Let the A.P. be a, a + d, a + 2d, ...., a + (2n – 1)d, a + 2nd
Then, the progression of odd terms is a, a + 2d, a + 4d, ...., a + 2nd.
This progression has (n + 1) terms.
Its sum = \(\frac{n+1}{2}\) [a+ a+ 2nd] = \(\frac{n+1}{2}\)[2a + 2nd] = (n+1)(a+nd)
The progression of even terms is a + d, a + 3d, ...... a + (2n – 1)d.
This progression has n terms.
Its sum = \(\frac{n}{2}\) [(a+d) + (a + (2n-1) d] = \(\frac{n}{2}\)[2a +2nd] = n(a+nd)
\(\therefore\) \(\frac{Sum \, of\, odd\,terms}{Sum\,of\,even\,terms}= \frac{(n+1)(a+nd)}{n(a+nd)}\)
= \(\frac{n+1}{n}\)
