Mathematics - Sequence & Series Question with Solution | TestHub

Given:

$c_{\mathrm{k}}=a_{\mathrm{k}}+b_{\mathrm{k}}$ and $a_1=b_1=4$

Also,

${\mathrm{a}}_2=4{\mathrm{r}}_1$ and $a_3=4r_1^2$

$b_2=4r_2$ and  $b_3=4r_2^2$

Now,

$c_2=a_2+b_2=5$

$\Rightarrow 4r_1+4r_2=5$

$\Rightarrow r_1+r_2=\frac{5}{4}$

And,

$c_3=a_3+b_3=\frac{13}{4}$

$\Rightarrow r_1^2+r_2^2=\frac{13}{16}$

$\Rightarrow {\left(r_1+r_2\right)}^2-2r_1{r}_2=\frac{13}{16}$

$\Rightarrow \frac{25}{16}-2r_1{r}_2=\frac{13}{16}$

$\Rightarrow 2r_1{r}_2=\frac{12}{16}$

$\Rightarrow r_1{r}_2=\frac{3}{8}$

$\Rightarrow 8r_1\left(\frac{5}{4}-r_1\right)3$

$\Rightarrow 10r_1-8r_1^2=3$

$\Rightarrow 8r_1^2-10r_1+3=0$

$\Rightarrow r_1=\frac{10\pm \sqrt{100-96}}{16}$

$\Rightarrow r_1=\frac{3}{4}, \frac{1}{2}$

$\Rightarrow r_2=\frac{1}{2},\frac{3}{4}$ 

Now,

$\sum _{\mathrm{k}=1}^{\infty }{\mathrm{c}}_{\mathrm{k}}-\left(12a_6+8b_4\right)$

$=\left(c_1+c_2+c_3+....\right)+\left[12\times \left\{4\times \left(\frac{1}{2^5}\right)\right\}+8\left\{4\times {\left(\frac{3}{4}\right)}^3\right\}\right]$

$=\left(a_1+a_2+a_3+....\right)+\left(b_1+b_2+b_3+....\right)+\left[\left(\frac{3}{2}\right)+\left(\frac{27}{2}\right)\right]$

$=\frac{4}{1-r_1}+\frac{4}{1-r_2}-15$

$=24-15=9$