JEE Main 2024 (08 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let the positive integers be written in the form : $7 ⋱ 42815396 10 ⋱$ If the $k^{th}$ row contains exactly $k$ numbers for every natural number $k$ , then the row in which the number 5310 will be, is

Q: Why can't I just divide $5310$ by the row number to find the answer?

That is a common trap! Division assumes a constant rate of change, but this pyramid grows triangularly. The number of elements in each row increases by one every time, meaning the growth is quadratic, not linear. You must use the sum formula $S_n = \frac{n(n+1)}{2}$ to account for this acceleration.

Q: Why do we use the inequality $\frac{(n-1)n}{2} < 5310 \leq \frac{n(n+1)}{2}$?

This is the 'sandwich' technique. We know $5310$ must fall after the end of the previous row (the $(n-1)^{\text{th}}$ row) but before or exactly at the end of the current row (the $n^{\text{th}}$ row). It is the most robust way to define a range without needing to solve complex equations.

Q: Is there a risk in using the approximation $n^2 \approx 10620$?

In JEE, time is currency. While the quadratic formula is exact, the approximation gets you to the neighborhood instantly. As long as you verify the boundaries (as we did with $102$ and $103$), the approximation is a perfectly valid and efficient strategy that saves precious minutes.

Enter Numerical Value:

Visualized Solution

Analyzing the Number Pyramid

The numbers are arranged in a triangular grid.
Row $1$ has $1$ number.
Row $2$ has $2$ numbers.
Row $k$ has exactly $k$ numbers.

Focusing on the Last Number

To find the row of a specific number, we need reference points.
The most predictable reference point is the last number of each row.

Pattern of the Last Numbers

Row $1$ ends with $1$ .
Row $2$ ends with $3$ ( $1 + 2$ ).
Row $3$ ends with $6$ ( $1 + 2 + 3$ ).
Row $4$ ends with $10$ ( $1 + 2 + 3 + 4$ ).

Formula for the Last Number

The last number of row $n$ is the sum of the first $n$ natural numbers.
$S_{n} = 1 + 2 + 3 + \dots + n$
$S_{n} = \frac{n ( n + 1 )}{2}$

Locating $5310$

Let the number $5310$ be in the $n^{th}$ row.
It must be strictly greater than the last number of row $(n - 1)$ .
It must be less than or equal to the last number of row $n$ .

Setting up the Inequality

$Last number of row (n - 1) < 5310 \leq Last number of row n$
$\frac{( n - 1 ) n}{2} < 5310 \leq \frac{n ( n + 1 )}{2}$

Approximating $n$

Solving the exact quadratic inequality is tedious.
Let's use an approximation: $\frac{n ( n + 1 )}{2} \approx \frac{n ^{2}}{2}$
$\frac{n ^{2}}{2} \approx 5310$

Calculating the Approximate Square

Multiply by $2$ :
$n^{2} \approx 5310 \times 2$
$n^{2} \approx 10620$

Estimating the Square Root

We need to find $10620$ .
We know $10 0^{2} = 10000$ .
Let's check $10 3^{2}$ : $103 \times 103 = 10609$ .
So, $n$ is very close to $103$ .

Testing the Boundary for $n = 102$

Let's find the last number of row $102$ .
$S_{102} = \frac{102 \times 103}{2}$
$S_{102} = 51 \times 103 = 5253$

Testing the Boundary for $n = 103$

Now, let's find the last number of row $103$ .
$S_{103} = \frac{103 \times 104}{2}$
$S_{103} = 103 \times 52 = 5356$

Final Conclusion

We have: $5253 < 5310 \leq 5356$
This perfectly matches our condition: $S_{102} < 5310 \leq S_{103}$
Therefore, the number $5310$ is located in the $10 3^{rd}$ row.

00:00 / 00:00

The Sigma Insight: Sum of Special Series

Solution Diagram

Analyzing the Structure

The pyramid is organized such that the

n^{th}

row contains exactly

n

integers. This structure is governed by the sequence of triangular numbers.

