Question

If S₁ is the sum of an arithmetic progression of 'n' odd number of terms and S₂ the sum of the terms of the series in odd places, then \[\frac{S_1}{S_2} =\]

 

Options

• \[\frac{2n}{n + 1}\]

   

• \[\frac{n}{n + 1}\]

   

• \[\frac{n + 1}{2n}\]

   

• \[\frac{n + 1}{n}\]

   

Answer 1

In the given problem, we are given S₁ as the sum of an A.P of ‘n’ odd number of terms and S₂ the sum of the terms of the series in odd places.

We need to find  `(S_1)/(S_2)`

Now, let a₁, a₂…. a_n be the n terms of A.P

Where n is odd

Let d be the common difference of the A.P

Then,

`S_1 = n /2 [ 2a_1 + ( n - 1) d]`             ............(1)

And  S₂  be the sum of the terms of the places in odd places,

Where, number of terms = `( n + 1) /2` 

Common difference = 2d

So,

`S_2 =  ((n + 1)/2 )/2 [2a_1 + ((n+1)/2 - 1) 2d]`

`S_2 = ( n+1)/4 [2a_1 + ((n-1)/2)2d]`

`S_2 = ( n +1)/4 [ 2a _1 + (n-1)d ]`              .............(2) 

Now,

`(S_1)/(S_2) = (n/2[2a_1 + (n-1)d])/((n+1)/4[2a_1 + (n-1)d])`

`(S_1)/(S_2) = (4n)/(2(n +1))`

`(S_1)/(S_2) = (2n)/(n + 1)`

Thus,  `(S_1)/(S_2) = (2n)/(n + 1)`