Correct answer is D. \(\frac{n+1}{n}\)
S1 = \(\frac{n}{2}\)(2a + (n–1) d)
Out of these odd numbers of terms, there are \(\frac{n+1}{2}\) terms in odd places
S2 = \(\frac{n+1}{2 \times2}\)(2a + ( \(\frac{n+1}{2}\)–1) d)
Common difference of two odd places is 2d
Now,