Asked by sonaisha | 10th Mar, 2009, 12:00: AM

Expert Answer:

Let there be 2n+1 terms in the A.P.

 we know that 2n+1 is an odd number  where n is a non negative integer.

Now out of these 2n+1 terms,

 n+1 terms are going to be in odd positions  like 1,3,5... 2n+1and n terms are going to be in even positions like 2,4,6... 2n.

So  if the first term of the given a and the common difference is d

then t

the terms in odd positions are a,a+2d,a+4d...a+(2n)d

the terms in even positins are a+d,a+3d,a+5d...a+(2n-1)d

So ,

sum of the terms in  odd positions


=a+{a+2d}+{a+4d}+...+{a+(2n)d}..(these are  total n+1 terms)




sum of the terms in even positions=E(say)

=(a+d)+(a+3d)+...+(a+[2n-1]d)...(these are n terms)



So required ratio=




Hence proved.



Answered by  | 10th Mar, 2009, 11:27: AM

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