Key Concepts and Formulas

• Sum of a Geometric Progression (GP): For a GP with first term $A$, common ratio $r$, and $n$ terms, the sum $S_n$ is given by: $S_n = \frac{A(r^n - 1)}{r - 1} \quad \text{when } r \ne 1$
• Sum of an Arithmetic Progression (AP): For an AP with first term $A$, common difference $D$, and $n$ terms, the sum $S_n$ is given by: $S_n = \frac{n}{2}(2A + (n-1)D)$

Step-by-Step Solution

Step 1: Deconstruct the given series into two separate progressions. The given series is $S = \{x + ka\} + \{x^2 + (k + 2)a\} + \{x^3 + (k + 4)a\} + \{x^4 + (k + 6)a\} + \dots$ for 9 terms. We can split this sum into two parts: a sum of powers of $x$ and a sum of terms involving $a$. $S = (x + x^2 + x^3 + \dots + x^9) + (ka + (k+2)a + (k+4)a + \dots + \text{9th term involving } a)$ Let $S_1 = x + x^2 + x^3 + \dots + x^9$ and $S_2 = ka + (k+2)a + (k+4)a + \dots + \text{9th term involving } a$. Thus, $S = S_1 + S_2$.

Step 2: Calculate the sum of the first part ($S_1$), which is a Geometric Progression. The series $S_1 = x + x^2 + x^3 + \dots + x^9$ is a Geometric Progression with: First term ($A$) = $x$ Common ratio ($r$) = $x$ Number of terms ($n$) = 9 Since $x \ne 1$, we use the GP sum formula: $S_1 = \frac{A(r^n - 1)}{r - 1} = \frac{x(x^9 - 1)}{x - 1} = \frac{x^{10} - x}{x - 1}$

Step 3: Calculate the sum of the second part ($S_2$), which is related to an Arithmetic Progression. The series $S_2 = ka + (k+2)a + (k+4)a + \dots$ can be rewritten by factoring out $a$: $S_2 = a [k + (k+2) + (k+4) + \dots]$ The terms inside the bracket form an Arithmetic Progression with: First term ($A'$) = $k$ Common difference ($D$) = $(k+2) - k = 2$ Number of terms ($n$) = 9 Using the AP sum formula for the terms inside the bracket: $S_{\text{AP part}} = \frac{n}{2}(2A' + (n-1)D) = \frac{9}{2}(2k + (9-1)2) = \frac{9}{2}(2k + 8 \times 2) = \frac{9}{2}(2k + 16)$ $S_{\text{AP part}} = \frac{9}{2} \times 2(k + 8) = 9(k + 8)$ Therefore, $S_2 = a \times S_{\text{AP part}} = 9a(k + 8)$.

Step 4: Combine the sums of $S_1$ and $S_2$ to get the total sum $S$. $S = S_1 + S_2 = \frac{x^{10} - x}{x - 1} + 9a(k + 8)$ To compare this with the given expression for $S$, we express $S$ with a common denominator $(x-1)$: $S = \frac{x^{10} - x}{x - 1} + \frac{9a(k + 8)(x - 1)}{x - 1} = \frac{x^{10} - x + 9a(k + 8)(x - 1)}{x - 1}$

Step 5: Equate the derived sum with the given sum and solve for $k$. The problem states that $S = \frac{x^{10} - x + 45a(x - 1)}{x - 1}$. Comparing our derived sum with the given sum: $\frac{x^{10} - x + 9a(k + 8)(x - 1)}{x - 1} = \frac{x^{10} - x + 45a(x - 1)}{x - 1}$ Since the denominators are equal and $x \ne 1$, the numerators must be equal. Also, the terms $x^{10} - x$ are identical on both sides. Therefore, the remaining terms must be equal: $9a(k + 8)(x - 1) = 45a(x - 1)$ Given that $a \ne 0$ and $x \ne 1$, we can divide both sides by $a(x - 1)$: $9(k + 8) = 45$ Divide by 9: $k + 8 = \frac{45}{9}$ $k + 8 = 5$ Subtract 8 from both sides: $k = 5 - 8$ $k = -3$

Common Mistakes & Tips

• Incorrectly identifying AP/GP parameters: Double-check the first term, common difference/ratio, and number of terms for each progression.
• Algebraic errors with common denominators: Ensure that when combining $S_1$ and $S_2$, the term $9a(k+8)$ is correctly multiplied by $(x-1)$.
• Cancellation conditions: Remember that division by $a$ and $(x-1)$ is only valid because the problem states $a \ne 0$ and $x \ne 1$.

Summary

The problem involves finding the sum of a series that can be decomposed into a Geometric Progression and an Arithmetic Progression. By calculating the sum of each progression separately, combining them, and equating the result to the given sum expression, we can form an equation to solve for the unknown constant $k$. The derived sum for the series is $\frac{x^{10} - x + 9a(k + 8)(x - 1)}{x - 1}$, and equating this to the given sum $\frac{x^{10} - x + 45a(x - 1)}{x - 1}$ leads to the equation $9a(k + 8)(x - 1) = 45a(x - 1)$. After simplifying and solving for $k$, we find $k = -3$.

The final answer is $\boxed{-3}$.