Let a_ 1 , a_ 2 , , a_ 10 be an AP with common difference -3 and … - Vidyadip

MCQ

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

✓
$2021$
B
$1234$
C
$2227$
D
$2119$

✓

Answer

Correct option: A.

$2021$

a
$c_{2}=a_{2}+b_{2}=a_{1}-3+2 b_{1}=12$

$a_{1}+2 b_{1}=15....(1)$

$c_{3}=a_{3}+b_{3}=a_{1}-6+4 b_{1}=13$

$a_{1}+4 b_{1}=19....(2)$

from $(1)\, \,(2) b_{1}=2, a_{1}=11$

$\sum_{k=1}^{10} c_{k}=\sum_{k=1}^{10}\left(a_{k}+b_{k}\right)=\sum_{k=1}^{10} a_{k}+\sum_{k=1}^{10} b_{k}$

$=\frac{10}{2}(2 \times 11+9 \times(-3))+\frac{2\left(2^{10}-1\right)}{2-1}$

$=5(22-27)+2(1023)$

$=2046-25=2021$

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