Question

Let $\left\{a_k\right\}$ and $\left\{b_k\right\}, k \in N$, be two G.P.s with common ratio $r_1$ 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 $c _3=\frac{13}{4}$ then $\displaystyle\sum_{ k =1}^{\infty} c _{ k }-\left(12 a _6+8 b _4\right)$ is equal to

Answer 1

Given that
$c_k=a_k+b_k$ and $a_1=b_1=4$
also $a_2=4 r_1 \,\,\, a_3=4 r_1^2 $
$b_2=4 r_2 \,\,\, b_3=4 r_2^2$
Now $c_2=a_2+b_2=5$ and $c_3=a_3+b_3=\frac{13}{4}$
$\Rightarrow r _1+ r _2=\frac{5}{4}$ and $r _1^2+ r _2^2=\frac{13}{16}$
Hence $r_1 r_2=\frac{3}{8} \,\,\,\, $ which gives $r_1=\frac{1}{2} \,\,\,\, \& \,\,\,\,r_2=\frac{3}{4}$
$\displaystyle\sum_{ k =1}^{\infty} c _{ k }-\left(12 a _6+8 b _4\right)$
$ =\frac{4}{1-r_1}+\frac{4}{1-r_2}-\left(\frac{48}{32}+\frac{27}{2}\right) $
$=24-15=9$