If `S_1` is the sum of an AP of 'n' odd number of terms and `S_2` be the sum of the terms of series in odd places of the same AP then `S_1/S_2` =

Explanation

Let A.P. be `a_(1),a_(2),a_(3),…a_(n),` where n is odd.
`thereforeS_(1)=n/2(a_(1)+a_(n))`
Even numbered terms are `a_(2),a_(4),a_(6),…a_(n-1)`
So, there are `(n-1)/2` even numbered terms.
`thereforeS_(2)=((n-1)/2)/2(a_(2)+a_(n-1))=(n-1)/4(a_(1)+a_(n))`
`therefore(S_(1))/(S_(2))=(2n)/(n-1)`