Question

If S_n denote the sum of the first n terms of an A.P. If S_2n = 3S_n, then S_3n : S_n is equal to

विकल्प

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Answer 1

Here, we are given an A.P. whose sum of n terms is S_n and `S_(2m) = 3S_n`.

We need to find `(S_(3m))/(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_(3m) = (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]`                    ................(3) 

Now, `S_(2n) = 3S_n`

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

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

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

On further solving, we get,

2(2a + 2nd - d ) = 3 (2a + nd - d)

   4a + 4nd - 2d = 6a + 3nd - 3d

                      2a = nd  +  d                    .....................(4) 

So,

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

Taking `n/2` common, we get,

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

        `=(3(nd + d + 3nd - d))/((nd + d + nd - d))`               (Using 4)

        `= (3(4nd))/(2nd)`

         =  6

Therefore, `S_(3n)/(S_n) = 6`