The last number of the

n^{th}

row, which we call the "fence post," is defined by the sum of the first

n

integers:

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

The Master Inequality

To locate the row containing the number

5310

, we must identify the integer

n

such that

5310

falls between the end of the previous row and the end of the current row. This is expressed by the following inequality:

\frac{( n - 1 ) n}{2} < 5310 \leq \frac{n ( n + 1 )}{2}

Estimation Strategy

In the context of a competitive exam, we avoid tedious algebraic solving by using estimation. We approximate the expression as:

\frac{n ^{2}}{2} \approx 5310 \Rightarrow n^{2} \approx 10620

Since

10 0^{2} = 10000

, we know that

n

must be slightly larger than

100

. Testing values near

103

, we observe that

10 3^{2} = 10609

, which provides a very close approximation.

Final Verification

We now verify the boundaries by calculating the exact values for

n = 102

and

n = 103

For

n = 102

S_{102} = \frac{102 \times 103}{2} = 5253

For

n = 103

S_{103} = \frac{103 \times 104}{2} = 5356

Since

5253 < 5310 \leq 5356

, the number

5310

must reside within the

10 3^{rd}

row.

The number $5310$ is located in row $103$ .

Common Traps & FAQs

Why can't I just divide

5310

by the row number to find the answer?

That is a common trap! Division assumes a constant rate of change, but this pyramid grows triangularly. The number of elements in each row increases by one every time, meaning the growth is quadratic, not linear. You must use the sum formula

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

to account for this acceleration.

Why do we use the inequality

\frac{( n - 1 ) n}{2} < 5310 \leq \frac{n ( n + 1 )}{2}

This is the 'sandwich' technique. We know

5310

must fall after the end of the previous row (the

(n - 1)^{th}

row) but before or exactly at the end of the current row (the

n^{th}

row). It is the most robust way to define a range without needing to solve complex equations.

Is there a risk in using the approximation

n^{2} \approx 10620

In JEE, time is currency. While the quadratic formula is exact, the approximation gets you to the neighborhood instantly. As long as you verify the boundaries (as we did with

102

and

103

), the approximation is a perfectly valid and efficient strategy that saves precious minutes.

Animated Solution for Mathematics - Sequence and Series: Let the positive integers be written in the form : $7 ⋱ 42815396 10 ⋱$ If the $k^{th}$ row contains exactly $k$ numbers for every natural number $k$ , then the row in which the number 5310 will be, is

Enter Numerical Value:

Visualized Solution

Analyzing the Number Pyramid

Focusing on the Last Number

Pattern of the Last Numbers

Formula for the Last Number

Locating $5310$

Setting up the Inequality

Approximating $n$

Calculating the Approximate Square

Estimating the Square Root

Testing the Boundary for $n = 102$

Testing the Boundary for $n = 103$

Final Conclusion

The Sigma Insight: Sum of Special Series

Analyzing the Structure

The Master Inequality

Estimation Strategy

Final Verification

Similar Questions

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{1} + a_{2} + \dots + a_{n} = \frac{n ^{2} + 3 n}{( n + 1 ) ( n + 2 )}$ . If $28 \sum_{k = 1}^{10} \frac{1}{a _{k}} = p_{1} p_{2} p_{3} \dots p_{m}$ , where $p_{1}, p_{2}, \dots, p_{m}$ are the first $m$ prime numbers, then $m$ is equal to

Let $S_{k} = \frac{1 + 2 + 3 + ... + k}{k}$ . If $S_{1}^{2} + S_{2}^{2} + ... + S_{10}^{2} = \frac{5}{12} A$ , then A is equal to :

Let $α = 1^{2} + 4^{2} + 8^{2} + 1 3^{2} + 1 9^{2} + 2 6^{2} + \dots$ upto 10 terms and $β = \sum_{n = 1}^{10} n^{4}$ . If $4 α - β = 55 k + 40$ , then $k$ is equal to

For $k \in N$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k ^{2}} + \frac{13}{k ^{3}} + \frac{19}{k ^{4}} + \dots$ is 10, then the value of $k$ is

Let $S_{n} = \sum_{k = 1}^{4 n} (- 1)^{\frac{k ( k + 1 )}{2}} k^{2}$ . Then $S_{n}$ can take value(s)

The sum $\sum_{k = 1}^{20} (1 + 2 + 3 + .... + k)$ is _____________.

