Q.Solve the equation $3x^3-26x^2+52x-24=0$ if its roots form a geometric progression.

When a cubic's three roots are known in advance to be in A.P., write them as $a-d,\ a,\ a+d$. Their sum is $3a$, so Vieta's relation $S_1=3a$ fixes $a=S_1/3$ immediately; substituting the three roots into $S_3=a(a^2-d^2)$ (or into $S_2$) then gives a linear equation for $d^2$, and all three roots follow.

For roots in G.P., write them as $a/r,\ a,\ ar$. Their product is $a^3$, so Vieta's relation $S_3=a^3$ fixes $a$ directly (a perfect cube in a well-posed problem); substituting into $S_1=a\big(\tfrac1r+1+r\big)$ then gives a quadratic in $r$.

For roots in H.P., the roots themselves do not sit in an additive progression, but their reciprocals do. Since the equation with roots $1/\alpha,1/\beta,1/\gamma$ is obtained from the original by reversing its coefficients (the reciprocal-roots transformation, see Transformation of Equations), the H.P. condition on $\alpha,\beta,\gamma$ becomes an A.P. condition on the roots of the reversed equation. The middle term of three numbers in A.P. equals one-third their sum, so the middle reciprocal root is pinned down at once; substituting it back into the reversed equation produces a polynomial identity purely in the original coefficients -- this is exactly how a relation such as $2b^3-9abc+27c^2=0$ (for $x^3+ax^2+bx+c=0$ with roots in H.P.) is derived.