Key Concepts and Formulas

1. Arithmetic Progression (AP): The $k^{th}$ term of an AP is given by $a_k = a_1 + (k-1)d$, where $a_1$ is the first term and $d$ is the common difference. The sum of the first $n$ terms is $S_n^A = \frac{n}{2} [2a_1 + (n-1)d]$.
2. Geometric Progression (GP): The $k^{th}$ term of a GP is given by $b_k = b_1 r^{k-1}$, where $b_1$ is the first term and $r$ is the common ratio. The sum of the first $n$ terms is $S_n^G = \frac{b_1(r^n - 1)}{r-1}$ for $r \neq 1$.
3. Sum of a Combined Sequence: If $c_k = a_k + b_k$, then the sum of the first $n$ terms of $c_k$ is $\sum_{k=1}^{n} c_k = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k$.

Step-by-Step Solution

Step 1: Define the terms of the AP and GP. We are given that $a_1, a_2, \dots, a_{10}$ is an AP with common difference $d = -3$. So, the $k^{th}$ term is $a_k = a_1 + (k-1)(-3)$. We are also given that $b_1, b_2, \dots, b_{10}$ is a GP with common ratio $r = 2$. So, the $k^{th}$ term is $b_k = b_1 (2)^{k-1}$. The sequence $c_k$ is defined as $c_k = a_k + b_k$.

Step 2: Use the given values of $c_2$ and $c_3$ to form equations. We are given $c_2 = 12$ and $c_3 = 13$. For $k=2$: $c_2 = a_2 + b_2 = 12$. Substituting the formulas for $a_2$ and $b_2$: $a_2 = a_1 + (2-1)(-3) = a_1 - 3$ $b_2 = b_1 (2)^{2-1} = 2b_1$ So, the equation for $c_2$ becomes: $(a_1 - 3) + 2b_1 = 12$, which simplifies to $a_1 + 2b_1 = 15$. (Equation 1)

For $k=3$: $c_3 = a_3 + b_3 = 13$. Substituting the formulas for $a_3$ and $b_3$: $a_3 = a_1 + (3-1)(-3) = a_1 - 6$ $b_3 = b_1 (2)^{3-1} = 4b_1$ So, the equation for $c_3$ becomes: $(a_1 - 6) + 4b_1 = 13$, which simplifies to $a_1 + 4b_1 = 19$. (Equation 2)

Step 3: Solve the system of linear equations for $a_1$ and $b_1$. We have two equations with two unknowns:

1. $a_1 + 2b_1 = 15$
2. $a_1 + 4b_1 = 19$

Subtract Equation 1 from Equation 2: $(a_1 + 4b_1) - (a_1 + 2b_1) = 19 - 15$ $2b_1 = 4$ $b_1 = 2$

Substitute the value of $b_1$ into Equation 1: $a_1 + 2(2) = 15$ $a_1 + 4 = 15$ $a_1 = 11$

So, the first term of the AP is $a_1 = 11$ and the first term of the GP is $b_1 = 2$.

Step 4: Calculate the sum of the first 10 terms of the AP. The sum of the first 10 terms of the AP is $S_{10}^A = \frac{10}{2} [2a_1 + (10-1)d]$. Substituting $a_1 = 11$, $d = -3$, and $n = 10$: $S_{10}^A = 5 [2(11) + (9)(-3)]$ $S_{10}^A = 5 [22 - 27]$ $S_{10}^A = 5 [-5]$ $S_{10}^A = -25$

Step 5: Calculate the sum of the first 10 terms of the GP. The sum of the first 10 terms of the GP is $S_{10}^G = \frac{b_1(r^{10} - 1)}{r-1}$. Substituting $b_1 = 2$, $r = 2$, and $n = 10$: $S_{10}^G = \frac{2(2^{10} - 1)}{2-1}$ $S_{10}^G = \frac{2(1024 - 1)}{1}$ $S_{10}^G = 2(1023)$ $S_{10}^G = 2046$

Step 6: Calculate the sum of the first 10 terms of the sequence $c_k$. The sum is $\sum_{k=1}^{10} c_k = S_{10}^A + S_{10}^G$. $\sum_{k=1}^{10} c_k = -25 + 2046$ $\sum_{k=1}^{10} c_k = 2021$

Common Mistakes & Tips

• Algebraic Errors: Be careful when solving the system of equations for $a_1$ and $b_1$. Small mistakes here will propagate through the entire solution.
• Formula Application: Ensure you are using the correct formulas for AP and GP sums and term definitions. Double-check the values of $n$, $a_1$, $d$, $b_1$, and $r$.
• Calculation of Powers: Accurately calculate powers of numbers, especially $2^{10}$ in this case.

Summary

The problem requires us to find the sum of the first 10 terms of a sequence $c_k$ which is formed by adding the corresponding terms of an arithmetic progression ($a_k$) and a geometric progression ($b_k$). We are given the common difference of the AP and the common ratio of the GP, along with the values of $c_2$ and $c_3$. By using these values, we set up a system of linear equations to solve for the first terms of the AP ($a_1$) and GP ($b_1$). Once $a_1$ and $b_1$ are found, we can use the standard summation formulas for AP and GP to calculate their respective sums over the first 10 terms. The total sum of $c_k$ is then the sum of these two individual sums.

The sum of the AP is $S_{10}^A = -25$. The sum of the GP is $S_{10}^G = 2046$. Therefore, $\sum_{k=1}^{10} c_k = -25 + 2046 = 2021$.

The final answer is \boxed{2021}.