Solve it in 1 minute!

Let the First term in the sequence be $a$ and common ratio be $r$ and sum of infinite GP is given by $\large{\frac{a}{1-r}}$ (see wiki ). Now at the question we have

$\large{a+ar+ar^2+ar^3+\ldots = \frac{a}{1-r} = 3 \\ {(\frac{a}{1-r})}^{2}=9 \\ \frac{a^2}{(1-r)^2} = 9 \quad \quad (1)}$

$\large{a^2+a^2r^2+a^2r^4+ \ldots=\frac{a^2}{1-r^2}=81 \quad \quad (2)}$

Now Dividing $(1)$ by $(2)$

$\large{\frac { { a }^{ 2 } }{ { \left( 1-r \right) }^{ 2 } } \times \frac { 1-{ r }^{ 2 } }{ { a }^{ 2 } } =\frac{1}{9}\\ \frac { \left( 1+r \right) \left( 1-r \right) }{ { \left( 1-r \right) }^{ 2 } } =\frac{1}{9}\\ 9+9r=1-r\\ |r|=\frac { 4 }{ 5 } }$

This gives $p+q=9$ .

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Well i took 2:10 to solve it