Arithmetic progression

Arithmetic progression

Let's see the sequence of numbers below:

 {1, 3, 5,7,9,...}

{2,4,6,8,10,...}

{5, 10, 15, 20,...}

There is a common special property in each of the above sequences: the difference between any two consecutive terms of the sequence is the same.

 This special type of sequence of numbers is known as Arithmetic progression (A.P.).

Definition:

A sequence of numbers in which the difference between every pair of consecutive terms are the same (constant); is called an Arithmetic progression and the difference is called as common difference.

As for example,  the sequence of numbers 

{ 2,5,8,11,14,17,... } is an arithmetic progression , having a common difference 3.

The n-th term of an A.P.

If "a" be the first term and "d" be the common difference of an arithmetic progression  having the n-th term t(n); then,

   t(n)=a+(n-1)d.

example:
    find out the 7-th term of the arithmetic progression {5, 12, 19,...}.
     Here, the first term is 5.
      The common difference is 7.
       so, the 7-th term t(7) is= 5+{(7-1)×7}
                                                = 5+42
                                                = 47.

The sum of first n terms of an A.P.

If  "a" , "d", t(n) are the first term, common difference and n-th term of an arithmetic progression respectively ; then , the sum of first n terms denoted by s(n) is given by:

  s(n)= (n/2) ×{a+t(n)}

 or, (n/2)×{2a+(n-1)d}, where , we use, t(n)= a+(n-1)d.

example:
    calculate the sum of the arithmetic progression {2, 5, 8,...,152}.
       Here, at first we are to find out the number of terms in the givrn arithmetic progression.
          Here, t(n)= 152 and a=2, d=3.
                so, 2+(n-1)×3 =152
                 or, n-1= 50
                  or, n=51.
   So, the number of terms in the given  arithmetic progression is 51.
 Now, the sum of the serirs is
   =(51/2)×(2+152)
   =51×77
   =3927.
  Therefore, the sum of the given arithmetic progression is 3927.

Properties of an A.P.

(1) If we add or subtract a constant term with each term of an arithmetic progression ; the new sequence will form a new arithmetic progression.

(2) If we multiply or divide a constant term with each term of an arithmetic progression , the new sequence of numbers will form a new arithmetic progression.

(3) If the sum of three terms of an arithmetic progression is given then we can consider the terms as, a-d, a, a+d.

(4) If the sum of four terms of an arithmetic progression is given then we can consider the terms as, a-3d, a-d, a+d, a+3d.

(5) In an arithmetic progression the sum of equidistant terms from the begining and ending sides is equals to the sum of the first and last term of the arithmetic progression.

Arithmetic mean

 If three terms (consecutive)  are in arithmetic progression; the middle term of them is called the arithmetic mean.

i.e., if a, b  be the terms of an arithmetic progression , and x be their arithmetic mean (A.M.) then, a, b,x are in arithmetic progression.

i.e, x-a= b-x

or, x=(a+b)/2.

Thus, The arithmetic mean  of two terms in an arithmetic progression  is the half of their sum.

As for example, in the arithmetic progression  {3,6,9,12,...};

6 is the arithmetic mean of 3 and 9.


Note:
   (1)  If the number of tetms of an arithmetic progression is even , then there are two middle terms. The middle terms are the (n/2)-th term and (n/2 +1)-th term of the
arithmetic progression.
    (2) If the number of terms of an arithmetic progression is odd, then there is only one middle term. The middle term is the {(n+1)/2}-th term of the arithmetic progression.


     example:
       find out the middle term or terms of the arithmetic progression {3,7,11,...,95}.
        Here, the first term is 3 , common difference is 4 and the n-th term is 95.
      So, 3+(n-1)×4 = 95
       or, n-1=23
       or, n= 24.
 Thus,  the number of terms of the given arithmetic progression is even. So, there are two middle terms. The middle terms are the 12-th and 13-th terms of the arithmetic progression.

Note:

      The sum of first n natural numbers is :

S(n)=1+2+...+n = (n/2)(n+1).
       The sum of squares of first n natural numbers is:
 S(n)=1²+2²+3²+...+n² =(n/6)(n+1)(2n+1).
        The sum of cubes of first n natural numbers is:
S(n)=1³+2³+3³+...+n³={(n/2)(n+1)}².

If you find out any incorrect information or know anything more about this , please write it in the comment section!