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(A) Given $c_{2} = a_{2} + b_{2} = (a_{1} - 3) + (2b_{1}) = 12$,so $a_{1} + 2b_{1} = 15 \dots (1)$.Given $c_{3} = a_{3} + b_{3} = (a_{1} - 6) + (4b_{1

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$2021$

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(A) Given $c_{2} = a_{2} + b_{2} = (a_{1} - 3) + (2b_{1}) = 12$,so $a_{1} + 2b_{1} = 15 \dots (1)$.Given $c_{3} = a_{3} + b_{3} = (a_{1} - 6) + (4b_{1}) = 13$,so $a_{1} + 4b_{1} = 19 \dots (2)$.Subtracting $(1)$ from $(2)$,we get $2b_{1} = 4$,which implies $b_{1} = 2$.Substituting $b_{1} = 2$ into $(1)$,we get $a_{1} + 4 = 15$,so $a_{1} = 11$.Now,$\sum_{k=1}^{10} c_{k} = \sum_{k=1}^{10} a_{k} + \sum_{k=1}^{10} b_{k}$.The sum of the $AP$ is $S_{a} = \frac{10}{2} [2(11) + (10-1)(-3)] = 5(22 - 27) = 5(-5) = -25$.The sum of the $GP$ is $S_{b} = \frac{b_{1}(r^{10} - 1)}{r - 1} = \frac{2(2^{10} - 1)}{2 - 1} = 2(1024 - 1) = 2(1023) = 2046$.Therefore,$\sum_{k=1}^{10} c_{k} = -25 + 2046 = 2021$.

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 $\sum_{k=1}^{10} c_{k}$ is equal to:

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