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

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.P.is 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

O(say)

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

=(n+1)/2[a+a+(2n)d]

=(n+1)[a+nd]

simly,

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

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

=n/2[(a+d)+(a+[2n-1]d)

=n[a+nd]

So required ratio=

O/E

=n+1[a+nd]/n[a+nd]

=n+1/n

Hence proved.

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

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