If (a + 2b + c), (a - c) and (a - 2b + c) are in continued proportion, prove that b is the mean proportional between a and c.

Since, the numbers are in continued proportion,

$\therefore \dfrac{(a + 2b + c)}{(a - c)} = \dfrac{(a - c)}{(a - 2b + c)} \\[1em] \Rightarrow (a + 2b + c)(a - 2b + c) = (a - c)^2 \\[1em] \Rightarrow (a^2 - 2ab + ac + 2ab - 4b^2 + 2bc + ac - 2bc + c^2) = (a^2 + c^2 - 2ac) \\[1em] \Rightarrow a^2 + c^2 + 2ac - 4b^2 = a^2 + c^2 - 2ac \\[1em] \Rightarrow 4b^2 = 4ac \\[1em] \Rightarrow b^2 = ac.$

Since, b² = 4ac, hence proved that b is the mean proportional between a and c.