JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + ...$ upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is $2026 S_{2025},$ then the absolute difference betwen $2 0^{t h}$ and $1 5^{t h}$ terms of the A.P. is equal to

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Progressive Hints

Hint 1

Hint 1: Look closely at the denominators of the series: 2, 6, 12, 20... These are not random numbers. They are products of consecutive integers: 1x2, 2x3, 3x4, 4x5. This is the hidden geometric reality you must visualize to unlock the pattern.

Hint 2

Hint 2: Once you identify the general term as

T_{k} = \frac{1}{k ( k + 1 )}

, you need to use the method of partial fractions. Rewrite the numerator as

(k + 1) - k

to split the term into a difference of two simpler fractions. This is the classic 'telescoping' technique.

Hint 3

Hint 3: After you simplify the sum to

S_{n} = 1 - \frac{1}{n + 1}

, calculate

S_{2025}

. Then, equate the sum of the A.P. terms,

S_{6}^{'} = 9 p

, to the expression

2026 \cdot S_{2025}

. Remember, the absolute difference between the

2 0^{t h}

and

1 5^{t h}

terms of an A.P. is simply

∣5 d ∣

, which is

∣5 p ∣

Visualized Solution

Analyzing the Series $S_{n}$

Given series: $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots$
Observe the denominators: $2 = 1 \times 2$ , $6 = 2 \times 3$ , $12 = 3 \times 4$ , $20 = 4 \times 5$ .

Defining the General Term $T_{k}$

General term $T_{k} = \frac{1}{k ( k + 1 )}$
Here, $k$ ranges from $1$ to $n$ .

Applying the Method of Difference

Rewrite the numerator: $T_{k} = \frac{( k + 1 ) - k}{k ( k + 1 )}$
Split the fraction: $T_{k} = \frac{k + 1}{k ( k + 1 )} - \frac{k}{k ( k + 1 )}$
Simplified form: $T_{k} = \frac{1}{k} - \frac{1}{k + 1}$

Summing the Telescoping Series

$S_{n} = \sum_{k = 1}^{n} (\frac{1}{k} - \frac{1}{k + 1})$
Expanding the sum: $S_{n} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n + 1})$

Evaluating $S_{2025}$

After cancellation: $S_{n} = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$
Substitute $n = 2025$ into the formula: $S_{2025} = \frac{2025}{2025 + 1}$
Result: $S_{2025} = \frac{2025}{2026}$

Arithmetic Progression Parameters

A.P. parameters: First term $a = - p$ , Common difference $d = p$
Sum of first $n$ terms: $S_{n}^{'} = \frac{n}{2} [2 a + (n - 1) d]$

Calculating the Sum of A.P. Terms

For $n = 6$ : $S_{6}^{'} = \frac{6}{2} [2 (- p) + (6 - 1) p]$
Simplify: $S_{6}^{'} = 3 [- 2 p + 5 p]$
Final sum in terms of $p$ : $S_{6}^{'} = 3 [3 p] = 9 p$

Equating the Expressions

Given condition: $S_{6}^{'} = 2026 \cdot S_{2025}$
Substitute values: $9 p = 2026 \cdot \frac{2025}{2026}$
Simplify: $9 p = 2025$

Solving for $p$

Solve for $p$ : $9 p = 45 \Rightarrow p = 5$
Therefore, $a = - 5$ and $d = 5$

Difference of A.P. Terms

$2 0^{t h}$ term: $T_{20} = a + 19 d$
$1 5^{t h}$ term: $T_{15} = a + 14 d$
Absolute difference: $∣ T_{20} - T_{15} ∣ = ∣ (a + 19 d) - (a + 14 d) ∣$
Simplify: $∣5 d ∣ = ∣5 p ∣$

Final Answer

Since $p = 5$ , $∣5 p ∣ = 5 \times 5 = 25$

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

Analyzing the Symphony of Series

Imagine you are standing at the edge of a mathematical landscape, looking at a series that seems chaotic at first glance:

S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots

It looks intimidating, but in the world of JEE, chaos is just order waiting to be discovered. Let us decode this together.

