Mathematics - Sequence & Series Question with Solution | TestHub

Given,

${\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,\dots$. be a GP of increasing positive numbers,

And the product of fourth and sixth terms is $9$

So, ${\mathrm{a}}_4·{\mathrm{a}}_6=9\Rightarrow {\left({\mathrm{a}}_5\right)}^2=9\Rightarrow {\mathrm{a}}_5=3$

Also given the sum of fifth and seventh terms is $24$,

So, ${\mathrm{a}}_5+{\mathrm{a}}_7=24$

$\Rightarrow {\mathrm{a}}_5+{\mathrm{a}}_5{\mathrm{r}}^2=24$

$\Rightarrow \left(1+{\mathrm{r}}^2\right)=8\Rightarrow \mathrm{r}=\sqrt{7}$

Now using ${\mathrm{a}}_5=3\Rightarrow a_1{r}^4=3\Rightarrow a_1=\frac{3}{r^4}=\frac{3}{49}$

Then, ${\mathrm{a}}_1{\mathrm{a}}_9+{\mathrm{a}}_2{\mathrm{a}}_4{\mathrm{a}}_9+{\mathrm{a}}_5+{\mathrm{a}}_7$

$=\frac{3}{49}\times \frac{3}{49}\times {\left(\sqrt{7}\right)}^8+\frac{3}{49}\times \sqrt{7}\times \frac{3}{49}\times {\left(\sqrt{7}\right)}^3\times \frac{3}{49}\times {\left(\sqrt{7}\right)}^8+\frac{3}{49}\times {\left(\sqrt{7}\right)}^4+\frac{3}{49}\times {\left(\sqrt{7}\right)}^6$

$=9+27+3+21=60$