Sequence and series

Asked by coolameet08 | 14th Feb, 2010, 11:38: AM

Expert Answer:

S1 = (2n+1)(a + a + (2n+1-1))d)/2

Irrespective of n being even or odd, 2n is always even, so the odd number of terms will be 2n/2 = n,

and if we have total 2n+1 terms the odd terms will be n+1, since the last term will always be odd.

Also series of odd terms is also a AP with common difference 2d.

S2 = (n+1)(a + a + (n+1-1)(2d)/2

S1/S2 = (2n+1)/(n+1)




Answered by  | 14th Feb, 2010, 12:00: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.