Question

If S_r denotes the sum of the first r terms of an A.P. Then , S_3n: (S_2n − S_n) is

Options

• n

• 3n

• 3

• none of these

Answer 1

Here, we are given an A.P. whose sum of r terms is S_r. We need to find  `(S_(3n))/(S_(2n) - S_n)`.

Here we use the following formula for the sum of n terms of an A.P.

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

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, first we find S_3n,

`S_(3n) = (3n)/2 [ 2a + (3n - 1)d]`

       `=(3n)/2 [2a + 3nd - d ]`               .................(1) 

Similarly,

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

      `= (2n)/2 [2a + 2nd -d]`              .................(2)

Also,

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

    `=n/2 [2a + nd - d ]`

So, using (1), (2) and (3), we get,

`(S_(3n))/(S_(2n) - S_n) = ((3n)/2 [2a + 3nd - d])/((2n)/2 [ 2a + 2nd - d ] - n/2 [ 2a + nd - d ])`

Taking `n/2` common, we get,

`(S_(3n))/(S_(2n) - S_n) =(3[2a + 3nd - d])/(2[2a + 2nd - d ]- [2a  + nd - d])`

                 `=(3[2a + 3nd - d])/(4a + 4nd - 2d - 2a  - nd + d)`

                `=(3[2a + 3nd - d])/(2a + 3nd - d)`

                 = 3

Therefore, `(S_(3n))/(S_(2n)- S_n )= 3 `