Let S be the sum of the first 9 terms of the series: {x+ka} + {x² + (k+2)a} + {x³ + (k+4)a} + {x⁴ +

Q: Wait, why do we split the series into two? Can't we just use the formula for the sum of an Arithmetico-Geometric Series?

While you could technically treat this as an Arithmetico-Geometric series, splitting it is much cleaner and less prone to calculation errors. By separating the powers of $x$ (the G.P.) from the multiples of $a$ (the A.P.), you isolate the variables, making the comparison with the given expression straightforward and intuitive.

Q: Why did we split the fraction $\frac{x^{10}-x+45a(x-1)}{x-1}$?

Great question! We split it to match the structure of our calculated sum. Our calculated sum was $S_1 + S_2$. By splitting the given expression into $\frac{x^{10}-x}{x-1} + \frac{45a(x-1)}{x-1}$, we immediately see that the first part is exactly our $S_1$. This allows us to equate the remaining parts ($S_2$ and $45a$) directly.

Q: Is it mathematically sound to divide by $9a$ to find $k$?

Yes, it is perfectly sound! The problem explicitly states that $a \neq 0$. In algebra, we can only divide by an expression if we are certain it is non-zero. Since $a$ is non-zero, $9a$ is also non-zero, allowing us to divide both sides safely to isolate $k$.

Question

Let $S$ be the sum of the first 9 terms of the series: ${x+ka} + {x^2 + (k+2)a} + {x^3 + (k+4)a} + {x^4 + (k+6)a} +...$ where $a 
eq 0$ and $x 
eq 1$ If $S = \frac{x^{10}-x+45a(x-1)}{x-1}$, then $k$ is equal to :

Options:

(A) -3

(B) 1

(C) -5

(D) 3

Sigma Percentile · Accepted Answer

### The Art of Decomposition: Unraveling the Mixed Series

Imagine you are standing before a complex, intimidating structure. At first glance, it looks like a chaotic jumble of terms: $(x+ka) + (x^2 + (k+2)a) + (x^3 + (k+4)a) + \dots$ for nine terms.

It is easy to feel overwhelmed, but in the world of JEE Advanced, intimidation is just a sign that you need to change your perspective. The secret to mastering such problems is not brute force, but the elegant art of decomposition.

### Phase 1: The Divide and Conquer Strategy

When you look at the series, don't see one big, scary expression. Look for the rhythm.

Notice that every single term is a sum of two distinct components: a power of $x$ and a multiple of $a$. We have $x, x^2, x^3, \dots$ and we have $ka, (k+2)a, (k+4)a, \dots$.

This is a classic mixed series. Our strategy is simple: divide and conquer. We will split this into two separate series, $S_1$ and $S_2$, and solve them individually.

### Phase 2: The Geometric Progression ($S_1$)

Let us isolate the $x$ terms: $S_1 = x + x^2 + x^3 + \dots + x^9$. This is a beautiful, classic Geometric Progression (G.P.).

The first term is $x$, and the common ratio is $x$. We know the sum formula for a G.P. with $n$ terms is:

$$S_n = \frac{a(r^n - 1)}{r - 1}$$

Substituting our values, where $a=x$, $r=x$, and $n=9$, we get:

$$S_1 = \frac{x(x^9 - 1)}{x - 1} = \frac{x^{10} - x}{x - 1}$$

This is the first piece of our puzzle.

### Phase 3: The Arithmetic Progression ($S_2$)

Now, let us turn our attention to the $a$ terms: $S_2 = ka + (k+2)a + (k+4)a + \dots$ for nine terms. Look at the coefficients of $a$: $k, k+2, k+4, \dots$.

This is an Arithmetic Progression (A.P.) where the first term $A = ka$ and the common difference $D = 2a$. The sum formula for an A.P. is:

$$S_n = \frac{n}{2} [2A + (n-1)D]$$

With $n=9$, we calculate:

$$S_2 = \frac{9}{2} [2(ka) + (9-1)(2a)]$$

Simplifying the inside, we get $\frac{9}{2} [2ka + 16a]$. Factoring out the $2$, we get $9(ka + 8a)$, which simplifies to $9a(k+8)$.

### Phase 4: The Synthesis and The Reveal

We have our two components. The total sum $S$ is:

$$S = S_1 + S_2 = \frac{x^{10} - x}{x - 1} + 9a(k+8)$$

Now, look at the expression given in the problem:

$$S = \frac{x^{10}-x+45a(x-1)}{x-1}$$

If we split this fraction, we get:

$$\frac{x^{10}-x}{x-1} + \frac{45a(x-1)}{x-1}$$

Since $x 
eq 1$, we can cancel the $(x-1)$ term, leaving us with $\frac{x^{10}-x}{x-1} + 45a$.

Now, the magic happens. We equate our calculated sum with the given sum:

$$\frac{x^{10}-x}{x-1} + 9a(k+8) = \frac{x^{10}-x}{x-1} + 45a$$

The complex G.P. terms cancel out perfectly, leaving us with the simple equation $9a(k+8) = 45a$.

Since $a 
eq 0$, we divide by $9a$ to find $k+8 = 5$, which leads us to the final answer: **$k = -3$**.

Animated Solution for Mathematics - Sequence and Series: Let $S$ be the sum of the first 9 terms of the series: $x + k a + x^{2} + (k + 2) a + x^{3} + (k + 4) a + x^{4} + (k + 6) a + ...$ where $a \neq = 0$ and $x \neq = 1$ If $S = \frac{x ^{10} - x + 45 a ( x - 1 )}{x - 1}$ , then $k$ is equal to :

Select Answer:

Visualized Solution

Analyze the Series Structure

Split into Two Series

Identify the Geometric Progression $S_{1}$

Apply the G.P. Sum Formula

Identify the Arithmetic Progression $S_{2}$

Apply the A.P. Sum Formula

Simplify the A.P. Sum $S_{2}$

Combine Both Sums

Analyze the Given Expression

Simplify the Given Expression

Compare and Equate

Solve for $k$

The Sigma Insight: Sum of Special Series

The Art of Decomposition

Phase 1

Phase 2

Phase 3

Phase 4

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