__MY NOTES ON: Arithmetic Progression (AP) and Geometric Progression (GP)__

__Progression, Sequence or Series – What’s the Difference?__
·
Numbers or numerical

**are given as:**__terms in progression or sequence__
T

_{1}, T_{2}, T_{3}, …, T_{n.}
·
Numbers or numerical

**are given as:**__terms in series__
o
T

_{1}+ T_{2}+ T_{3}+ …+ T_{n }; or,
o
T

_{m}+ T_{(m+1)}+…+ T_{n}
·
Notations for sum of terms in series:

o

**S**; (or, any letter or symbol defined in the question); or
o

**∑**(sigma)
·
Notations for sum of first n terms in series:

o

**S**_{n}; (or, any letter or symbol defined in the question); or
o

·
Notations for sum of m
=

^{th}to n^{th}terms in series:**S**_{n}–**S**_{(m-1)}

__AP: Arithmetic Progression__
o
What is an arithmetic progression (AP)?

An AP is a sequence of numerical terms with
a common difference between any two consecutive terms. Thus, if the progression:

T

_{1}, T_{2}, T_{3}, …, T_{n}, is such that
T

_{n}– T_{(n-1)}= T_{3}- T_{2}= T_{2}- T_{1}= d (common difference);
then, the
progression is an AP.

o
Notations:

o
1

^{st}term = a
o
Common difference = d (pl. see above)

o
n

^{th}term = T_{n}
o
Deriving Basic Formula for n

^{th}Term of AP, T_{n}:__1__

^{st}term__2__

^{nd}term

__3__

^{rd}term

__4__

^{th}term

__n__

^{th}term
T

_{1}T_{2}T_{3}T_{4}T_{n}
a a + d a +

**2**d a +**3**d a +**(n – 1)**d
Thus, the basic formula for n

^{th}Term of AP is:**T**………….Basic Equation (BE)

_{n}= a + (n - 1)d

Four other formulae/methods may be used to find

**T**:_{n}**T**= dn + (a - d)….................... (BE Expanded)

_{n}**T**= dn + c (where c = a - d)…. (BE in Linear Format)

_{n}**T**= T

_{n}_{n-1}+ d………………….. (Preceding Term + d)

**T**=

_{n}**S**

_{n}–

**S**

_{(n-1)}……………….. (Difference of Sums)

o
Examples of APs:

o
1, 5, 9,…, T

_{n}: a = 1, d = 4, T_{n}= 1 + 4(n -1) = 4n – 3
o
2, -2, -6,…, T

_{n}: a = 2, d = -4, T_{n}= 2 – 4(n – 1) = 6 – 4n
o
2, 5, 8,…, T

_{n}: a = 2, d = 3, T_{n}= 2 + 3(n – 1) = 3n -1
o
Sum of Arithmetic Series (Finite Portion of AP):

o
Sum of first n terms of AP:

§
=

**S**_{n}= (n/2)(a + T_{n}); or,
= (n/2)[2a + (n – 1)d]

__Deriving Sum of First n Terms Formula__:

Select 2 identical arithmetic series and
proceed as shown before…

o
Sum of m

^{th}term to n^{th}term of AP:
·
=

**S**_{n}–**S**_{(m-1) }(apply preceding formula)
o Mean
of Arithmetic Series, n:

n =

**S**_{n}/n = (1/n)(n/2)(a + T_{n}) = (a + T_{n})/2
or, n =

**S**_{n}/n = (1/n)(n/2)[2a + (n – 1)d] = (1/2)[2a + (n – 1)d]
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__GP: Geometric Progression or Sequence__
A GP is a sequence of numerical terms with a common ratio (or,
multiplier) between any two consecutive terms. Thus, if the progression:

T

_{1}, T_{2}, T_{3}, …, T_{n}, is such that
T

_{n}/ T_{(n-1)}= T_{3}/ T_{2}= T_{2}/ T_{1}= r (common ratio);
then, the
progression is an GP.

o
Notations:

o
1

^{st}term = a
o
Common ratio = r (pl. see above)

o
n

^{th}term = T_{n}
o
Deriving Basic Formula for n

^{th}Term of GP, T_{n}:__1__

^{st}term__2__

^{nd}term

__3__

^{rd}term

__4__

^{th}term

__n__

^{th}term
T

_{1}T_{2}T_{3}T_{4}T_{n}
a ar ar

^{2}ar^{3}ar^{(n – 1)}
Thus, the basic formula for n

^{th}Term of GP is:**T**; or,

_{n}= ar^{(n – 1)}**T**

_{n}= (a/r)r^{n}
(Many students find it useful to use the 2

^{nd}eqn to find n when T_{n}is given:
For example:

The geometric progression 6, -12, 24, …, 6144 consists of n
terms. Find the value of n.

