JEE Advanced 2022

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $l_{1}, l_{2}, \dots, l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_{1}$ , and let $w_{1}, w_{2}, \dots, w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_{2}$ , where $d_{1} d_{2} = 10$ . For each $i = 1, 2, \dots, 100$ , let $R_{i}$ be a rectangle with length $l_{i}$ , width $w_{i}$ and area $A_{i}$ . If $A_{51} - A_{50} = 1000$ , then the value of $A_{100} - A_{90}$ is _____________.

Q: Wait, why does the $l_1w_1$ term just disappear? Is it possible that it's not zero?

That is a brilliant observation! It is not that $l_1w_1$ is zero; it is that it is a constant term present in every single area $A_i$. When we calculate the difference $A_i - A_j$, we are essentially subtracting the same constant from itself, which results in zero. It is a beautiful example of how structure simplifies complexity—we do not need to know the individual values of $l_1$ or $w_1$ to find the difference!

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Do not get overwhelmed by the variables

l_{1}, w_{1}, d_{1}, d_{2}

. Instead, visualize the area

A_{i}

as a quadratic function of the index

i

. The key is to see that

A_{i}

is a product of two linear expressions, which naturally forms a quadratic structure. Look for the pattern in the expansion.

Hint 2

The Arsenal: You have two arithmetic progressions. When you multiply them, the constant term

l_{1} w_{1}

appears. When you calculate the difference between two areas,

A_{i} - A_{j}

, this constant term will always cancel out. Use the difference of squares identity,

a^{2} - b^{2} = (a - b) (a + b)

, to simplify the calculations involving the squared terms.

Hint 3

The Execution: Define

K = l_{1} d_{2} + w_{1} d_{1}

. Your area formula becomes

A_{i} = l_{1} w_{1} + (i - 1) K + 10 (i - 1)^{2}

. Use the given

A_{51} - A_{50} = 1000

to solve for

K

. Once you have

K

, the expression

A_{100} - A_{90}

becomes a straightforward calculation.

Visualized Solution

Define the Arithmetic Progressions

Let the length of the rectangle $R_{i}$ be $l_{i}$ .
$l_{i} = l_{1} + (i - 1) d_{1}$
Let the width of the rectangle $R_{i}$ be $w_{i}$ .
$w_{i} = w_{1} + (i - 1) d_{2}$

Define the Area $A_{i}$

Area of the rectangle $R_{i}$ is $A_{i} = l_{i} \cdot w_{i}$
Substitute the expressions for $l_{i}$ and $w_{i}$ :
$A_{i} = (l_{1} + (i - 1) d_{1}) (w_{1} + (i - 1) d_{2})$

Expand the Area Product

Expand the product:
$A_{i} = l_{1} w_{1} + (i - 1) l_{1} d_{2} + (i - 1) w_{1} d_{1} + (i - 1)^{2} d_{1} d_{2}$
Group the terms with $(i - 1)$ :
$A_{i} = l_{1} w_{1} + (i - 1) (l_{1} d_{2} + w_{1} d_{1}) + (i - 1)^{2} d_{1} d_{2}$

Substitute $d_{1} d_{2} = 10$ and Simplify

Given that $d_{1} d_{2} = 10$ .
Let's define a constant $K = l_{1} d_{2} + w_{1} d_{1}$ to simplify.
Substitute these into the area equation:
$A_{i} = l_{1} w_{1} + (i - 1) K + 10 (i - 1)^{2}$

Setup $A_{51}$ and $A_{50}$

We are given the condition: $A_{51} - A_{50} = 1000$
Substitute $i = 51$ into the area formula:
$A_{51} = l_{1} w_{1} + 50 K + 10 (50)^{2}$
Substitute $i = 50$ into the area formula:
$A_{50} = l_{1} w_{1} + 49 K + 10 (49)^{2}$

Subtract $A_{50}$ from $A_{51}$

Subtract $A_{50}$ from $A_{51}$ :
$A_{51} - A_{50} = (50 K - 49 K) + 10 (5 0^{2} - 4 9^{2})$
The $l_{1} w_{1}$ terms cancel out.
$K + 10 (5 0^{2} - 4 9^{2}) = 1000$

Evaluate Difference of Squares

Use the identity $a^{2} - b^{2} = (a - b) (a + b)$ :
$5 0^{2} - 4 9^{2} = (50 - 49) (50 + 49) = (1) (99) = 99$
Substitute this back into the equation:
$K + 10 (99) = 1000$
$K + 990 = 1000$

Solve for $K$

Solve for $K$ :
$K = 1000 - 990$
$K = 10$
Recall that $K = l_{1} d_{2} + w_{1} d_{1} = 10$

Setup $A_{100}$ and $A_{90}$

We need to find the value of $A_{100} - A_{90}$ .
Substitute $i = 100$ :
$A_{100} = l_{1} w_{1} + 99 K + 10 (99)^{2}$
Substitute $i = 90$ :
$A_{90} = l_{1} w_{1} + 89 K + 10 (89)^{2}$

