Arithmetic Progression

Arithmetic progressions

A series is said to be arithmetic progressions if the difference between any two consecutive terms is constant.

i.e. If T_n is the nth term and T_n+1 are two consecutive terms of the series, then

T_n+1 – T_n = d

Important relations for Arithmetic progressions

     1. Nth term of the series, T_n = a + (n-1)d

                        Where a is the first term of the series and d is the common difference and n is the nth term

     2.   Sum of the arithmetic series for n terms is

S = n[2a +(n-1)d]/2

Where a is the first term
 n is the number of terms
d is the common difference

     3.    Sum of the A.M. for n terms when first term, last term and the number of terms are given is

S =n(a+l)/2

Where, n is the number of terms in the series

a is the first term 
l is the last term of the series

     4.    The common difference d = T_n –T_n-1

Arithmetic Mean

1. Arithmetic mean of two quantities A, B is defined as   (A+B)/2
2. Arithmetic mean of n quantities is defined as (a1 + a2 + a3 + ....+ an) /n

Important notes 

      1.    Sum of first n natural number =  n(n+1)/2

Inserting ‘n’ Arithmetic means between two numbers

Let’s say a₁, a₂, a₃, a₄,…, a_n are the required ‘n’ means to be inserted between

A and B

Then the series would look like A, a₁, a₂, a₃, a₄,…, a_n,,B

We get a series with the first as A and last term as B.

Now we just have to calculate the common difference d between any two consecutive terms of the series.

Using the two formulas for the sum of the A.P. series and equating them to get

step1 :   (A+B) /2 = n(2A +(n-1)d)/2
step2 :  (A+B) = 2A + (n-1)d

s    step3:    B-A = (n-1)d

      step4:    d = (B-A)/(n-1) ---------------(a)

For inserting n means between and B we will calculate the common difference using above formula (a).

Now, knowing the values of A, B and n we would get the required numbers of arithmetic means between any two numbers A and B.

1. The first term is A
2. The second term is A + (B-A)/(n-1)
3. The third term is    A + 2(B-A)/(n-1)
.......
    the (n-1)th term is A +(n-2)(B-A)/(n-1)
then the nth term is B.