Let a_k and b_k , k ∈ mathbbN, be two G.P.s with common ratio r_1 and r_2 respectively such that a_1= b_1=4 and _1< _2. Let _ k= a_ k+ b_ k, k ∈…

Given:

${c}_{k}={a}_{k}+{b}_{k}$ck​=ak​+bk​ and ${a}_{1}={b}_{1}=4$a1​=b1​=4

Also,

${a}_{2}=4{r}_{1}$a2​=4r1​ and ${a}_{3}=4{r}_{1}^{2}$a3​=4r12​

${b}_{2}=4{r}_{2}$b2​=4r2​ and ${b}_{3}=4{r}_{2}^{2}$b3​=4r22​

Now,

$${c}_{2}={a}_{2}+{b}_{2}=5$$c2​=a2​+b2​=5

$$\Rightarrow 4{r}_{1}+4{r}_{2}=5$$⇒4r1​+4r2​=5

$$\Rightarrow {r}_{1}+{r}_{2}=\frac{5}{4}$$⇒r1​+r2​=45​

And,

$${c}_{3}={a}_{3}+{b}_{3}=\frac{13}{4}$$c3​=a3​+b3​=413​

$$\Rightarrow {r}_{1}^{2}+{r}_{2}^{2}=\frac{13}{16}$$⇒r12​+r22​=1613​

$$\Rightarrow {({r}_{1}+{r}_{2})}^{2}-2{r}_{1}{r}_{2}=\frac{13}{16}$$⇒(r1​+r2​)2−2r1​r2​=1613​

$$\Rightarrow \frac{25}{16}-2{r}_{1}{r}_{2}=\frac{13}{16}$$⇒1625​−2r1​r2​=1613​

$$\Rightarrow 2{r}_{1}{r}_{2}=\frac{12}{16}$$⇒2r1​r2​=1612​

$$\Rightarrow {r}_{1}{r}_{2}=\frac{3}{8}$$⇒r1​r2​=83​

$$\Rightarrow 8{r}_{1}(\frac{5}{4}-{r}_{1})3$$⇒8r1​(45​−r1​)3

$$\Rightarrow 10{r}_{1}-8{r}_{1}^{2}=3$$⇒10r1​−8r12​=3

$$\Rightarrow 8{r}_{1}^{2}-10{r}_{1}+3=0$$⇒8r12​−10r1​+3=0

$$\Rightarrow {r}_{1}=\frac{10\pm \sqrt{100-96}}{16}$$⇒r1​=1610±100−96​​

$$\Rightarrow {r}_{1}=\frac{3}{4},\frac{1}{2}$$⇒r1​=43​,21​

$$\Rightarrow {r}_{2}=\frac{1}{2},\frac{3}{4}$$⇒r2​=21​,43​

Now,

$$\sum _{k=1}^{\infty }{c}_{k}-(12{a}_{6}+8{b}_{4})$$k=1∑∞​ck​−(12a6​+8b4​)

$$=({c}_{1}+{c}_{2}+{c}_{3}+....)+[12\times {4\times (\frac{1}{{2}^{5}})}+8{4\times {(\frac{3}{4})}^{3}}]$$=(c1​+c2​+c3​+....)+[12×4×(251​)+84×(43​)3]

$$=({a}_{1}+{a}_{2}+{a}_{3}+....)+({b}_{1}+{b}_{2}+{b}_{3}+....)+[(\frac{3}{2})+(\frac{27}{2})]$$=(a1​+a2​+a3​+....)+(b1​+b2​+b3​+....)+[(23​)+(227​)]

$$=\frac{4}{1-{r}_{1}}+\frac{4}{1-{r}_{2}}-15$$=1−r1​4​+1−r2​4​−15

$$=24-15=9$$=24−15=9