Subtract $A_{90}$ from $A_{100}$

Subtract $A_{90}$ from $A_{100}$ :
$A_{100} - A_{90} = (99 K - 89 K) + 10 (9 9^{2} - 8 9^{2})$
$A_{100} - A_{90} = 10 K + 10 (9 9^{2} - 8 9^{2})$

Evaluate Difference of Squares Again

Substitute $K = 10$ :
First term: $10 K = 10 (10) = 100$
Second term using $a^{2} - b^{2}$ :
$9 9^{2} - 8 9^{2} = (99 - 89) (99 + 89) = (10) (188) = 1880$
Expression becomes: $100 + 10 (1880)$

Final Calculation

Final calculation:
$A_{100} - A_{90} = 100 + 18800$
Final Answer: $A_{100} - A_{90} = 18900$
Key Takeaway: Grouping complex terms into a single constant simplifies the algebra significantly.

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

Solution Diagram

Analyzing the Setup

Imagine you are standing in a vast, open field, and before you, stretching toward the horizon, are one hundred rectangles. Each one is slightly different from the last, yet they are all bound by a hidden, rhythmic order.

We are dealing with two arithmetic progressions: one for the lengths,

l_{i}

, and one for the widths,

w_{i}

. As we move from the first rectangle to the hundredth, the dimensions grow with a steady, predictable pulse.

The length of the

i

-th rectangle is given by

l_{i} = l_{1} + (i - 1) d_{1}

, and the width is

w_{i} = w_{1} + (i - 1) d_{2}

. When we calculate the area

A_{i} = l_{i} \cdot w_{i}

, we are multiplying two linear expressions.

The Quadratic Trap

Expanding the product

A_{i} = (l_{1} + (i - 1) d_{1}) (w_{1} + (i - 1) d_{2})

reveals a structure containing a constant term, a term linear in

(i - 1)

, and a term quadratic in

(i - 1)

A_{i} = l_{1} w_{1} + (i - 1) (l_{1} d_{2} + w_{1} d_{1}) + (i - 1)^{2} d_{1} d_{2}

We are given that

d_{1} d_{2} = 10

. Let us define a new constant,

K = l_{1} d_{2} + w_{1} d_{1}

. Our area formula now simplifies to:

A_{i} = l_{1} w_{1} + (i - 1) K + 10 (i - 1)^{2}

This reduction transforms a complex expression into a simple quadratic in terms of

(i - 1)

The Power of Subtraction

We are given that

A_{51} - A_{50} = 1000

. Applying our formula, where for

i = 51

we have

(i - 1) = 50

and for

i = 50

we have

(i - 1) = 49

, the constant

l_{1} w_{1}

vanishes:

A_{51} - A_{50} = (50 K - 49 K) + 10 (5 0^{2} - 4 9^{2}) = 1000

This simplifies to

K + 10 (5 0^{2} - 4 9^{2}) = 1000

. Using the difference of squares identity,

5 0^{2} - 4 9^{2} = (50 - 49) (50 + 49) = 99

Substituting this back, we find

K + 10 (99) = 1000

, which yields

K = 10

. We have successfully unlocked the secret of the progression.

The Final Leap

Now, we calculate the target value

A_{100} - A_{90}

. Applying the same logic, the constant

l_{1} w_{1}

cancels out:

A_{100} - A_{90} = (99 K - 89 K) + 10 (9 9^{2} - 8 9^{2})

Substituting

K = 10

into the expression, we obtain:

A_{100} - A_{90} = 10 (10) + 10 (9 9^{2} - 8 9^{2})

Using the difference of squares identity again,

9 9^{2} - 8 9^{2} = (99 - 89) (99 + 89) = 10 \times 188 = 1880

. Our final calculation is:

A_{100} - A_{90} = 100 + 10 (1880) = 100 + 18800 = 18900

The final result is 18900.

Common Traps & FAQs

Wait, why does the

l_{1} w_{1}

term just disappear? Is it possible that it's not zero?

That is a brilliant observation! It is not that

l_{1} w_{1}

is zero; it is that it is a constant term present in every single area

A_{i}

. When we calculate the difference

A_{i} - A_{j}

, we are essentially subtracting the same constant from itself, which results in zero. It is a beautiful example of how structure simplifies complexity—we do not need to know the individual values of

l_{1}

w_{1}

to find the difference!

Can I just expand everything and solve for

l_{1}, w_{1}, d_{1}, d_{2}

individually?

Technically, you could try, but you would quickly find yourself in a trap. You have four unknowns (

l_{1}, w_{1}, d_{1}, d_{2}

) but only one equation (

d_{1} d_{2} = 10

) and one difference condition. This is a classic JEE trap designed to waste your time. The problem is constructed so that you only need the value of the combined constant

K = l_{1} d_{2} + w_{1} d_{1}

. Always look for the 'lumped' variable before diving into individual components.