When you attempt this question and you will find logarithm useful and
with your solid foundation in logarithm, you can also solve 2013 Jan P2 Q9 (e)
easily – please see end of these notes:

o
Examples of GPs:

o
1, 3, 9, 27, …, T

_{n}(= ar^{(n – 1)}= (a/r)r^{n}, where**r = 3, T**_{n}**à +∞, all T**s**same sign as**);*a*
o
1, -3, 9, -27, …, T

_{n}(= ar^{(n – 1)}= (a/r)r^{n}, where**r = -3, T alternates in sign**);
o
3, 3, 3, 3, …, T

_{n}(= ar^{(n – 1)}= (a/r)r^{n}, where**r = 1, same value for all Ts**);
o
3/10, 3/100, 3/1000, …, T
=

_{n}[= ar^{(n – 1)}= (a/r)r^{n}, where**r = 1/10 or 0.1**,**T**_{n}**à 0**(i.e. T_{n}decays or decreases exponentially towards zero) - this type of GP where -1 < r < 1, or ׀r׀ < 1, is known as a**convergent GP**where the sum of its series will yield a constant value**S**_{∞}= a/(1 + r)];
o
-3, 3, -3, 3, …, T

_{n}(= ar^{(n – 1)}= (a/r)r^{n}, where**r = -1, constant modulus value for all T**s i.e. ׀T׀ = constant value (= 3 in this GP) but T alternates in sign – this is known as alternating sequence with constant modulus value)
From the above GPs, it can be seen that the behaviour of GP depends on
the value of r (the common ratio). Always, r ≠ 0. If:

o
r is +ve: then all terms will be of the same
sign as a (the first term);

(see GP: 1, 3, 9, 27, …, T

_{n}(= ar^{(n – 1)}= (a/r)r^{n}, where**r = 3, T**_{n}**à +∞, all T**s**same sign as**)*a*
o
r is –ve: then terms will alternate in sign;

(see GP: 1, -3, 9, -27, …, T

_{n}(= ar^{(n – 1)}= (a/r)r^{n}, where**r = -3, T alternates in sign**)
o
r > 1: then Ts à + ∞, if a is +ve; Ts à
- ∞, if a is –ve;

(see GP: 1, 3, 9, 27, …, T

_{n}(= ar^{(n – 1)}= (a/r)r^{n}, where**r = 3, T**_{n}**à +∞, all T**s**same sign as**)*a*
o
r = 1: then all Ts are of the same value – a
constant value GP;

(see GP: 3, 3, 3, 3, …, T

_{n}(= ar^{(n – 1)}= (a/r)r^{n}, where**r = 1, same value for all Ts**)
o
-1 < r < 1; or, ׀r׀ < 1 (and r ≠ 0): then T
=

_{n}decays exponentially towards zero i.e. as n à**+∞, T**_{n}**à 0**: All such GPs are known as**convergent GP**s where the sum of each series yields a constant value**S**_{∞}= a/(1 + r);
(see GP: 3/10, 3/100, 3/1000, …, T

_{n}[= ar^{(n – 1)}= (a/r)r^{n}, where**r = 1/10 or 0.1**,**T**_{n}**à 0**)
o
r = -1: then, this is an alternating GP with a
constant modulus value of

*a*.
(see
GP: -3, 3, -3, 3, …, T

_{n}(= ar^{(n – 1)}= (a/r)r^{n}, where**r = -1, constant modulus value for all T**s i.e. ׀T׀ = constant value (= 3 in this GP) but T alternates in sign.)
o
Sum of Finite Portion of a Geometric Series:

o
Sum of first n terms of GP:

§
=

**S**_{n}= a[(r^{n}- 1)/(r -1)]; or,
= a[(1 – r

^{n})/(1 – r)]__Steps in Deriving Sum of First n Terms for a GP__:

1.
Select 2 identical geometric series

**S**_{n}
2.
Multiply the 2

^{nd}series with r
3.
Now, subtract as follows:

a.
rS

_{n}– S_{n}; or
b.
S

_{n}- rS_{n}
4.
The subtraction will lead you to these results:

a.
rS

_{n}– S_{n}= ar^{n}– a
S

_{n}(r -1) = a(r^{n}– 1)

__S____; (__

_{n}= a[(r^{n}- 1)/(r -1)]**ideal for r > 1**); or

b.
S

_{n}- rS_{n}= a - ar^{n}
S

_{n}(1 –r) = a(1 – r^{n})

__S____;__

_{n}= a[(1 – r^{n})/(1 – r)]__(__

**ideal for 0 < r < 1**)

o
Sum of m

^{th}term to n^{th}term of GP:
·
=

**S**_{n}–**S**_{(m-1) }(depending on r, use the above formulae)
o
Sum to Infinity of a Convergent Series, where -1
< r < 1 or ׀r׀ < 1:

Use:
=

**S**_{n}= a[(1 – r^{n})/(1 – r)]
When n à ∞, r

^{n}à 0, (1 – r^{n}) à 1**S**

_{n}à a(1)/(1 – r)

**S**

_{n}à a(1)/(1 – r)

**S**

_{∞}= a/(1 + r) =

Therefore,

**S**

_{∞}= a/(1 + r);

o Geometric
Mean of 3 Consecutive GP Terms: a, b and c:

Geometric Mean = b

and, b

^{2}= ac; or b =
Q: Prove that b

^{2}= ac.
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__Try These 2 Basic Questions on AP and GP:__
Q1. A finite
portion of an arithmetic series is given as: T

_{m}+ T_{(m+1)}, …, T_{n}. There are 22 terms in this finite portion and the difference in value of T_{n}and T_{m }is 252.
a)
Find the common difference, d, of this number series. (12)

b)
If T

_{n}is 260, find the sum of this finite portion. (2948)
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Q2.
In a geometric series, the 7

^{th}term is 54 and, the 10^{th}term is 2. Find:
a)
the common ratio, r, of this series. (1/3)

b)
i) the 6

^{th}term (162)
ii) the 1

^{st}term (39,366)
c)
the sum to infinity of this series. (59,049)

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