Let \ a_k\ and \ b_k\ , k ∈ ℕ , be two G.P.s with common ratio r_1...

Given:

${c}_{k} = {a}_{k} + {b}_{k}$ and ${a}_{1} = {b}_{1} = 4$

Also,

${a}_{2} = 4{r}_{1}$ and ${a}_{3} = 4{r}_{1}^{2}$

${b}_{2} = 4{r}_{2}$ and ${b}_{3} = 4{r}_{2}^{2}$

Now,

${c}_{2} = {a}_{2} + {b}_{2} = 5$

$⇒ 4{r}_{1} + 4{r}_{2} = 5$

$⇒ {r}_{1} + {r}_{2} = \dfrac{5}{4}$

And,

${c}_{3} = {a}_{3} + {b}_{3} = \dfrac{13}{4}$

$⇒ {r}_{1}^{2} + {r}_{2}^{2} = \dfrac{13}{16}$

$⇒ {\left({r}_{1} + {r}_{2}\right)}^{2} - 2{r}_{1}{r}_{2} = \dfrac{13}{16}$

$⇒ \dfrac{25}{16} - 2{r}_{1}{r}_{2} = \dfrac{13}{16}$

$⇒ 2{r}_{1}{r}_{2} = \dfrac{12}{16}$

$⇒ {r}_{1}{r}_{2} = \dfrac{3}{8}$

$⇒ 8{r}_{1}\left(\dfrac{5}{4} - {r}_{1}\right)3$

$⇒ 10{r}_{1} - 8{r}_{1}^{2} = 3$

$⇒ 8{r}_{1}^{2} - 10{r}_{1} + 3 = 0$

$⇒ {r}_{1} = \dfrac{10 \pm \sqrt{100 - 96}}{16}$

$⇒ {r}_{1} = \dfrac{3}{4}, \dfrac{1}{2}$

$⇒ {r}_{2} = \dfrac{1}{2}, \dfrac{3}{4}$

Now,

$\mathop{∑}_{k = 1}^{\infty}{c}_{k} - \left(12{a}_{6} + 8{b}_{4}\right)$

$= \left({c}_{1} + {c}_{2} + {c}_{3} + . . . .\right) + \left[12 \times \left\{4 \times \left(\dfrac{1}{{2}^{5}}\right)\right\} + 8\left\{4 \times {\left(\dfrac{3}{4}\right)}^{3}\right\}\right]$

$= \left({a}_{1} + {a}_{2} + {a}_{3} + . . . .\right) + \left({b}_{1} + {b}_{2} + {b}_{3} + . . . .\right) + \left[\left(\dfrac{3}{2}\right) + \left(\dfrac{27}{2}\right)\right]$

$= \dfrac{4}{1 - {r}_{1}} + \dfrac{4}{1 - {r}_{2}} - 15$

$= 24 - 15 = 9$