Why is the difference of squares identity so important here? Can't I just calculate

5 0^{2}

and

4 9^{2}

You certainly could, but in the high-pressure environment of the JEE Advanced, speed and accuracy are your best friends. Calculating

5 0^{2} = 2500

and

4 9^{2} = 2401

is prone to simple arithmetic errors. Using

(50 - 49) (50 + 49) = 1 \times 99 = 99

is not only faster but also elegant. It reduces the cognitive load, allowing you to focus on the physics and logic of the problem rather than tedious multiplication.

Enter Numerical Value:

Visualized Solution

Define the Arithmetic Progressions

Define the Area $A_{i}$

Expand the Area Product

Substitute $d_{1} d_{2} = 10$ and Simplify

Setup $A_{51}$ and $A_{50}$

Subtract $A_{50}$ from $A_{51}$

Evaluate Difference of Squares

Solve for $K$

Setup $A_{100}$ and $A_{90}$

Subtract $A_{90}$ from $A_{100}$

Evaluate Difference of Squares Again

Final Calculation

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

Analyzing the Setup

The Quadratic Trap

The Power of Subtraction

The Final Leap

Similar Questions

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

Suppose $a_{1}, a_{2}, \dots, a_{n}, \dots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is $5 : 17$ and $110 < a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -

Let $a_{1}, a_{2}, a_{3}, \dots$ be in an arithmetic progression of positive terms. Let $A_{k} = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2}$ . 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

Let $S_{1}$ be the sum of first $2 n$ terms of an arithmetic progression. Let $S_{2}$ be the sum of first $4 n$ terms of the same arithmetic progression. If $(S_{2} - S_{1})$ is $1000$ , then the sum of the first $6 n$ terms of the arithmetic progression is equal to:

Let $a_{1}, a_{2}, \dots, a_{21}$ be an $A P$ such that $\sum_{n = 1}^{20} \frac{1}{a _{n} a _{n + 1}} = \frac{4}{9}$ . If the sum of this $A P$ is $189$ , then $a_{6} a_{16}$ is equal to :

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6 : 11$ and the seventh term lies in between $130$ and $140$ , then the common difference of this A.P. is \dots.

Let $a_{1}, a_{2}, a_{3}, \dots$ be an $A . P .$ If $\frac{a _{1} + a _{2} + \dots + a _{10}}{a _{1} + a _{2} + \dots + a _{p}} = \frac{100}{p ^{2}}, p \neq = 10$ , then $\frac{a _{11}}{a _{10}}$ is equal to:

Let $a_{1}, a_{2}, .... a_{n}$ be a given A.P. whose common difference is an integer and $S_{n} = a_{1} + a_{2} + .... + a_{n}$ If $a_{1} = 1, a_{n} = 300$ and $15 < n < 50$ , then the ordered pair $(S_{n - 4}, a_{n - 4})$ is equal to

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is $15 : 7$ , then $S_{15} - S_{5}$ is equal to:

Let $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ be an arithmetic progression with $a_{1} = 3$ and $S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100$ . For any integer $n$ with $1 \leq n \leq 20$ , let $m = 5 n$ . If $S_{m} / S_{n}$ does not depend on $n$ , then $a_{2}$ is \dots.

Enter Numerical Value:

Visualized Solution

Define the Arithmetic Progressions

Define the Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ai​

Expand the Area Product

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d1​d2​=10 and Simplify

Setup .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A51​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A50​

Subtract .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A50​ from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A51​

Evaluate Difference of Squares

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

Setup .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A100​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A90​

Subtract .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A90​ from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A100​

Evaluate Difference of Squares Again

Final Calculation

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

Analyzing the Setup

The Quadratic Trap

The Power of Subtraction

The Final Leap

Similar Questions

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,…,a21​ be an AP such that ∑n=120​an​an+1​1​=94​. If the sum of this AP is 189, then a6​a16​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​,… be an A.P. If a1​+a2​+⋯+ap​a1​+a2​+⋯+a10​​=p2100​,p=10, then a10​a11​​ is equal to:

Define the Area $A_{i}$

Substitute $d_{1} d_{2} = 10$ and Simplify

Setup $A_{51}$ and $A_{50}$

Subtract $A_{50}$ from $A_{51}$

Solve for $K$

Setup $A_{100}$ and $A_{90}$

Subtract $A_{90}$ from $A_{100}$

Let $a_{1}, a_{2}, \dots, a_{21}$ be an $A P$ such that $\sum_{n = 1}^{20} \frac{1}{a _{n} a _{n + 1}} = \frac{4}{9}$ . If the sum of this $A P$ is $189$ , then $a_{6} a_{16}$ is equal to :

Let $a_{1}, a_{2}, a_{3}, \dots$ be an $A . P .$ If $\frac{a _{1} + a _{2} + \dots + a _{10}}{a _{1} + a _{2} + \dots + a _{p}} = \frac{100}{p ^{2}}, p \neq = 10$ , then $\frac{a _{11}}{a _{10}}$ is equal to: