Let a_1,a_2,a_3, be a G.P.

Question

Let $a_1,a_2,a_3,\ldots$ be a G.P. of increasing positive numbers. Let the sum of its 6th and 8th terms be $2$ and the product of its 3rd and 5th terms be $\dfrac19$. Then $6(a_2+a_4)(a_4+a_6)$ is equal to:

Accepted Answer

$a_3a_5 = a^2r^6 = a_4^2 = \dfrac19 \Rightarrow a_4=\dfrac13$ (positive terms).

$a_6+a_8 = a_6(1+r^2) = 2$, and $a_6=a_4r^2=\dfrac{r^2}{3}$, so $\dfrac{r^2}{3}(1+r^2)=2 \Rightarrow r^4+r^2-6=0 \Rightarrow r^2=2$ (taking the positive root).

$$a_2 = \frac{a_4}{r^2} = \frac16, \qquad a_6 = \frac{r^2}{3} = \frac23.$$

$$a_2+a_4 = \frac16+\frac13=\frac12, \qquad a_4+a_6 = \frac13+\frac23=1.$$

$$6(a_2+a_4)(a_4+a_6) = 6\times\frac12\times1 = 3.$$

Solutions · 1