Review question

# If AP terms $a_4,a_8, a_{16}$ are in GP, are $a_3, a_6, a_{12}$ in GP too? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5345

## Solution

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 $2$.