JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 20)
Let a1, a2, ......., a10 be an AP with common difference $$-$$ 3 and b1, b2, ........., b10 be a GP with common ratio 2. Let ck = ak + bk, k = 1, 2, ......, 10. If c2 = 12 and c3 = 13, then $$\sum\limits_{k = 1}^{10} {{c_k}} $$ is equal to _________.
Answer
2021
Explanation
$$a_{1}, a_{2}, a_{3}, \ldots, a_{10}$$ are in AP common difference $$=-3$$
$$b_{1}, b_{2}, b_{3}, \ldots, b_{10}$$ are in GP common ratio $$=2$$
Since, $$c_{k}=a_{k}+b_{k}, k=1,2,3 \ldots \ldots, 10$$
$$\therefore c_{2} =a_{2}+b_{2}=12$$
$$ c_{3} =a_{3}+b_{3}=13$$
Now, $$\mathrm{C}_{3}-\mathrm{C}_{2}=1$$
$$ \begin{array}{ll} \Rightarrow & \left(a_{3}-a_{2}\right)+\left(b_{3}-b_{2}\right) \neq 1 \Rightarrow-3+\left(2 b_{2}-b_{2}\right) \neq 1 \\ \Rightarrow & b_{2}=4 \\ \therefore & a_{2}=8 \end{array} $$
So, AP is $$11,8,5, \ldots$$.
Now, $$\sum_{k=1}^{10} C_{k}=\sum_{k=1}^{10} a_{k}+\sum_{k=1}^{10} b_{k}$$
$$ \begin{aligned} &=\left(\frac{10}{2}\right)[22+9(-3)]+2\left(\frac{2^{10}-1}{2-1}\right) \\ &=5(22-27)+2(1023)=2046-25 \\ &=2021 \end{aligned} $$
$$b_{1}, b_{2}, b_{3}, \ldots, b_{10}$$ are in GP common ratio $$=2$$
Since, $$c_{k}=a_{k}+b_{k}, k=1,2,3 \ldots \ldots, 10$$
$$\therefore c_{2} =a_{2}+b_{2}=12$$
$$ c_{3} =a_{3}+b_{3}=13$$
Now, $$\mathrm{C}_{3}-\mathrm{C}_{2}=1$$
$$ \begin{array}{ll} \Rightarrow & \left(a_{3}-a_{2}\right)+\left(b_{3}-b_{2}\right) \neq 1 \Rightarrow-3+\left(2 b_{2}-b_{2}\right) \neq 1 \\ \Rightarrow & b_{2}=4 \\ \therefore & a_{2}=8 \end{array} $$
So, AP is $$11,8,5, \ldots$$.
Now, $$\sum_{k=1}^{10} C_{k}=\sum_{k=1}^{10} a_{k}+\sum_{k=1}^{10} b_{k}$$
$$ \begin{aligned} &=\left(\frac{10}{2}\right)[22+9(-3)]+2\left(\frac{2^{10}-1}{2-1}\right) \\ &=5(22-27)+2(1023)=2046-25 \\ &=2021 \end{aligned} $$
Comments (0)
