Arithmetic Sequence or Arithmetic Progression

An arithmetic sequence or Progression (abbreviated as A.P) is a sequence in which each term after the first is obtained by adding to the preceding term, a fixed number which is called the common difference.

In other words, quantities are said to be in Arithmetic Sequence, when they increase or decrease by a by common difference. Thus each of the following series forms an arithmetic progression.

  1. 3,7,1,15, \ldots
  2. 8,2, - 4, - 10, \ldots
  3. {a_1},{a_1} + d,{a_1} + 2d,{a_1} + 3d, \ldots

 

The common difference is found by subtracting any term of the series from that which follows it. In the first of the above examples, the common difference is 4; in the second it is  - 6; in the third it is d.

But 1,2,4,8,16, \ldots is not an A.P. Here the second term minus first term is 1, while the third term minus the second is 2, the difference so obtained does not remain the same.

The nth term of an Arithmetic Progression:

Let {a_1}be the first term and d be the constant difference. Then the second term is {a_1} + d, the third term is {a_1} + 2d. In each of these terms, the coefficient of d is 1 less than the number of term. Similarly, the 10th term is{a_1} + 9d. The nth term is the \left( {n - 1} \right)th term after the first term and is obtained after d has added \left( {n - 1} \right)times in succession. Hence, if {a_n}represents the nth term, then

{a_n} = {a_1} + \left( {n - 1} \right)d

Example:

Find the seventh term of an A.P in which the first term is 11 and the common difference is 4.

Solution:
            The seventh term may be designed as {a_7}, we use
                                     {a_n} = {a_1} + \left( {n  - 1} \right)d
            as the formula and substitute for the variables to find{a_7};
                        Here n = 7,   {a_1} = 11,   d = 4
                        \therefore    {a_n} = 11 + \left( {7 - 1} \right)4 = 11 + 6 \times 4 = 35
            Thus, the required seventh term is 35.

Example:

Find the 13thterm of the following arithmetic progression. 7,17,27, \ldots

Solution:
                        Here n = 13,   {a_1} = 7,   d = 17 - 7 = 10
                                    \therefore {a_n} = {a_1} + \left( {n  - 1} \right)d
                        Gives   {a_{13}} = 7 + \left( {13  - 1} \right)\left( {10} \right) = 7 + 120 = 127
            Thus, the required 13th term is 127.

Example:

Find the 20th term of an A.P. whose 3rd term is 7 and the 8th term is 17.

Solution:
            Using   {a_n} = {a_1} + \left( {n  - 1} \right)d, we have

                        {a_3} = {a_1} + \left( {3 - 1} \right)d             \Rightarrow 7 = {a_1} +  2d   ------- (1)
                        {a_8} = {a_1} + \left( {8 - 1} \right)d             \Rightarrow 17 = {a_1} +  7d ------- (2)

            Subtracting (1) and (2), we get
                        10 = 5d            \Rightarrow d = 2
            Putting the value d = 2 in (1), we obtain
                        7 = {a_1} + 2\left( 2 \right)           \Rightarrow 7 = {a_1} + 4   \Rightarrow {a_1} = 3
            Putting {a_1} = 3, d = 2, n = 20 in {a_n} = {a_1} + \left( {n  - 1} \right)d we get
                        {a_{20}} = 3 + \left( {20 - 1} \right)2 = 3 + 79 \times 2
                               = 3 + 38 = 41

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