Mathematical Progressions

Arithmetical Progression.

Each term differs by a constant value.  Examples:

1, 3, 5, 7, 9, 11, …., general term 1+2r

33, 22, 11, 0, -11, -22, …, general term 33-11r

a, a+d, a+2d, a+3d, a+4d, …., general term a+rd

Consider the last series whose general term is a+rd

First term is a, second is a+d,  third term is a+2d, rth term is a+(r-1)d and last term where the number of terms is n is a+(n-1)d.

Sum of arithmetical series

Let Sn be the sum of series whose general term is a+(r-1)d and has n terms.

Sn = a +a+d + a+2d + a+3d + a+4d + …. +  a+(n-2)d + a+(n-1)d    . . . .    (1)

Which is identical to

Sn = a+(n-1)d  + a+(n-2)d + …. + a+4d + a+3d + a+2d + a+d + a   . . . .    (2)

Add (1) & (2)

2Sn = 2a +(n-1)d  + 2a+(n-1)d + …. + 2a +(n-1)d  + 2a +(n-1)d  + 2a +(n-1)d

      = n(2a +(n-1)d)

 Sn = n(a+½(n-1)d)  

or 

 Sn = an +½n(n-1)d      . . . .   (3)

Example

Let a=3, d=2, n=5, substitute in (3)

S5 = 3x5 + ½x5x4x2 = 15 + 20 = 35

Check

S5 = 3 + 5 + 7 + 9 + 11 = 35


Proof by induction

Consider equation (3).

a and d are fixed and n  is variable.

For n = 1

        S1 = a

S1 = a +½x0d                = a

For n = 2

        S1 = a + a+d                = 2a+d

S1 = 2a +½x2x1d        = 2a+d

For n = 3

        S1 = a + a+d + a+2d        = 3a+3d

S1 = 3a +½x3x2d        = 3a+3d

Assume (3) it is true for all values of n.

         Sn = an + ½n(n-1)d

So        Sn+1 = an + ½n(n-1)d + a+nd

              = a(n+1) + ½n2d - ½nd + nd

              = a(n+1) + ½n2d + ½nd

              = a(n+1) + ½n(n+1)d

Rearranging

Sn+1 = a(n+1) + ½(n+1)nd         . . . . (4)

It has been shown that (3) is true for n=1,2,3 and by (4) it must be true for n=4, and therefore for all positive values of n.


Geometric Series or Geometric Progression

Each term is multiplied by a constant factor, e.g.

        a, ar, ar2, ar3, ….., arn-1

Where,

        initial value = a, constant factor = r and no. of terms = n.

        

Sn =  a + ar + ar2 + ar3 +  …..  + arn-1   ……  (6)

So

        rSn =  ar + ar2 + ar3 + ar4 …..  + arn     ……  (7)

Subtract (7) from (6)

        

Sn (1-r) = a - arn

        Sn  =  a(1 - rn)