[SOLVED] Arithmetic Progression

[anemone]

Let $a_1,a_2,\,\cdots,\,a_{2n}$ be an arithmetic progression of positive real numbers with common difference $d$. Let
(1) $a_1^2+a_3^2+\cdots+a_{2n-1}^2=x$
(2) $a_2^2+a_4^2+\cdots+a_{2n}^2=y$
(3) $a_n+a_{n+1}=z$
Express $d$ in terms of $x,\,y,\,z,\,n$.

 

[kaliprasad]

Spoiler: My Solution

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 $d \sum_{k=1}^{2n} (a_{k}) = y-x\cdots(3)$


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


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


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

So $dnz = y-x$

Or $d = \frac{y-x}{nz}$