2021/096) Given an arithmetic series a with a common difference d $a_1 + a_{2n+1}= z$ and $\sum_{k=1}^{n} a_{2k-1}^2 = x\cdots(1)$ $\sum_{k=1}^{n} a_{2k}^2 = y\cdots(2)$ Show $d = \frac{y-x}{2nz}$

We are given

$\sum_{k=1}^{n} a_{2k-1}^2 = x\cdots(1)$

$\sum_{k=1}^{n} a_{2k}^2 = y\cdots(2)$

subtract (1) from (2) to get

$\sum_{k=1}^{n} (a_{2k}^2- a_{2k-1}^2)  = y-x$

Or $\sum_{k=1}^{n} (a_{2k}- a_{2k-1})(a_{2k} + a_{2k-1})   = y-x$

But $(a_{2k}- a_{2k-1}= d$ common difference so we get

$\sum_{k=1}^{n} d(a_{2k} + a_{2k-1})   = y-x$

or $d \sum_{k=1}^{n} (a_{2k} + a_{2k-1})   = y-x$

Or $d \sum_{k=1}^{2n} (a_{k})   = y-x\cdots($

as $a_k = a_1 + (k-1) d$ for any k so we have

Now $a_k + a_{2n+1-k} = a_1 + (k-1)d + a_1 + (2n+1-k-1)d = 2a_1 + (2n-1) d = = a_1 + a_1 + (2n-1) d = a_1 + a_{2n}$

so $a_n + a_{n+1}d = a_1 + a_{2n} = z$

so $a_k + a_{2n+1-k} = z$

so 

 $d \sum_{k=1}^{2n} (a_{k})$

  $= d \sum_{k=1}^{n} (a_{k} + a_{2n+1-k})$

  $= d \sum_{k=1}^{n} z$

  = 2dnz

 

  So $2dnz = y-x$