If `S_1` be the sum of `(2n+1)` term of an A.P. and `S_2` be the sum of its odd terms then prove that `S_1: S_2=(2n+1):(n+1)dot`

Explanation

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

`S_n=a_n+(n(n-1))/2*d`

`S_1=a+(2n+11)+(2n+1-1)d/2=[2n+1][a+nd=-(1)`

Number of terms in `S_2=n+1`

`S_2=(n+1)a+((n+1)(n+1-1))/2(2d)=(n+1)a+(n+1)nd`

`=(n+1)(a+nd)`

`S_1/S_2=([2n+1][a+nd])/([n+1][a+nd])=(2n+1)/(n+1)`.