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Thursday, 14 July 2016

Arithmetic Progression / Geometric Progression

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: = SnS(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 = SnS(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:

·      = SnS(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]

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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:
            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:

·      = SnS(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)

 Sn 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, b2 = ac; or b =

Q: Prove that b2 = 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: 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)

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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)

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Binomial Series / Binomial Theorem

My Personal Notes on Binomial Series / Binomial Theorem:

1.      Meaning of Binomial Series:
·        A binomial is a mathematical expression which has 2 different terms, such as in: (a + b)0; (x + y)1; (x + 2y)2; (1 + x)-1; (1 – 2x)-1;…………. (1 + 3x2)(-1/4); (1 + x)5; (1 + x/√3)10; (1 + 3x)(1/5); (1 – 3x/4)(1/3);…
·        Terms in series are given as:
o   T1 + T2 + T3 + …+ Tn  ; or,
o   Tm + T(m+1) +…+ Tn
·        Binomial series refers to the series of terms created from the expansion of a binomial to its power (or, exponent or index).

2.      Binomial Expansion Leads to Binomial Series with Binomial Coefficient for Each Term in the Series:
Expanding these binomial leads to these series of term(s) and coefficients:
                                    Terms                                          Bi. Coefficients
·        (x + y)0:       1x0y0 = 1                                                   1
·        (x + y)1:                 x + y                                            1     1
·        (x + y)2:            x2 + 2xy + y2                                  1    2     1
·        (x + y)3:        x3 + 3ax2y + 3axy2 + y3                    1   3     3    1
·        (x + y)4:    x4 + 4x3y + 6x2y2 + 4xy3 +y4            1    4    6    4    1
·        (x + y)5: …                                                      1   5     10   10   5   1
·        (x + y)6: …                                                    1   6  15   20   15  6   1  
·        (x + y)n: xn + (n/1!)x(n-1)y1 + ((n)(n-1)/2!)x(n-2)y2 + …+ yn

As can be seen from the above:
a.       Binomial expansion to power n produces (n + 1) number of terms in the binomial series;
b.      For each and every term in the series, the sum of powers of x and y is n
c.       The 1st ‘nomial’ (or, term) x  is in descending order of power from n à 0;
d.      The 2nd ‘nomial’ y is in ascending order of power from 0 à n; the power of the 2nd nomial plus 1 gives the position of the term in the series. Thus, in (x + y)n, term containing 2nd nomial y  to the power 2 i.e. y2 is the 3rd term of the binomial series (3 = 2 +1).
e.       The binomial coefficients form what is called ‘Pascal Triangle’ (Pascal: 1623 – 1662) or ‘Yang Triangle’(Yang Hui: 1261):
                                                               i.      Each line of the Δ starts and ends with 1
                                                             ii.      Each line is symmetrical about its middle
                                                            iii.      Each line of binomial coefficients (BC) is obtainable from the binomial coefficients in the line above it – by adding the top-flanking BC (binomial coefficients).
                                                           iv.      Hence, when the power of expansion n is not too large, say, n < 8, Pascal or Yang Δ can be used to quickly find out the binomial coefficients of the terms

3.      Usefulness and Limitation of Pascal or Yang Triangle
a.       Useful to get binomial coefficients when power n is small (say, n < 8)
b.      Difficult to do so when power n is high

4.      Limitation of Pascal Triangle Overcome by Binomial Theorem
a.       Pascal discovered (in fact, Yang Hui discovered about 450 years earlier than Pascal) what is known as Binomial Theorem as follows: They discovered that the binomial coefficient (BC) of any term in a binomial series, say, xayb term in (x + y)n can be expressed as:

BC = n!/(a!b!)

Where,
·        a + b = n
·        n! is read as n-factorial
·        n! = n(n -1)(n – 2)…(3)(2)(1)
1! = 1
2! = 2 x 1
3! = 3 x 2 x 1
4! = 4 x 3 x 2 x 1
5! = 5 x 4 x 3 x 2 x 1
6! = 6 x 5 x 4 x 3 x 2 x 1
·        0! is defined as: 0! = 1

Since,               a + b = n,
therefore:          a = (n – b)       ;or,       a! = (n – b)!
or,                    b = (n – a)       ;or,       b! = (n – a)!  

Hence,      BC = n!/(a!b!)        Or,    BC = nCa = nCb
= n!/[(n-b)!b!] =
= n!/[a!(n-a)!] =
                
b.      Discovery of Binomial Theorem:
Now, look at the BC in this series:

(x + y)6 = x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + y6

You will discover that:
·          The BC in 3rd term (b = 2) = 15 = 6!/(4!2!)
·          The BC in 4th term (b = 3) = 20 = 6!/(3!3!)
·          The BC in 5th term (b = 4) = 15 = 6!/(2!4!)
·          The BC in 6th term (b = 5) = 6 = 6!/(1!5!)
·          The BC in 7th term (b = 6) = 1 = 6!/(0!6!)
·          In general, the BC of xry(n-r) term in (x + y)n = n!/(r!(n-r)!) = nCr =

·          Thus, generally, binomial expansion can be expressed as:
o  (x + y)n: nC0xny0 + nC1x(n-1)y1 + nC2x(n-2)y2 +… + nC(n-1)xy(n-1) + nCnx0 yn; or
o  (x + y)n: xn + x(n-1)y1 + x(n-2)y2 +… + xy(n-1) + yn;
o  (x + y)n: xn + n!/((n-1)!1!)x(n-1)y1 + n!/(n-2)!2!)x(n-2)y2 + …+ yn
o  (x + y)n: xn + (n/1!)x(n-1)y1 + ((n)(n-1)/2!)x(n-2)y2 + …+ yn

·          Special case where 1st nomial x = 1:
o  (1 + y)n: 1 + nC1y1 + nC2y2 +… + nC(n-1)y(n-1) + nCnyn; or
o  (1 + y)n: 1 + y1 + y2 +… + y(n-1) + yn;
o  (1 + y)n: 1 + n!/((n-1)!1!)y1 + n!/(n-2)!2!)y2 + …+ yn
o  (1 + y)n: 1 + (n/1!)y1 + ((n)(n-1)/2!)y2 + n(n-1)(n-2)/3!y3…+ yn

c.       Conditions for binomial series (1 + x)n  to be convergent:
                                                               i.      1st nomial = 1,
                                                             ii.      modulus of 2nd nomial, lxl < 1,
                                                            iii.      n is rational
    hence, these binomial series are convergent:
·        (1 + x)n : IF lxl < 1, in other words:  -1 < x < 1
·        (1 – 2x)-1: IF l-2xl < 1 or lxl < ½ i.e -1/2 < x < ½
·        (1 + 3x2)(-1/2): IF l3x2l < 1; lxl < 1/3(1/2);  -1/3(1/2) < x < 1/3(1/2)


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