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`

Explanation

Let a be the first term and d be the common difference of the A.P.
Then the A.P. will be
` a, a+d,a+2d ,…, a + 2nd `
Sum of its odd terms
`= a + (a + 2d) + (a + 4d) + …` to (n+1) terms
`(n+1)/(2) [2a + (n+1-1) 2d]`
` = (n + 1 ) (a + nd)` ...(i)
Sum of even terms ` = (a + d) + (a + 3d ) + ...` to n terms
` = (n)/(2) [2(a + d) + (n - 1) 2d ] `
` = n (a + nd) ` (ii)
from (i) and (ii)
`("Sum of odd terms ")/("Sum of even terms")= ((n + 1)(a + nd))/(n(a + nd)) = (n+1)/(n)` Hence proved