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if, S₁ is the sum of an arithmetic progression of 'n' odd number of terms and S₂ is the sum of the terms of the series in odd places, then S₁/S₂=??

$L e t a b e t h e f i r s t t e r m a n d d b e t h e c o m m o n d i f f e r e n c e o f a n A . P . S o, t h e s e r i e s a r e, a, a + d, a + 2 d, ........ a + (n - 1) d, .... N o w, S_{1} = S u m o f n o d d n u m b e r o f t e r m s = \frac{n}{2} [2 a + (n - 1) d] A n d, T h e t e r m s o f t h e s e r i e s i n o d d p l a c e s a r e, a, a + 2 d, a + 4 d, ....... a + (n - 1) d, ... S o, f i r s t t e r m = a, a n d c o m m o n d i f f e r e n c e = 2 d N o w, S_{2} = S u m o f t h e t e r m s o f t h e s e r i e s i n o d d p l a c e s = \frac{\frac{n + 1}{2}}{2} [2 a + (\frac{n + 1}{2} - 1) 2 d] = \frac{n + 1}{4} [2 a + (n - 1) d] T h u s, \frac{S_{1}}{S_{2}} = \frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{n + 1}{4} [2 a + (n - 1) d]} \Rightarrow \frac{S_{1}}{S_{2}} = \frac{2 n}{n + 1}$

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