Let a1, a2, a3, ... be in an arithmetic progression of positive terms. Let Ak = a1² - a2² + a3² - a4

Q: Why can't we just solve for $a_1$ and $d$ using $A_3$ and $A_5$ without deriving the formula?

You absolutely can! But deriving the formula $A_k = -kd [ 2a_1 + (2k-1)d ]$ is like building a master key. It saves you from tedious algebra and reduces the chance of calculation errors. In JEE Advanced, efficiency is as important as accuracy. Think of it as sharpening your axe before chopping the tree.

Q: Why did we reject the negative root for $d$?

Great catch! The problem explicitly states that the terms of the AP are positive. If $d$ were negative, the terms would eventually become negative as $n$ increases. We must always respect the constraints given in the problem statement—they are the boundaries of our mathematical universe.

Q: Is there a faster way to calculate $A_7$?

Once you have $a_1 = 1$ and $d = 3$, you have the 'DNA' of the sequence. Using the formula we derived is the fastest, most elegant way. Don't try to write out all 14 terms; trust the algebra and the pattern you've uncovered!

Question

Let $a_1, a_2, a_3, \dots$ be in an arithmetic progression of positive terms. Let $A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2k-1}^2 - a_{2k}^2$. If $A_3 = -153, A_5 = -435$ and $a_1^2 + a_2^2 + a_3^2 = 66$, then $a_{17} - A_7$ is equal to

Sigma Percentile · Accepted Answer

### The Symphony of Sequences: Unlocking the Alternating Sum

Welcome, future engineer. Today, we aren't just solving a problem; we are peeling back the layers of an arithmetic progression to reveal a hidden, elegant structure.

When you first look at the expression for $A_k$:

$$A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2k-1}^2 - a_{2k}^2$$

It might look like a chaotic mess of squares. But in the world of JEE Advanced, chaos is just order waiting to be discovered.

### Phase 1: The Power of Pairing

Imagine you are standing before this series. Your instinct might be to expand every term, but that is a trap. Instead, look at the pairs: $(a_1^2 - a_2^2)$, $(a_3^2 - a_4^2)$, and so on.

Does this remind you of something? It is the classic difference of squares identity: $x^2 - y^2 = (x-y)(x+y)$.

When we apply this to our series, each pair transforms. Since we are dealing with an arithmetic progression where the common difference is $d$, we know that $a_{i+1} - a_i = d$. Therefore, $a_i - a_{i+1} = -d$.

Suddenly, the complexity collapses. Each pair becomes $-d(a_i + a_{i+1})$. We have successfully linearized the problem!

### Phase 2: The General Term Transformation

Now, we move to the summation. We have:

$$A_k = -d \sum_{i=1}^k (a_{2i-1} + a_{2i})$$

This looks much friendlier, doesn't it? We know the general term of an AP is $a_n = a_1 + (n-1)d$.

Let's substitute this into our sum. The term $a_{2i-1}$ becomes $a_1 + (2i-2)d$, and $a_{2i}$ becomes $a_1 + (2i-1)d$. Adding them together, we get $2a_1 + (4i-3)d$.

Don't let the summation notation intimidate you. We are simply summing $2a_1$ and $(4i-3)d$ from $i=1$ to $k$.

The sum of $2a_1$ is simply $2ka_1$. The sum of $(4i-3)$ is a standard arithmetic series sum, which simplifies to $k(2k-1)$. Putting it all together, we arrive at our master formula:

$$A_k = -kd [ 2a_1 + (2k-1)d ]$$

This is the 'DNA' of our series.

### Phase 3: Solving the System

With our formula in hand, the problem becomes a game of substitution. We are given $A_3 = -153$ and $A_5 = -435$. By plugging $k=3$ and $k=5$ into our formula, we generate a system of two equations.

For $k=3$: 
$$d(2a_1 + 5d) = 51$$

For $k=5$: 
$$d(2a_1 + 9d) = 87$$

Subtracting these equations is where the magic happens. The $2a_1d$ terms cancel out, leaving us with $4d^2 = 36$. This gives us $d^2 = 9$.

Since the problem guarantees positive terms, we confidently choose $d=3$. Substituting this back, we find $a_1 = 1$. We have cracked the code!

### Phase 4: The Final Victory

We verify our findings with the condition $a_1^2 + a_2^2 + a_3^2 = 66$. With $a_1=1$ and $d=3$, our terms are $1, 4, 7$. Indeed, $1^2 + 4^2 + 7^2 = 1 + 16 + 49 = 66$. It fits perfectly.

Finally, we calculate $a_{17} - A_7$. We find $a_{17} = a_1 + 16d = 1 + 48 = 49$.

Using our formula for $A_7$, we get:

$$A_7 = -7(3) [ 2(1) + (13)(3) ] = -21 [ 2 + 39 ] = -21(41) = -861$$

The final step is a simple subtraction: $49 - (-861) = 910$.

See? What started as a terrifying series of squares ended as a beautiful, logical progression. You didn't just solve a problem; you mastered the pattern. The final answer is **910**.

Enter Numerical Value:

Visualized Solution

Defining the Arithmetic Progression

Simplifying the General Term using $x^{2} - y^{2}$

Expressing the Sum in terms of $a_{1}$ and $d$

Evaluating the Summation

Setting up Equations for $A_{3}$ and $A_{5}$

Solving for Common Difference $d$

Solving for First Term $a_{1}$

Verification with Third Condition

Calculating $a_{17}$ and $A_{7}$

Final Calculation: $a_{17} - A_{7}$

The Sigma Insight: Arithmetic Progression (A.P.)

The Symphony of Sequences

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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Enter Numerical Value:

Visualized Solution

Defining the Arithmetic Progression

Simplifying the General Term using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−y2

Expressing the Sum 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; } d

Evaluating the Summation

Setting up Equations for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A5​

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

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