parmenides51 wrote: a) We have a geometric sequence of $3$ terms. If the sum of these terms is $26$ , and their sum of squares is $364$ , find the terms of the sequence. Let $x,xy,xy^2$ be the three numbers We have : $x(1+y+y^2)=26$ $x^2(1+y^2+y^4)=364$ Second line is $x^2(y^2+y+1)(y^2-y+1)=364$ And so $x(y^2-y+1)=14$ Dividing first line by last one, we get $\frac{y^2+y+1}{y^2-y+1}=\frac{13}7$ Which is $y^2-\frac{10}3y+1=0$ and so $y=\frac 13$ or $y=3$ And so (using $x=\frac{26}{1+y+y^2}$) the two solutions $\boxed{2,6,18}$ and $\boxed{18,6,2}$

parmenides51 wrote: b) Suppose that $a,b,c,u,v,w$ are positive real numbers , and each of $a,b,c$ and $u,v,w$ are geometric sequences. Suppose also that $a+u,b+v,c+w$ are an arithmetic sequence. Prove that $a=b=c$ and $u=v=w$ Just write the two sequences $x,xy,xy^2$ and $z,zt,zt^2$ with $x,y,z,t>0$ And you get $(x+z)+(xy^2+zt^2)=2(xy+zt)$ Which is $x(y-1)^2+z(t-1)^2=0$, and so $y=t=1$ Q.E.D.