$$\{ {a_k}\} $$ be a G.P. with $${a_1} = 4,r = {r_1}$$

And

$$\{ {b_k}\} $$ be G.P. with $${b_1} = 4,r = {r_2}$$ $$({r_1} < {r_2})$$

Now

$${C_k} = {a_k} + {b_k}$$

$${c_1} = 4 + 4 = 8$$ and $${c_2} = 5$$

$${a_2} + {b_2} = 5$$

$$\therefore$$ $${r_1} + {r_2} = {5 \over 4}$$

and $${c_3} = {{13} \over 4} \Rightarrow r_4^2 + r_2^2 = {{13} \over {16}}$$

$$\therefore$$ $${{25} \over {16}} - 2{r_1}{r_2} = {{13} \over {16}} \Rightarrow 2{r_1}{r_2} = {3 \over 4}$$

$$\therefore$$ $${r_2} - {r_1} = \sqrt {{{25} \over {16}} - {3 \over 2}} = {1 \over 4}$$

$$\therefore$$ $${r_2} = {3 \over 4},{r_1} = {1 \over 2}$$

$$\therefore$$ $${a_6} = 4 \times {1 \over {{2^5}}} = {1 \over 8},{b_4} = 4 \times {{27} \over {64}} = {{27} \over {16}}$$

and $$\sum\limits_{K = 1}^\infty {{C_K} = 4\left[ {{1 \over {1 - {1 \over 2}}} + {1 \over {1 - {3 \over 4}}}} \right] = 24} $$

$$\therefore$$ $$\sum\limits_{K = 1}^\infty {{C_K} - (12{a_6} + 8{b_4}) = 09} $$