Let $a_{n}$ be $n^{t h}$ term of the series $5 + 8 + 14 + 23 + 35 + 50 + \dots$ and $S_{n} = \sum_{k = 1}^{n} a_{k}$ . Then $S_{30} - a_{40}$ is equal to

The sum to 10 terms of the series $\frac{1}{1 + 1 ^{2} + 1 ^{4}} + \frac{2}{1 + 2 ^{2} + 2 ^{4}} + \frac{3}{1 + 3 ^{2} + 3 ^{4}} + \dots$ is:-

For positive integers n, if $4 a_{n} = (n^{2} + 5 n + 6)$ and $S_{n} = \sum_{k = 1}^{n} (\frac{1}{a _{k}})$ , then the value of $507 S_{2025}$ is:

If $1 + (1 - 2^{2} \cdot 1) + (1 - 4^{2} \cdot 3) + (1 - 6^{2} \cdot 5) + ...... + (1 - 2 0^{2} \cdot 19) = α - 220 β$ , then an ordered pair $(α, β)$ is equal to :

Enter Numerical Value:

Visualized Solution

Analyzing the Number Pyramid

Focusing on the Last Number

Pattern of the Last Numbers

Formula for the Last Number

Locating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5310

Setting up the Inequality

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

Calculating the Approximate Square

Estimating the Square Root

Testing the Boundary for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=102

Testing the Boundary for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=103

Final Conclusion

The Sigma Insight: Sum of Special Series

Analyzing the Structure

The Master Inequality

Estimation Strategy

Final Verification

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sk​=k1+2+3+...+k​. If S12​+S22​+...+S102​=125​A, then A is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α=12+42+82+132+192+262+… upto 10 terms and β=∑n=110​n4. If 4α−β=55k+40, then k is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For k∈N, if the sum of the series 1+k4​+k28​+k313​+k419​+… is 10, then the value of k is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sn​=∑k=14n​(−1)2k(k+1)​k2. Then Sn​ can take value(s)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum ∑k=120​(1+2+3+....+k) is _____________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let an​ be nth term of the series 5+8+14+23+35+50+… and Sn​=∑k=1n​ak​. Then S30​−a40​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum to 10 terms of the series 1+12+141​+1+22+242​+1+32+343​+… is:-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For positive integers n, if 4an​=(n2+5n+6) and Sn​=∑k=1n​(ak​1​), then the value of 507S2025​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1+(1−22⋅1)+(1−42⋅3)+(1−62⋅5)+......+(1−202⋅19)=α−220β, then an ordered pair (α,β) is equal to :

Locating $5310$

Approximating $n$

Testing the Boundary for $n = 102$

Testing the Boundary for $n = 103$

Let $S_{k} = \frac{1 + 2 + 3 + ... + k}{k}$ . If $S_{1}^{2} + S_{2}^{2} + ... + S_{10}^{2} = \frac{5}{12} A$ , then A is equal to :

Let $α = 1^{2} + 4^{2} + 8^{2} + 1 3^{2} + 1 9^{2} + 2 6^{2} + \dots$ upto 10 terms and $β = \sum_{n = 1}^{10} n^{4}$ . If $4 α - β = 55 k + 40$ , then $k$ is equal to

For $k \in N$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k ^{2}} + \frac{13}{k ^{3}} + \frac{19}{k ^{4}} + \dots$ is 10, then the value of $k$ is

Let $S_{n} = \sum_{k = 1}^{4 n} (- 1)^{\frac{k ( k + 1 )}{2}} k^{2}$ . Then $S_{n}$ can take value(s)

The sum $\sum_{k = 1}^{20} (1 + 2 + 3 + .... + k)$ is _____________.

Let $a_{n}$ be $n^{t h}$ term of the series $5 + 8 + 14 + 23 + 35 + 50 + \dots$ and $S_{n} = \sum_{k = 1}^{n} a_{k}$ . Then $S_{30} - a_{40}$ is equal to

The sum to 10 terms of the series $\frac{1}{1 + 1 ^{2} + 1 ^{4}} + \frac{2}{1 + 2 ^{2} + 2 ^{4}} + \frac{3}{1 + 3 ^{2} + 3 ^{4}} + \dots$ is:-

For positive integers n, if $4 a_{n} = (n^{2} + 5 n + 6)$ and $S_{n} = \sum_{k = 1}^{n} (\frac{1}{a _{k}})$ , then the value of $507 S_{2025}$ is:

If $1 + (1 - 2^{2} \cdot 1) + (1 - 4^{2} \cdot 3) + (1 - 6^{2} \cdot 5) + ...... + (1 - 2 0^{2} \cdot 19) = α - 220 β$ , then an ordered pair $(α, β)$ is equal to :