Let a1 , a2 ,....,a10 be an AP with common difference –3 and b1 , b2 ,...., b10 be a GP with common ratio 2.

Question:

Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{10}$ 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 $\sum_{\mathrm{k}=1}^{10} c_{\mathrm{k}}$ is equal to______

Solution:

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

$a_{1}+2 b_{1}=15$  .....(1)

$\alpha^{2}-\alpha+2 \lambda=0$        ......(2)

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

$\sum_{\mathrm{k}=1}^{10} \mathrm{c}_{\mathrm{k}}=\sum_{\mathrm{k}=1}^{10}\left(\mathrm{a}_{\mathrm{k}}+\mathrm{b}_{\mathrm{k}}\right)=\sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}+\sum_{\mathrm{k}=1}^{10} \mathrm{~b}_{\mathrm{k}}$

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

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

$=2046-25=2021$

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