Question

Let $a_{1}, a_{2}, \ldots, a_{10}$ be an $AP$ with common difference $-3$ and $b_{1}, b_{2}, \ldots ., b_{10}$ be a $GP$ with common ratio $2$ . Let $c _{ k }= a _{ k }+ b _{ k }, k =1,2, \ldots, 10 .$ If $c _{2}=12$ and $c_{3}=13$, then $\displaystyle\sum_{ k =1}^{10} C _{ k }$ is equal to

Answer 1

$c_{2}=a_{2}+b_{2}$
$=a_{1}-3+2 b_{1}=12$
$a_{1}+2 b_{1}=15\,\,\, ...(1)$
$c_{3}=a_{3}+b_{3}$
$=a_{1}-6+4 b_{1}=13$
$a_{1}+4 b_{1}=19\,\,\, ...(2)$
from $(1) \,\&\, (2)$
$b_{1}=2, a_{1}=11$
$\displaystyle\sum_{ k =1}^{10} c _{ k }=\displaystyle\sum_{ k =1}^{10}\left( a _{ k }+ b _{ k }\right)=\displaystyle\sum_{ k =1}^{10} a _{ k }+\displaystyle\sum_{ k =1}^{10} b _{ k } $
$=\frac{10}{2}(2 \times 11+9 \times(-3))+\frac{2\left(2^{10}-1\right)}{2-1}$
$=5(22-27)+2(1023) $
$=2046-25=2021$