terms of GPs :
if |r| > 1
if |r| <
1
Limiting sums of GPs (when 
)
: 
Sigma Notation
:
Applications of
arithmetic and geometric series.
Applications of arithmetic
series
Applications of arithmetic series
include problems of the type A clerk is employed at an initial salary of $10
200 per annum. After each year of service he receives an increment of $900. What
is his salary in his ninth year of service, and what will be his total earnings
for the first nine years?
Applications of geometric series
Applications of geometric series
include the following types.
(i)
Superannuation
The compound interest
formula
An = P(1 + r/100)n,
where P is the principal
(initial amount), r% the rate of interest per period, and
An the amount accumulated after n periods, should be
understood.
The formula requires a calculator.
The following superannuation problem can then be undertaken. A man invests
$1000 at the beginning of each year in a superannuation fund. Assuming interest
is paid at 8% per annum on the investment, how much will his investment amount
to, after 30 years?
Obviously the first $1000 is
invested at 8% compound interest for 30 years, the next $1000 for 29 years, and
the last $1000 for 1 year.
Thus his investment after 30 years
is (in dollars)
1000 (1.0830 +
1.0829 +
+ 1.08).
This is a geometric series of 30
terms, with first term 1080 and common ratio 1.08, so that this sum
is
1080 x (1.08 30 - 1)
=
1080 x 9
062657
----------------------
-------------------
1.08 - 1
0.08
= $122 346, to the nearest
dollar.
(ii) Time
Payments
A woman borrows $3000 at 1.5% per
month reducible interest and pays it off in equal monthly instalments. What
should her instalments be in order to pay off the loan at the end of 4
years?
Let $An be the amount owing after n
months and $M be the
repayment.
Using the compound interest
formula, after one month and paying the first instalment $M, she will owe
A1 = 3000 x
1.015 M.
Similarly,
A2 = A1 x
1.015 M
=
(3000 x
1.015
M) x
1.015
M
= 3000 x
1.0152
M x
1.015
M
= 3000 x
1.0152
M
(1.015 + 1)
A3 = A2 x
1.015 M
=
(3000 x
1.0152
M
(1.015 + 1)) x
1.015
M
= 3000 x
1.0153
M
(1.015 + 1) x
1.015 M
= 3000 x
1.0153
M
(1.0152 +
1.015 + 1)
and so on.
Generalising
A48 = 3000 x
(1.015)48 M (1.01547 +
1.01546 + . . . + 1.015 + 1).
The part in brackets (1 + 1.015 +
+ 1.01547) is a
geometric series, so
A48 = 3000 x
(1.015)48 M ( 
).
But A48 = 0
(nothing owing after 48 months). Rearranging
\
M(1.01548 - 1)
3000 x
1.01548,
------------
=
1.015 - 1
so that
M = 3000 x
1.01548 x 0.015
------------------------
(1.01548 - 1)
= 88.12
The instalment amount should be
$88.12.
Students should understand the
difference between the reducible interest rate, and the rate published by
finance companies. The published rate in this case is the equivalent simple
interest rate on
$3000 for 4 years, ie
R= 100 I
100 (88.12 x 48 - 3000)
------
=
--------------------------- =
10.248 % pa.
Pn
3000 x 4
(iii) Applications to recurring
decimals
Recurring decimals should be
expressed as rational numbers, eg
0.121212
= 12/100 + 12/10000 +
12/1000000
= 12/100 (1 + 1/102
+ 1/104
)
= 12/100 (1(1 -
(1/102)∞
)
----------------
(1 -
(1/102)
= 12/100 (1 (1 - 0))
-----------
99/100
= 12/100 x 100/99
=
12/99
= 4/33
Problem:
.
The thing in brackets is a GP with
a = 0.045 and r = 0.01.
Find its limiting sum to express
as a
fraction.