Let a _ 1 , a _ 2 , a _ 3 , . be a GP of increasing positive...

Given,

${a}_{1}, {a}_{2}, {a}_{3}, \ldots$. be a GP of increasing positive numbers,

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

So, ${a}_{4} \cdot {a}_{6} = 9 ⇒ {\left({a}_{5}\right)}^{2} = 9 ⇒ {a}_{5} = 3$

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

So, ${a}_{5} + {a}_{7} = 24$

$⇒ {a}_{5} + {a}_{5}{r}^{2} = 24$

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

Now using ${a}_{5} = 3 ⇒ {a}_{1}{r}^{4} = 3 ⇒ {a}_{1} = \dfrac{3}{{r}^{4}} = \dfrac{3}{49}$

Then, ${a}_{1}{a}_{9} + {a}_{2}{a}_{4}{a}_{9} + {a}_{5} + {a}_{7}$

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

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