Let $\{ {a_k}\}$ and $\{ {b_k}\} ,k \in N$, be two G.P.s with common ratios ${r_1}$ and ${r_2}$ respectively such that ${a_1} = {b_1} = 4$ and ${r_1} < {r_2}$. Let ${c_k} = {a_k} + {b_k},k \in N$. If ${c_2} = 5$ and ${c_3} = {{13} \over 4}$ then $\sum\limits_{k = 1}^\infty {{c_k} - (12{a_6} + 8{b_4})}$ is equal to __________.

<p>$\{ {a_k}\}$ be a G.P. with ${a_1} = 4,r = {r_1}$</p> <p>And</p> <p>$\{ {b_k}\}$ be G.P. with ${b_1} = 4,r = {r_2}$ $({r_1} < {r_2})$</p> <p>Now</p> <p>${C_k} = {a_k} + {b_k}$</p> <p>${c_1} = 4 + 4 = 8$ and ${c_2} = 5$<p> <p>${a_2} + {b_2} = 5$</p> <p>$\therefore$ ${r_1} + {r_2} = {5 \over 4}$</p> <p>and ${c_3} = {{13} \over 4} \Rightarrow r_4^2 + r_2^2 = {{13} \over {16}}$</p> <p>$\therefore$ $${{25} \over {16}} - 2{r_1}{r_2} = {{13} \over {16}} \Rightarrow 2{r_1}{r_2} = {3 \over 4}$$</p> <p>$\therefore$ ${r_2} - {r_1} = \sqrt {{{25} \over {16}} - {3 \over 2}} = {1 \over 4}$</p> <p>$\therefore$ ${r_2} = {3 \over 4},{r_1} = {1 \over 2}$</p> <p>$\therefore$ $${a_6} = 4 \times {1 \over {{2^5}}} = {1 \over 8},{b_4} = 4 \times {{27} \over {64}} = {{27} \over {16}}$$</p> <p>and $$\sum\limits_{K = 1}^\infty {{C_K} = 4\left[ {{1 \over {1 - {1 \over 2}}} + {1 \over {1 - {3 \over 4}}}} \right] = 24} $$</p> <p>$\therefore$ $\sum\limits_{K = 1}^\infty {{C_K} - (12{a_6} + 8{b_4}) = 09}$</p>