Let a_1, a_2, a_3, ....

Question

Let a$_1$, a$_2$, a$_3$, .... 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 $$\frac{1}{9}$$. Then $6(a_2+a_4)(a_4+a_6)$ is equal to

Accepted Answer

Given the conditions :

${a}_{6}+{a}_{8}=2\Rightarrow a{r}^{5}+a{r}^{7}=2$
${a}_{3}\cdot {a}_{5}=\frac{1}{9}\Rightarrow {a}^{2}\cdot {r}^{2}\cdot {r}^{4}=\frac{1}{9}\Rightarrow a{r}^{3}=\frac{1}{3}$

From this, we can form the equation $\frac{{r}^{2}}{3}+\frac{{r}^{4}}{3}=2$, which simplifies to ${r}^{4}+{r}^{2}=6$.

This can be factored to give $({r}^{2}-2)({r}^{2}+3)=0$, yielding ${r}^{2}=2$ (since ${r}^{2}$ cannot be $-3$ for real $r$).

So, we have $r=\sqrt{2}$.

Substituting $r=\sqrt{2}$ into the equation $ar=\frac{1}{6}$, we get $a=\frac{1}{6\sqrt{2}}$.

Now, we find the value of $6({a}_{2}+{a}_{4})({a}_{4}+{a}_{6})$:

$6({a}_{2}+{a}_{4})({a}_{4}+{a}_{6})=6(ar+a{r}^{3})(a{r}^{3}+a{r}^{5})$

$=6(\frac{1}{6\sqrt{2}}+\frac{1}{3\sqrt{2}})(\frac{1}{3\sqrt{2}}+\frac{2}{3\sqrt{2}})$

$=6\cdot \frac{1}{2}\cdot 1=3$.

Solutions · 1