Decoding the Telescoping Series

Look at the denominators:

2

1 \times 2

6

2 \times 3

12

3 \times 4

, and

20

4 \times 5

. The general term is:

T_{k} = \frac{1}{k ( k + 1 )}

We use the method of difference, a favorite tool of examiners. We rewrite the numerator as

(k + 1) - k

, which allows us to split the fraction:

T_{k} = \frac{k + 1}{k ( k + 1 )} - \frac{k}{k ( k + 1 )} = \frac{1}{k} - \frac{1}{k + 1}

When we sum these, we get:

S_{n} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n + 1})

Watch as the middle terms vanish, leaving only:

S_{n} = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}

For

n = 2025

, we obtain:

S_{2025} = \frac{2025}{2026}

The Bridge to Arithmetic Progressions

Now, we pivot to the second act. We have an A.P. with first term

a = - p

and common difference

d = p

. The sum of the first six terms is:

S_{6}^{'} = \frac{6}{2} [2 (- p) + (6 - 1) p] = 3 [- 2 p + 5 p] = 9 p

The problem states this sum is equal to

2026 \cdot S_{2025}

. Substituting our previous result, we get:

9 p = 2026 \cdot \frac{2025}{2026}

The

2026

cancels out beautifully, leaving

9 p = 2025

, which simplifies to

9 p = 45

. Thus, we find $p = 5$ .

The Final Elegance

We are asked for the absolute difference between the

2 0^{t h}

and

1 5^{t h}

terms. Using the general term formula

T_{n} = a + (n - 1) d

T_{20} = a + 19 d

T_{15} = a + 14 d

Subtracting them gives:

∣ T_{20} - T_{15} ∣ = ∣ (a + 19 d) - (a + 14 d) ∣ = ∣5 d ∣ = ∣5 p ∣

Since

p = 5

, the final result is:

$∣5 \times 5∣ = 25$

See how the complexity dissolved? That is the power of breaking problems down. You have mastered this!

Common Traps & FAQs

Wait, why do we call this a 'telescoping' series? What does that even mean?

Think of a handheld telescope that collapses into itself. In this series, when you write

T_{k} = \frac{1}{k} - \frac{1}{k + 1}

, the terms cancel out like the segments of a telescope sliding into one another. The

- \frac{1}{2}

cancels with

+ \frac{1}{2}

, the

- \frac{1}{3}

with

+ \frac{1}{3}

, and so on, leaving only the first and last pieces standing. It is a beautiful way to simplify a complex sum!

I got

S_{n} = \frac{n}{n + 1}

, but why is the sum of the A.P.

9 p

? Did I mess up the formula?

Not at all! You are likely on the right track. The sum of the first

n

terms of an A.P. is

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

. With

n = 6

a = - p

, and

d = p

, we get

\frac{6}{2} [2 (- p) + 5 p] = 3 [3 p] = 9 p

. Your math is solid; keep going!

Why do we need the absolute difference

∣ T_{20} - T_{15} ∣

? Can't I just calculate the terms individually?

You absolutely could calculate them individually, but that's the long way around! Notice that

T_{20} = a + 19 d

and

T_{15} = a + 14 d

. When you subtract them, the

a

cancels out immediately. It's a beautiful algebraic shortcut that saves you time and reduces the chance of calculation errors. Always look for these cancellations!

Select Answer:

Visualized Solution

Analyzing the Series $S_{n}$

Defining the General Term $T_{k}$

Applying the Method of Difference

Summing the Telescoping Series

Evaluating $S_{2025}$

Arithmetic Progression Parameters

Calculating the Sum of A.P. Terms

Equating the Expressions

Solving for $p$

Difference of A.P. Terms

Final Answer

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

Analyzing the Symphony of Series

Decoding the Telescoping Series

The Bridge to Arithmetic Progressions

The Final Elegance

Similar Questions

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Select Answer:

Visualized Solution

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

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

Applying the Method of Difference

Summing the Telescoping Series

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2025​

Arithmetic Progression Parameters

Calculating the Sum of A.P. Terms

Equating the Expressions

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

Difference of A.P. Terms

Final Answer

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

Analyzing the Symphony of Series

Decoding the Telescoping Series

The Bridge to Arithmetic Progressions

The Final Elegance

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Analyzing the Series $S_{n}$

Defining the General Term $T_{k}$

Evaluating $S_{2025}$

Solving for $p$

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

In an A.P., the sixth terms $a_{6} = 2$ . If the $a_{1} a_{4} a_{5}$ is the greatest, then the common difference of the A.P., is equal to

If the $1 0^{t h}$ term of an A.P. is $\frac{1}{20}$ and its $2 0^{t h}$ term is $\frac{1}{10}$ , then the sum of its first 200 term is

If the variance of the terms in an increasing A.P., $b_{1}, b_{2}, b_{3}, \dots, b_{11}$ is 90, then the common difference of this A.P. is