Three numbers are in an increasing geometric progression with common ratio r.

Question

Three numbers are in an increasing geometric progression with common ratio $r$. If the middle number is doubled, the new numbers are in an arithmetic progression with common difference $d$. If the fourth term of the G.P. is $3r^2$, then $r^2-d$ is equal to:

Accepted Answer

Let the terms be $\dfrac ar, a, ar$. With middle doubled: $\dfrac ar, 2a, ar$ in A.P.: $4a = ar+\dfrac ar \Rightarrow r+\dfrac1r=4 \Rightarrow r^2-4r+1=0 \Rightarrow r=2\pm\sqrt3.$

Fourth term of G.P.: $ar^2=3r^2 \Rightarrow a=3$. Taking $r=2+\sqrt3$: $d = 2a-\dfrac ar = 3\sqrt3$.

$$r^2-d = (2+\sqrt3)^2-3\sqrt3 = 7+4\sqrt3-3\sqrt3 = 7+\sqrt3.$$

Solutions · 1