parmenides51 wrote: Consider an arithmetic progression made up of $100$ terms. If the sum of all the terms of the progression is $150$ and the sum of the even terms is $50$, find the sum of the squares of the $100$ terms of the progression. Let $a$ be the first term of progression and $d$ be the common difference We have $150=\sum_{k=0}^{99}(a+kd)=100a+4950d$ and so $2a+99d=3$ So $d$ is odd and the even progression contains $50$ terms starting at $s=a$ if $a$ is even or $a+d$ if $a$ is odd. And we have $50=\sum_{k=0}^{49}(s+2kd)=50s+2450d$ and so $s+49d=1$ If $a$ is even, system $2a+99d=3$ and $a+49d=1$ gives $(a,d)=(-48,1)$ If $a$ is odd, system $2a+99d=3$ and $(a+d)+49d=1$ gives $(a,d)=(51,-1)$ And these two progressions are just the same (in reverse orders) And so in both cases $\sum_{k=0}^{99}(a+kd)^2=100a^2+9900ad+328350d^2=\boxed{83550}$