The fourth, eighth and sixteenth terms of an arithmetic progression are consecutive terms of a geometric progression. Show that the third, sixth and twelfth terms of the arithmetic progression are also consecutive terms of a geometric progression.

Let the first term of the arithmetic progression be \(a=a_1\) and the common difference be \(d\). Then the fourth, eighth and sixteenth terms are \[ a_4=a+3d,\, a_8=a+7d,\, a_{16}=a+15d. \] Since they are consecutive terms of an arithmetic progression, we must have that \[ \frac{a_8}{a_4}=r=\frac{a_{16}}{a_8}, \] where \(r\) is the common ratio of the geometric progression. Therefore \[ \frac{a+7d}{a+3d}=\frac{a+15d}{a+7d}, \] and therefore \[ (a+7d)^2=(a+15d)(a+3d). \] Multiplying out the brackets, we find that \[ a^2+14da+49d^2=a^2+18da+45d^2, \] and rearranging we find that \[ 4d^2-4da=0. \] Therefore \[ 4d(d-a)=0. \] If \(d=0\) then \(a_n=a\) for all \(n\). Therefore \(a_3=a_6=a_{12}\), which is part of a geometric progression with common ratio \(1\).

If \(d=a\), then \[ a_3=a+2d=3d,\, a_6=a+5d=6d,\, a_{12}=a+11d=12d, \] which is part of a geometric progression with common ratio \(2\).