Let a₁, a₂, ......., a₁₀ be an AP with common difference $-$ 3 and b₁, b₂, ........., b₁₀ be a GP with common ratio 2. Let c_k = a_k + b_k, k = 1, 2, ......, 10. If c₂ = 12 and c₃ = 13, then $\sum\limits_{k = 1}^{10} {{c_k}}$ is equal to _________.

$a_{1}, a_{2}, a_{3}, \ldots, a_{10}$ are in AP common difference $=-3$<br/><br/> $b_{1}, b_{2}, b_{3}, \ldots, b_{10}$ are in GP common ratio $=2$<br/><br/> Since, $c_{k}=a_{k}+b_{k}, k=1,2,3 \ldots \ldots, 10$<br/><br/> $\therefore c_{2} =a_{2}+b_{2}=12$<br/><br/> $c_{3} =a_{3}+b_{3}=13$<br/><br/> Now, $\mathrm{C}_{3}-\mathrm{C}_{2}=1$<br/><br/> $$ \begin{array}{ll} \Rightarrow & \left(a_{3}-a_{2}\right)+\left(b_{3}-b_{2}\right) \neq 1 \Rightarrow-3+\left(2 b_{2}-b_{2}\right) \neq 1 \\ \Rightarrow & b_{2}=4 \\ \therefore & a_{2}=8 \end{array} $$<br/><br/> So, AP is $11,8,5, \ldots$. <br/><br/> Now, $\sum_{k=1}^{10} C_{k}=\sum_{k=1}^{10} a_{k}+\sum_{k=1}^{10} b_{k}$<br/><br/> $$ \begin{aligned} &=\left(\frac{10}{2}\right)[22+9(-3)]+2\left(\frac{2^{10}-1}{2-1}\right) \\ &=5(22-27)+2(1023)=2046-25 \\ &=2021 \end{aligned} $$