Let $$a_1, a_2, \ldots, a_{10}$$ be an A.P. with common difference $$-3$$ and $$b_1, b_2, \ldots, b_{10}$$ be a G.P. 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 _________

We have two sequences.

The arithmetic progression is $$a_1,\,a_2,\,\ldots ,\,a_{10}$$ with common difference $$d=-3$$, so the general term is

$$a_k = a_1 + (k-1)d = a_1 + (k-1)(-3).$$

The geometric progression is $$b_1,\,b_2,\,\ldots ,\,b_{10}$$ with common ratio $$r = 2$$, hence

$$b_k = b_1 \, r^{\,k-1} = b_1 \, 2^{\,k-1}.$$

For every index $$k$$ we define $$c_k = a_k + b_k.$$ We are told that

$$c_2 = 12 \quad\text{and}\quad c_3 = 13.$$

First we write these two conditions in terms of $$a_1$$ and $$b_1$$.

For $$k = 2$$ we obtain

$$\begin{aligned} c_2 &= a_2 + b_2 \\ &= \bigl[a_1 + (2-1)(-3)\bigr] + \bigl[b_1 \, 2^{\,2-1}\bigr] \\ &= a_1 - 3 + 2\,b_1. \end{aligned}$$

Given $$c_2 = 12$$, this becomes

$$a_1 - 3 + 2\,b_1 = 12 \quad\text{(1)}.$$

For $$k = 3$$ we have

$$\begin{aligned} c_3 &= a_3 + b_3 \\ &= \bigl[a_1 + (3-1)(-3)\bigr] + \bigl[b_1 \, 2^{\,3-1}\bigr] \\ &= a_1 - 6 + 4\,b_1. \end{aligned}$$

Given $$c_3 = 13$$, we get

$$a_1 - 6 + 4\,b_1 = 13 \quad\text{(2)}.$$

Now we solve the simultaneous linear equations (1) and (2).

Subtract equation (1) from equation (2):

$$\bigl(a_1 - 6 + 4\,b_1\bigr) - \bigl(a_1 - 3 + 2\,b_1\bigr) = 13 - 12.$$

On the left, $$a_1$$ cancels and we are left with

$$-6 + 4\,b_1 + 3 - 2\,b_1 = 1.$$

This simplifies to

$$-3 + 2\,b_1 = 1,$$

so

$$2\,b_1 = 4 \quad\Longrightarrow\quad b_1 = 2.$$

Substituting $$b_1 = 2$$ back into equation (1) gives

$$a_1 - 3 + 2(2) = 12 \quad\Longrightarrow\quad a_1 - 3 + 4 = 12,$$

hence

$$a_1 + 1 = 12 \quad\Longrightarrow\quad a_1 = 11.$$

Now we want $$\displaystyle \sum_{k=1}^{10} c_k = \sum_{k=1}^{10} a_k \;+\; \sum_{k=1}^{10} b_k.$$

We compute the two sums separately.

Sum of the arithmetic progression. For an A.P. with first term $$a_1 = 11$$, common difference $$d = -3$$ and number of terms $$n = 10$$, the sum formula is

$$S_{\text{AP}} = \frac{n}{2}\,\Bigl[2a_1 + (n-1)d\Bigr].$$

Substituting the values,

$$\begin{aligned} S_{\text{AP}} &= \frac{10}{2}\,\Bigl[2(11) + 9(-3)\Bigr] \\ &= 5\,\Bigl[22 - 27\Bigr] \\ &= 5\,(-5) \\ &= -25. \end{aligned}$$

Sum of the geometric progression. For a G.P. with first term $$b_1 = 2$$, common ratio $$r = 2$$ and number of terms $$n = 10$$, the sum formula is

$$S_{\text{GP}} = b_1 \,\frac{r^{\,n}-1}{r-1}.$$

Substituting in the numbers,

$$\begin{aligned} S_{\text{GP}} &= 2\,\frac{2^{10} - 1}{2 - 1} \\ &= 2\,(1024 - 1) \\ &= 2 \times 1023 \\ &= 2046. \end{aligned}$$

Total sum. Adding the two results,

$$\sum_{k=1}^{10} c_k = S_{\text{AP}} + S_{\text{GP}} = -25 + 2046 = 2021.$$

So, the answer is $$2021$$.

Hence, the correct answer is Option A.