If there are `(2n+1)` terms in A.P., then prove that the ratio of the sum of odd terms and the sum of even terms is `(n+1): ndot`

Options

A. `(n)/(n+1)`
B. `(n^(2))/(n+1)`
C. `(n+1)/(n)`
D. `(n+1)/(2n)`

Answer

C

Explanation

Let the AP is
`a,a+d,a+2d,……,a+(2n-1)d,a+2nd`
Series of even terms.
`a+d,a+3d,…………,a+(2n-1)d,` has n terms
Sum of even number`=(n)/(2)[(a+d)+{a+(2n-1)d}]`
`=(n)/(2)[2a+2nd]=n[a+nd]`
Series of odd terms
`a,a+2d,a+4d,……......,a+2nd," has '(n+1)" terms."`
Sum of odd numbers`=(n+1)/(2)[a+(a+2nd)]`
`=(n+1)/(2)(2a+2nd)`
`=(n+1)(a+nd)`
So, the required ratio `=(n+1)/(n)`