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If there are $(2n+1)$ terms in A.P.,prove that the ratio of the sum of odd terms and the sum of even terms are in the ratio of $(n+1):n.$

The main confusion I am facing is that n is not specified here. I feel that It could be anything from a fraction to an irrational. Please help

Of course it's not me
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Arunabh
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  • It's clear from the context that $n\in \mathbb N$. – lulu Feb 15 '18 at 15:31
  • How can you say so? – Arunabh Feb 15 '18 at 15:33
  • $2n+1$ is meant to count the number of terms under consideration. Thus $2n+1$ is a natural number. I suppose one could argue that $n$ might be half of an odd natural number and if you really thought that was intended you could try to solve the problem in that case as well. But start with the case in which $2n+1$ is just the standard short hand for an odd number. – lulu Feb 15 '18 at 15:36
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    Let terms be $a_1,a_2,\dots,a_{2n+1}$. Sum of odd terms is $(n+1)(a_1+a_{2n+1})/2$. Sum of even terms is $n(a_2+a_{2n})/2$. But $a_1+a_{2n+1}=a_2+a_{2n}$. Hence $(n+1):n$. – almagest Feb 15 '18 at 16:29

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