MY NOTES ON: Arithmetic
Progression (AP) and Geometric Progression (GP)
Progression, Sequence or Series – What’s the Difference?
·
Numbers or numerical terms in progression or sequence are given as:
T1, T2, T3,
…, Tn.
·
Numbers or numerical terms in series are given as:
o
T1 + T2 + T3 +
…+ Tn ; or,
o
Tm + T(m+1) +…+ Tn
·
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
Sn;
(or, any letter or symbol defined in the question); or
o
·
Notations for sum of mth to nth terms in series:
= Sn
– 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:
T1,
T2, T3, …, Tn, is such that
Tn –
T(n-1) = T3 - T2 = T2 - T1
= d (common difference);
then, the
progression is an AP.
o
Notations:
o
1st term = a
o
Common difference = d (pl. see above)
o
nth term = Tn
o
Deriving Basic Formula for nth Term
of AP, Tn:
1st term 2nd term 3rd
term 4th term nth term
T1 T2 T3 T4 Tn
a a + d a + 2d a + 3d a + (n – 1)d
Thus, the basic formula for nth
Term of AP is:
Tn = a + (n - 1)d ………….Basic Equation (BE)
Four other formulae/methods may be used to find Tn :
Tn
= dn + (a - d)….................... (BE Expanded)
Tn
= dn + c (where c = a - d)…. (BE in Linear Format)
Tn
= Tn-1 + d………………….. (Preceding Term + d)
Tn
= Sn – S(n-1)……………….. (Difference of Sums)
o
Examples of APs:
o
1, 5, 9,…, Tn: a = 1, d = 4, Tn = 1 +
4(n -1) = 4n – 3
o
2, -2, -6,…, Tn: a = 2, d = -4, Tn = 2 –
4(n – 1) = 6 – 4n
o
2, 5, 8,…, Tn: a = 2, d = 3, Tn = 2 + 3(n – 1) = 3n -1
o
Sum of Arithmetic Series (Finite Portion of AP):
o
Sum of first n terms of AP:
§
= Sn = (n/2)(a + Tn); 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 mth term to nth
term of AP:
·
= Sn
– S(m-1) (apply preceding formula)
o Mean
of Arithmetic Series, n:
n = Sn/n = (1/n)(n/2)(a + Tn) = (a + Tn)/2
or, n = Sn/n
= (1/n)(n/2)[2a + (n – 1)d] = (1/2)[2a + (n – 1)d]
---------------------------------------------------------------------------------------------------------
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:
T1,
T2, T3, …, Tn, is such that
Tn /
T(n-1) = T3 / T2 = T2 / T1
= r (common ratio);
then, the
progression is an GP.
o
Notations:
o
1st term = a
o
Common ratio = r (pl. see above)
o
nth term = Tn
o
Deriving Basic Formula for nth Term
of GP, Tn:
1st term 2nd term 3rd
term 4th term nth term
T1 T2 T3 T4 Tn
a ar ar2 ar3 ar(n – 1)
Thus, the basic formula for nth
Term of GP is:
Tn = ar(n – 1) ;
or, Tn = (a/r)rn
(Many students find it useful to use the 2nd
eqn to find n when Tn 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, …, Tn (= ar(n – 1) = (a/r)rn, where r = 3, Tnà +∞, all Ts same sign as a);
o
1, -3, 9, -27, …, Tn (= ar(n –
1) = (a/r)rn, where r =
-3, T alternates in sign);
o
3, 3, 3, 3, …, Tn (= ar(n – 1)
= (a/r)rn, where r = 1, same
value for all Ts);
o
3/10, 3/100, 3/1000, …, Tn [= ar(n
– 1) = (a/r)rn, where r
= 1/10 or 0.1, Tn à 0 (i.e. Tn
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, …, Tn (= ar(n – 1)
= (a/r)rn, where r = -1,
constant modulus value for all Ts 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, …, Tn (= ar(n – 1) = (a/r)rn, where r = 3, Tnà +∞, all Ts same sign as a)
o
r is –ve: then terms will alternate in sign;
(see GP: 1, -3, 9, -27, …, Tn (= ar(n – 1) = (a/r)rn,
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, …, Tn (= ar(n – 1) = (a/r)rn, where r = 3, Tnà +∞, all Ts 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, …, Tn (= ar(n – 1) = (a/r)rn,
where r = 1, same value for all Ts)
o
-1 < r < 1; or, ׀r׀ < 1 (and r ≠ 0): then Tn decays exponentially
towards zero i.e. as n à +∞, Tn
à 0: All such GPs
are known as convergent GPs where
the sum of each series yields a constant value
= S∞ = a/(1 + r);
(see GP: 3/10, 3/100, 3/1000, …, Tn [= ar(n – 1) =
(a/r)rn, where r = 1/10 or
0.1, Tn à 0)
o
r = -1: then, this is an alternating GP with a
constant modulus value of a.
(see
GP: -3, 3, -3, 3, …, Tn (= ar(n – 1) = (a/r)rn,
where r = -1, constant modulus value for
all Ts 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:
§
= Sn = a[(rn - 1)/(r -1)]; or,
= a[(1 – rn)/(1 – r)]
Steps in Deriving
Sum of First n Terms for a GP:
1.
Select 2 identical geometric series Sn
2.
Multiply the 2nd series with r
3.
Now, subtract as follows:
a.
rSn – Sn; or
b.
Sn - rSn
4.
The subtraction will lead you to these results:
a.
rSn – Sn = arn – a
Sn(r -1) = a(rn – 1)
Sn = a[(rn - 1)/(r -1)]; (ideal for r > 1); or
b.
Sn - rSn = a - arn
Sn(1 –r) = a(1 – rn)
Sn = a[(1 – rn)/(1 – r)];
(ideal for 0 < r < 1)
o
Sum of mth term to nth
term of GP:
·
= Sn
– 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:
= Sn = a[(1 – rn)/(1
– r)]
When n à ∞, rn
à
0, (1 – rn) à 1
Sn
à
a(1)/(1 – r)
S∞ = a/(1 + r) =
Therefore,
o Geometric
Mean of 3 Consecutive GP Terms: a, b and c:
Geometric Mean = b
and, b2 = ac; or b =
Q: Prove that b2 = ac.
------------------------------------------------------------------------------------------------------------
Try These 2 Basic
Questions on AP and GP:
Q1. A finite
portion of an arithmetic series is given as: Tm + T(m+1),
…, Tn. There are 22 terms in this finite portion and the difference
in value of Tn and Tm is 252.
a)
Find the common difference, d, of this number series. (12)
b)
If Tn is 260, find the sum of this finite
portion. (2948)
-------------------------------------------------------------------------------------------------------
Q2.
In a geometric series, the 7th term is 54 and, the 10th
term is 2. Find:
a)
the common ratio, r, of this series. (1/3)
b)
i) the 6th term (162)
ii) the 1st term (39,366)
c)
the sum to infinity of this series. (59,049)
-----------------------------------------------------------------------------------------------------
----------------------------END-----------------------------
No comments:
Post a Comment