we know that the terms in an ap are

Question

We know that the terms in an A.P. are given by

a, a + d, a + 2d, a + 3d, ........ a + (n - 2)d, a + (n -  1)d

The sum of all these terms which is denoted by "S" is given by

S

=

n/2

{2a + (n - 1)d}

This is obtained as follows. We know that

S       =       (a) + (a + d) + (a + 2d) + (a + 3d) + ..... +

                   {(a + (n - 2)d)} + {(a + (n - 1)d)}

Now we reverse the order and write it as shown below.

S       =       (a + (n -1) d) + (a + (n - 2) d) + ......... +

                   (3d + a) +  (2d + a) + (d + a) + a

On adding the respective terms we get

2S     =       {a + a + (n - 1)d} + {a + d + a + (n - 2)d}

                   + ......... + {a + (n - 2)d  + a + d} +

                   {a + (n - 1)d + a} 

That is, we have:

2S     =       {2a + (n - 1)d} + {2a + d + (n - 2)d} +

                    ............... + {2a + d + (n - 2)d} +

                   {2a + (n - 1)d}

Further simplifying we obtain

2s     =       {2a + (n - 1)d} + {2a + d + nd - 2d} +.............. +

                   {2a + d + nd - 2d} + {2a + (n - 1)d}

On simplification we obtain

2s     =       {2a + (n - 1)d} + {2a + nd - d} + ......... +

                   {2a + nd - d} + {2a + (n - 1)d}

2s     =       {2a + (n - 1)d} + {2a + (n - 1)d} + ....... +

                   {2a + (n - 1)d} + {2a + (n - 1)d}

Since 2 + 2 + 2 + 2 = 2(1 + 1 + 1 + 1) = 2 x 4,

2a + (n - 1)d  multiplied n times will be  n.{2a + (n - 1)d}. Therefore,

         2s      =       n.{2a + (n - 1)d}

                            or

s

=

n/2

{2a + (n - 1)d}       ............. (a)

Since l = a + (n - 1)d,  equation (a) is also written as

s	=	n/2	{a + a + (n - 1)d}  or
s	=	n/2	{a + l}

Now we will find the sum of 20 terms when a = 5 and d = 2. Substituting these values in the formula, we obtain

s	=	20 /2	{2(5) + (20 - 1)2}
	=	480

This problem can also be solved by finding the last term which in this case happens to be T₂₀ and it is given by T₂₀ = 5 + (20 - 1)2 = 43. Therefore,

s	=	n /2	{a + l}
	=	20 /2	{5 + 43} = 480.

We observe that both these methods are essentially the same. With this background let us look at few more examples.

Example 

For the series given below, find the 23rd and the 27th terms.

                   38, 36, 34, .............

We are given the first term that is a = 38. The common difference d is given by 36 - 38 = -2.  The 23rd term is given by

         T₂₃    =       a + 22d

                   =       38 + 22(-2)  

                   =       38 - 44  = - 6  

Similarly the 27th term is given by

         T₂₇    =       a + 26d

                   =       38 + 26(-2)

                   =       38 - 52