Q.Three numbers are in Geometric Progression. Their product is $216$ and their sum is $26$. Find the numbers.

Take the three GP terms as $\frac{a}{r},\ a,\ ar$ so the product gives $a$ at once.

Choosing the symmetric form $\dfrac{a}{r},\ a,\ ar$ for three terms in GP makes the product simple.

Product: $\dfrac{a}{r} \cdot a \cdot ar = a^3 = 216 \Rightarrow a = 6$ (so the middle term is $6$).

Sum: $\dfrac{a}{r} + a + ar = 26$. Substituting $a = 6$: $\dfrac{6}{r} + 6 + 6r = 26 \Rightarrow \dfrac{6}{r} + 6r = 20$. Multiply through by $r$:

Factorising: $(3r - 1)(r - 3) = 0 \Rightarrow r = \dfrac13$ or $r = 3$. …

The quadratic $3r^2 - 10r + 3 = 0$ can also be solved by the formula: $r = \dfrac{10 \pm \sqrt{100 - 36}}{6} = \dfrac{10 \pm 8}{6}$, giving $r = 3$ or $ …

Writing the terms as $a, ar, ar^2$ instead of the symmetric form makes the product $a^3 r^3$ rather than a clean $a^3$, forcing a harder simultaneous solve. The $\frac{a}{r}, …