If 1/((20-a)(40-a)) + 1/((40-a)(60-a)) + ... + 1/((180-a)(200-a)) = 1/256, then the maximum value of

Q: Why do we multiply and divide by 20? Can't we just solve it directly?

Great question! We multiply and divide by $20$ because the difference between the factors in the denominator is exactly $20$. By doing this, we can express the numerator as the difference of the two factors, which allows us to split the fraction into two simpler parts. Without this step, the telescoping pattern remains hidden and the algebra becomes unmanageable.

Q: How do I know which terms cancel out in a telescoping series?

Think of it like a chain reaction. When you write out the terms, you will see that the second part of the first term is $\frac{-1}{40-a}$, and the first part of the second term is $\frac{1}{40-a}$. They are identical in magnitude but opposite in sign, so they cancel out perfectly. This continues for all middle terms, leaving only the very first and very last components of the series.

Q: Is there any restriction on the value of $a$?

Yes, absolutely! In the original expression, the denominators cannot be zero. Therefore, $20-a \neq 0$, $40-a \neq 0$, and so on, up to $200-a \neq 0$. This means $a$ cannot be any of the values $20, 40, 60, \dots, 200$. Our calculated values of $a=212$ and $a=8$ satisfy these conditions, so both are valid solutions.

Question

If $\frac{1}{(20-a)(40-a)} + \frac{1}{(40-a)(60-a)} + \dots + \frac{1}{(180-a)(200-a)} = \frac{1}{256}$, then the maximum value of $a$ is :

Options:

(A) 198

(B) 202

(C) 212

(D) 218

Sigma Percentile · Accepted Answer

### The Beauty of Telescoping Series: A Journey into Patterns

Imagine you are standing before a complex, intimidating series. It looks like a wall of fractions, each one seemingly unrelated to the next.

But as an elite JEE aspirant, you know that math is rarely about brute force; it is about finding the hidden rhythm, the underlying symmetry that makes the chaos collapse into simplicity.

Let us embark on this journey together to solve the problem:

$$ \frac{1}{(20-a)(40-a)} + \frac{1}{(40-a)(60-a)} + \dots + \frac{1}{(180-a)(200-a)} = \frac{1}{256} $$

### Phase 1: The Pattern Recognition

Look closely at the denominators. We have terms like $(20-a)(40-a)$, $(40-a)(60-a)$, and so on.

If you subtract the first factor from the second in any of these terms, what do you get? For the first term, $(40-a) - (20-a) = 20$. For the second, $(60-a) - (40-a) = 20$.

This constant difference of $20$ is the heartbeat of this problem. It is a signal that we are dealing with a **telescoping series**, a beautiful structure where terms cancel each other out like falling dominoes.

### Phase 2: The Telescoping Magic

To reveal this structure, we use a classic JEE technique: multiply and divide the entire series by the common difference, which is $20$.

Now, each term $T_k$ looks like:

$$ T_k = \frac{1}{20} \left[ \frac{20}{(20k-a)(20(k+1)-a)} ight] $$

We can rewrite the numerator $20$ as the difference of the denominator factors: $(20(k+1)-a) - (20k-a)$. This allows us to decompose each term into partial fractions:

$$ T_k = \frac{1}{20} \left( \frac{1}{20k-a} - \frac{1}{20(k+1)-a} ight) $$

### Phase 3: The Algebraic Cleanup

Now, let us expand the sum. When we write out the terms, we see:

$$ S = \frac{1}{20} \left[ \left( \frac{1}{20-a} - \frac{1}{40-a} ight) + \left( \frac{1}{40-a} - \frac{1}{60-a} ight) + \dots + \left( \frac{1}{180-a} - \frac{1}{200-a} ight) ight] $$

Notice the magic? The term $-\frac{1}{40-a}$ cancels with $+\frac{1}{40-a}$. The term $-\frac{1}{60-a}$ will cancel with the next one.

This continues until only the very first term, $\frac{1}{20-a}$, and the very last term, $-\frac{1}{200-a}$, remain. We are left with:

$$ S = \frac{1}{20} \left( \frac{1}{20-a} - \frac{1}{200-a} ight) $$

### Phase 4: The Quadratic Finale

With the series simplified, we take the LCM:

$$ S = \frac{1}{20} \left[ \frac{(200-a) - (20-a)}{(20-a)(200-a)} ight] = \frac{180}{20(20-a)(200-a)} = \frac{9}{(20-a)(200-a)} $$

We equate this to the given value $\frac{1}{256}$. Cross-multiplying gives us:

$$ (20-a)(200-a) = 9 	imes 256 = 2304 $$

Expanding this, we get $4000 - 200a - 20a + a^2 = 2304$, which simplifies to the quadratic equation:

$$ a^2 - 220a + 1696 = 0 $$

Solving this quadratic, we find the roots $a=212$ and $a=8$. The question asks for the maximum value of $a$, which is clearly **$212$**.

Animated Solution for Mathematics - Sequence and Series: If $\frac{1}{( 20 - a ) ( 40 - a )} + \frac{1}{( 40 - a ) ( 60 - a )} + \dots + \frac{1}{( 180 - a ) ( 200 - a )} = \frac{1}{256}$ , then the maximum value of $a$ is :

Select Answer:

Visualized Solution

Analyze the Series Pattern

Apply Method of Differences

Decompose into Partial Fractions

Expand the Telescoping Sum

Cancel the Terms

Simplify the Summation

Equate and Solve for $a$

Form the Quadratic Equation

Solve the Quadratic Equation

Identify the Maximum Value

The Sigma Insight: Sum of Special Series

The Beauty of Telescoping Series

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (20−a)(40−a)1​+(40−a)(60−a)1​+⋯+(180−a)(200−a)1​=2561​, then the maximum value of a is :