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

Answer 1

Answer: 9

Given that
$c_k=a_k+b_k$ and $a_1=b_1=4$
also $a_2=4r_1{a}_3=4r_1^2$
$b_2=4r_2{b}_3=4r_2^2$
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 $r_1=\frac{1}{2}\&r_2=\frac{3}{4}$
$\sum _{k=1}^{\infty }{c}_k-\left(12a_6+8b_4\right)$
$=\frac{4}{1-r_1}+\frac{4}{1-r_2}-\left(\frac{48}{32}+\frac{27}{2}\right)$
$=24-15=9$