Question

If S₁ is the sum of an arithmetic progression of 'n' odd number of terms and S₂ the sum of the terms of the series in odd places, then = \(\frac{S_1}{S_2}\)

A. \(\frac{2n}{n + 1}\)

B. \(\frac{n}{n+1}\)

C. \(\frac{n+1}{2n}\)

D. \(\frac{n+1}{n}\)

Answer 1

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,