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, …, T_n.

·        Numbers or numerical terms in series are given as:

o   T₁ + T₂ + T₃ + …+ 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:

            T₁, T₂, T₃, …, T_n, is such that

T_n – T_(n-1) = T₃ - T₂ = T₂ - T₁ = 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

  T₁                     T₂                     T₃                     T₄                     T_n

  a                    a + d                a + 2d              a + 3d              a + (n – 1)d

Thus, the basic formula for n^th Term of AP is:               

T_n        = a + (n - 1)d ………….Basic Equation (BE)

Four other formulae/methods may be used to find T_n :

T_n = dn + (a - d)….................... (BE Expanded)

T_n = dn + c (where c = a - d)…. (BE in Linear Format)

T_n = T_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:

            T₁, T₂, T₃, …, T_n, is such that

T_n / T_(n-1) = T₃ / T₂ = T₂ / T₁ = 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

  T₁                     T₂                     T₃                     T₄                     T_n

  a                      ar                     ar²                    ar³                  ar^{(n – 1)}

Thus, the basic formula for n^th Term of GP is:               

T_n = ar^{(n – 1)} ; or, T_n = (a/r)rⁿ

(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)rⁿ, where r = 3, T_nà +∞, all Ts same sign as a);

o   1, -3, 9, -27, …, T_n (= ar^{(n – 1)} = (a/r)rⁿ, where r = -3, T alternates in sign);   

o   3, 3, 3, 3, …, T_n (= ar^{(n – 1)} = (a/r)rⁿ, where r = 1, same value for all Ts);

o   3/10, 3/100, 3/1000, …, T_n [= ar^{(n – 1)} = (a/r)rⁿ, 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)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, …, T_n (= ar^{(n – 1)}  = (a/r)rⁿ, where r = 3, T_nà +∞, all Ts 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)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, …, T_n (= ar^{(n – 1)}  = (a/r)rⁿ, where r = 3, T_nà +∞, 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, …, T_n (= 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 T_n decays exponentially towards zero i.e. as n à +∞, T_n 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, …, T_n [= ar^{(n – 1)} = (a/r)rⁿ, 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)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:

§   = S_n    = 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 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 = arⁿ – a

S_n(r -1) = a(rⁿ – 1)                                       

S_n         = a[(rⁿ - 1)/(r -1)]; (ideal for r > 1); or

b.      S_n - rS_n = a - arⁿ

S_n(1 –r) = a(1 – rⁿ)

S_n         = a[(1 – rⁿ)/(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 – rⁿ)/(1 – r)]

           

When n à ∞,   rⁿ à 0, (1 – rⁿ) 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, 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: 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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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 + …+ T_n  ; or,

o   T_m + T_(m+1) +…+ T_n

·        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 1^st ‘nomial’ (or, term) x  is in descending order of power from n à 0;

d.      The 2^nd ‘nomial’ y is in ascending order of power from 0 à n; the power of the 2^nd nomial plus 1 gives the position of the term in the series. Thus, in (x + y)ⁿ, term containing 2^nd nomial y  to the power 2 i.e. y² is the 3^rd 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, x^ay^b 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 = ⁿC_a = ⁿC_b

= 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 3^rd term (b = 2) = 15 = 6!/(4!2!)

·          The BC in 4^th term (b = 3) = 20 = 6!/(3!3!)

·          The BC in 5^th term (b = 4) = 15 = 6!/(2!4!)

·          The BC in 6^th term (b = 5) = 6 = 6!/(1!5!)

·          The BC in 7^th term (b = 6) = 1 = 6!/(0!6!)

·          In general, the BC of x^ry^(n-r) term in (x + y)ⁿ = n!/(r!(n-r)!) = ⁿC_r =

·          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) + ⁿC_nx⁰ 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 1^st nomial x = 1:

o  (1 + y)ⁿ: 1 + ⁿC₁y¹ + ⁿC₂y² +… + ⁿC_(n-1)y^(n-1) + ⁿC_nyⁿ; 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.      1^st nomial = 1,

                                                             ii.      modulus of 2^nd 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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