Let a1, a2, ..., a10 be an AP with common difference -3 and b1, b2, ..., b10 be a GP with common rat

Q: Wait, why can't I just find $c_1, c_2, c_3$ and try to find a pattern for $c_k$?

That is a common trap! Because $c_k$ is the sum of an AP and a GP, it is neither an AP nor a GP itself. It is a hybrid sequence. Trying to find a simple pattern for $c_k$ will lead you in circles. Always decompose the sequence into its fundamental building blocks first.

Q: Do I really need to solve for $a_1$ and $b_1$ individually?

Yes, absolutely. The sum of the AP depends on its starting term $a_1$, and the sum of the GP depends on its starting term $b_1$. Without these two anchors, you cannot calculate the total sum of the ten terms.

Q: Why is the common difference $d = -3$ and not $3$?

The problem explicitly states the common difference is $-3$. Always read the signs carefully! A negative common difference means the AP is decreasing, which will naturally lead to a smaller (or even negative) sum for the AP part.

Q: Is there a shortcut for the sum of the GP if $r=2$?

While $r=2$ simplifies the denominator to $2-1=1$, there is no 'shortcut' that avoids the formula. The formula $S_{GP} = b_1(2^n - 1)$ is already the most efficient way to handle the exponential growth.

Q: What if I made a calculation error in the system of equations?

If your $a_1$ or $b_1$ values seem strange, re-check your substitution for $c_2$ and $c_3$. Remember that $a_2 = a_1 + d$ and $a_3 = a_1 + 2d$. If you accidentally used $a_3 = a_1 + 3d$, your entire system will collapse. Precision is the soul of JEE success!

Question

Let $a_1, a_2, \dots, a_{10}$ be an $AP$ with common difference $-3$ and $b_1, b_2, \dots, b_{10}$ be a $GP$ with common ratio $2$. Let $c_k = a_k + b_k, k = 1, 2, \dots, 10$. If $c_2 = 12$ and $c_3 = 13$, then $\sum_{k=1}^{10} c_k$ is equal to .

Sigma Percentile · Accepted Answer

### The Symphony of Sequences: A Mathematical Journey

Welcome, future engineer! Today, we are going to peel back the layers of a problem that might look like a simple sequence puzzle but is actually a beautiful study in the superposition of mathematical behaviors.

We are dealing with two distinct worlds: the Arithmetic Progression (AP), which marches forward with steady, linear steps, and the Geometric Progression (GP), which leaps ahead with exponential intensity.

When we combine them into $c_k = a_k + b_k$, we are essentially creating a hybrid sequence that inherits the properties of both.

### Phase 1: Decoding the DNA of the Sequences

Before we can calculate anything, we must define our players. We are given an AP with a common difference $d = -3$ and a GP with a common ratio $r = 2$.

Our goal is to find the sum of the first ten terms of the combined sequence $c_k$. Think of $a_k$ as a traveler moving at a constant velocity, and $b_k$ as a rocket accelerating exponentially.

To find the starting point of this journey, we need the first terms, $a_1$ and $b_1$. We are given two clues: $c_2 = 12$ and $c_3 = 13$.

By expressing these in terms of our unknowns, we get:

$$c_2 = a_2 + b_2 = (a_1 - 3) + 2b_1 = 12 \ \Rightarrow a_1 + 2b_1 = 15$$

$$c_3 = a_3 + b_3 = (a_1 - 6) + 4b_1 = 13 \ \Rightarrow a_1 + 4b_1 = 19$$

This is our system of equations. By subtracting the first from the second, we isolate the GP component: $2b_1 = 4$, which gives us $b_1 = 2$.

Substituting this back, we find $a_1 = 11$. We have successfully anchored our sequences!

### Phase 2: The Power of Linearity

Now, we need the sum $\sum_{k=1}^{10} c_k$. Here is where the beauty of summation comes in.

Because summation is a linear operator, we don't have to sum $c_k$ directly. We can sum the AP and the GP separately and then add the results:

$$\sum_{k=1}^{10} c_k = \sum_{k=1}^{10} a_k + \sum_{k=1}^{10} b_k$$

This is a powerful realization. It allows us to use the standard tools in our arsenal without getting bogged down in the complexity of the combined sequence.

### Phase 3: The Final Calculation

First, let's calculate the sum of the AP. Using the formula $S_{AP} = \frac{n}{2}[2a_1 + (n-1)d]$, with $n=10$, $a_1=11$, and $d=-3$, we get:

$$S_{AP} = \frac{10}{2}[2(11) + 9(-3)] = 5[22 - 27] = 5(-5) = -25$$

Next, we calculate the sum of the GP. Using $S_{GP} = \frac{b_1(r^n - 1)}{r - 1}$, with $b_1=2$ and $r=2$:

$$S_{GP} = \frac{2(2^{10} - 1)}{2 - 1} = 2(1024 - 1) = 2046$$

Finally, we combine these two results to find the total sum:

$$\sum_{k=1}^{10} c_k = -25 + 2046 = 2021$$

Look at that! Through careful decomposition and the application of fundamental principles, we have arrived at the solution.

Remember, in JEE Advanced, it is rarely about brute force; it is about recognizing the structure of the problem and using the right tools to dismantle it. You have done exactly that.

**The final answer is 2021.**

Enter Numerical Value:

Visualized Solution

Defining the Sequences

The Combined Sequence $c_{k}$

Analyzing the Second Term $c_{2}$

First Equation in Terms of $a_{1}$ and $b_{1}$

Analyzing the Third Term $c_{3}$

Second Equation in Terms of $a_{1}$ and $b_{1}$

Solving for $b_{1}$

Solving for $a_{1}$

Decomposing the Required Sum

Calculating the Sum of the AP

Calculating the Sum of the GP

Final Calculation

The Sigma Insight: Sum of Special Series

The Symphony of Sequences

Phase 1

Phase 2

Phase 3

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Enter Numerical Value:

Visualized Solution

Defining the Sequences

The Combined Sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ck​

Analyzing the Second Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c2​

First Equation in Terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​

Analyzing the Third Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c3​

Second Equation in Terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​