It looks as though it should be $\sum_{i=1}^na_{2i-1}^2$ etc, but $a_i$ is not defined for $i>n$.

There are $2n$ terms in total. Please correct the question.

Oh god the entire question you posted is wrong. The $n$ right after $\sum$ is the upper bound for the $i$. Please delete this post asap.

I am really very sorry The question has been edited

I get $d=(y-x)/nz$. Subtract the first equation from the second equation and you are done.

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Detailed Solution $a_2^2 + a_4^2 + .... + a_{2n}^2 = y$ $a_1^2 + a_3^2 + .... + a_{2n-1}^2 = x$ $(a_2^2 - a_1^2) + (a_4^2 - a_3^2) + .... + (a_{2n}^2 - a_{2n-1}^2) = y-x $ $(a_2 - a_1)(a_2 + a_1) + (a_4 - a_3)(a_4 + a_3) + .... + (a_{2n} - a_{2n-1})(a_{2n} + a_{2n-1}) =y-x$ As $a_{i+1} - a_i = D$ $D(a_1 + a_2 + a_3 + .... + a_{2n}) = y-x ......(1)$ Also $a_n + a_{n+1} = z$ $a_1 + (n-1)D + a_1 + nD = z$ $2a_1 + (2n-1)D = z$ This gives $\frac{2n}{2}(2a_1 + (2n-1)D) = nz$ LHS is the sum of 2n terms of the A.P. Using this in (1) $Dnz = y-x$ Thus $D = \frac{y-x}{nz}$

file:///C:/Users/arjun/OneDrive/0_Arjun/IMO/RMO%20past%20year%20papers/my%20solution,%20RMO%202014-2%20(paper%201).pdf Go to the above link to check my solution to the problem (I used a different idea which I think is more faster).

@above, attach bc it is a local file

Clearly $y-x=d*(\sum_{i=1}^{2n} a_i)\implies y-x=d*\frac{2n*z}{2}$ So $d=\frac{y-x}{nz}$