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# If there are (2n+1) terms in an arithmetic series, prove that the ratio of the sum of odd place terms to the sum of even place terms is (n

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Hello everyone, how are you today, hope you had an amazing Halloween!

I have this question i dont understand , i have looked up many websites but i dont understand their answers.

i would be grateful if any of you would explain it to me in detail.

If there are (2n+1) terms in an arithmetic series, prove that the ratio of the sum of odd place terms to the sum of even place terms is  (n+1) : n .

Nov 4, 2017

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Mmmmm....I'll try rosala ....!!!

Let's suppose that we have a partial series like this  where     a  = the first term   and d is the common difference between terms

a  + (a + d)   + (a + 2d)  + (a + 3d) + (a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)   + (a + 8d)

Let  n  be the number of even place terms

And    n + 1  the number of odd place terms

Then the sum of the even place terms is :

n *a   +   n^2* d     =  n (a + n*d)

And the sum of  the odd place terms is :

(n + 1) * a  + ( n+1) (n)* d   =  (n + 1) (a + n*d)     for  n ≥ 0

So....the ratio of the sum of the odd place terms to the even place terms is :

(n + 1) (a + n*d)               ( n + 1)

_____________   =         _______

n    (a + n*d)                       n

Nov 4, 2017
edited by CPhill  Nov 4, 2017
edited by CPhill  Nov 5, 2017