Strange numbers

Suppose that $a+b+c=\lambda$ .

If $\lambda = 0$ then $ab + ac + bc = 0$ as well, and hence $a,b,c$ are the roots of the equation $X^3 - abc = 0$ . Since $a,b,c$ are all real, this means that $a=b=c=\tfrac13\lambda$ , and hence $a=b=c=0$ . Thus we deduce that $\lambda \neq 0$ .

Thus we can write $ab + ac + bc = \lambda\alpha$ for some $\alpha$ , which implies that $abc = \alpha^3$ . Thus we deduce that $a,b,c$ are the roots of the equation $0 \; = \; X^3 - \lambda X^2 + \lambda\alpha X - \alpha^3 \; =\; (X - \alpha)(X^2 + \alpha X + \alpha^2 - \lambda X)$ and so, without loss of generality, $a = \alpha$ and $bc = \alpha^2$ . Certainly, then, $a,b,c$ are in geometric progression.

If we pick $a=1, b=2, c=4$ , then we have a solution of the equation where $a,b,c$ are in GP, but are not in AP, nor are they in HP.

The only thing we can say for certain is that the sequence is in GP.

Alternatively, me may look at the conditions necessary for the roots of a cubic to be in H.P, A.P and G.P respectively and compare it to the condition given.

$If\quad a,b,c\quad are\quad the\quad roots\quad of\quad the\quad cubic\quad p{ x }^{ 3 }+{ qx }^{ 2 }+rx+s,\quad then\quad the\quad given\quad condition\quad translates\quad to:\\ \\ \frac { { q }^{ 3 } }{ { r }^{ 3 } } =\frac { p }{ s } \\ \\ Consider\quad any\quad arbitary\quad cubic\quad f\left( x \right) =p{ x }^{ 3 }+{ qx }^{ 2 }+rx+s,\quad \quad For\quad roots\quad to\quad be\quad in:\\ \\ (1)\quad G.P:\quad Let\quad roots\quad be\quad \frac { \alpha }{ m } ,\alpha ,\quad \alpha m.\quad Then,\quad { \alpha }^{ 3 }=\frac { -s }{ p } \Rightarrow \quad \alpha =\sqrt [ 3 ]{ \frac { -s }{ p } } .\\ Since\quad \alpha \quad is\quad a\quad root\quad of\quad the\quad cubic,\quad f\left( \alpha \right) =0\Rightarrow \quad \boxed { \frac { { q }^{ 3 } }{ { r }^{ 3 } } =\frac { p }{ s } } \\ \\ \\ (2)\quad A.P:\quad Let\quad roots\quad be\quad \alpha -m,\quad \alpha \quad and\quad \alpha +m.\quad Then,\quad 3\alpha =\frac { -q }{ p } \Rightarrow \alpha =\frac { -q }{ 3p } \\ Since\quad \alpha \quad is\quad a\quad root\quad of\quad the\quad cubic,\quad f\left( \alpha \right) =0\Rightarrow \quad \boxed { 2{ q }^{ 3 }-9pqr+27s{ p }^{ 2 }=0 } \\ \\ \\ \\ (3)\quad H.P:\quad Consider\quad the\quad transformed\quad function\quad g\left( x \right) =\quad s{ x }^{ 3 }+{ rx }^{ 2 }+qx+p,\quad \\ which\quad has\quad roots\quad which\quad are\quad the\quad reciprocals\quad of\quad those\quad of\quad f\left( x \right) .\quad If\quad roots\\ of\quad f(x)\quad must\quad be\quad in\quad H.P,\quad then\quad roots\quad of\quad g(x)\quad must\quad be\quad in\quad A.P,\quad implying:\\ \boxed { 2{ r }^{ 3 }-9sqr+27p{ s }^{ 2 }=0 }$