Three positive numbers form an increasing G.P.

Question

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is:

Accepted Answer

Let $a,ar,a{r^2}$ are in $G.P.$

According to the question

$a,2ar,a{r^2}$ are in $A.P.$

$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$

$$ \Rightarrow 4r = 1 + {r^2}$$

$$ \Rightarrow {r^2} - 4r + 1 = 0$$

$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $$

Since $r > 1$

$$\therefore$$ $$\pi  = 2 - \sqrt 3 $$ is rejected

Hence, $$r = 2 + \sqrt 3 $$

Solutions · 1