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

Let the A.P. containing (2n + 1) terms bea, a + d, a + 2d, a + 3d, a + 4d, a + 5d, …, a + 2ndThe sum of odd terms of this A.P.=a+(a+2d)+(a+4d)+… to (n+1) terms =n+12[2a+(n+1−1)×2d]=(n+1)(a+nd)The sum of even terms of this A.P.=(a+d)+(a+3d)+(a+5d)+… to n terms =n2[2(a+d)+(n−1)×2d]=n(a+nd)Hence, the required ratio = n+1n