JEE Main 2021 (25 February Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $A_{1}, A_{2}, A_{3}, \dots$ be squares such that for each $n \geq 1$ , the length of the side of $A_{n}$ equals the length of diagonal of $A_{n + 1}$ . If the length of $A_{1}$ is 12 cm, then the smallest value of $n$ for which area of $A_{n}$ is less than one, is

Q: Wait, why is the diagonal of a square $\sqrt{2}$ times its side? I always forget this.

Think of a square with side $s$. If you draw a diagonal, you create a right-angled triangle with two sides of length $s$ and the hypotenuse as the diagonal $d$. By the Pythagorean theorem, $d^2 = s^2 + s^2 = 2s^2$. Taking the square root of both sides gives $d = \sqrt{2}s$. It is a fundamental geometric property that will serve you well in many problems!

Q: Why is the common ratio $1/\sqrt{2}$ and not $\sqrt{2}$? I keep getting them flipped.

Great question! The problem states that the side of $A_n$ equals the diagonal of $A_{n+1}$. So, $s_n = \sqrt{2} s_{n+1}$. To find the ratio for the sequence $s_1, s_2, s_3, \dots$, we need $s_{n+1}$ in terms of $s_n$. Dividing both sides by $\sqrt{2}$ gives $s_{n+1} = s_n / \sqrt{2}$. Since we are multiplying by $1/\sqrt{2}$ to get the next term, that is our common ratio $r$.

Q: Why do we square the side length to get the area? Is there a shortcut?

The area of a square is defined as $side \times side$, or $s^2$. There is no shortcut because the area is fundamentally the square of the side length. When you have an expression like $s_n = 12(1/\sqrt{2})^{n-1}$, squaring it is the most direct way to find the area. Just remember to square both the coefficient (12) and the geometric part $(1/\sqrt{2})^{n-1}$.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the squares as a nested sequence. The core geometric reality is that the side of one square is the diagonal of the next, which means the side length is shrinking by a factor of

1/ 2

at each step. Can you express

s_{n + 1}

in terms of

s_{n}

Hint 2

Hint 2 (The Arsenal): Recognize that the side lengths

s_{1}, s_{2}, s_{3}, \dots

form a Geometric Progression. Recall the formula for the

n

-th term of a G.P.,

s_{n} = a r^{n - 1}

, where

a

is the first term and

r

is the common ratio. What are

a

and

r

in this specific case?

Hint 3

Hint 3 (The Execution): The area of the

n

-th square is

A r e a_{n} = s_{n}^{2}

. Substitute your expression for

s_{n}

into this formula to get

A r e a_{n}

in terms of

n

. Then, set up the inequality

A r e a_{n} < 1

and solve for the smallest integer

n

Visualized Solution

Sequence of Squares

Let $A_{1}, A_{2}, A_{3}, \dots$ be a sequence of squares.
$s_{n}$ is the side length of square $A_{n}$ .
Initial condition: $s_{1} = 12$ .

Side and Diagonal Relation

Given condition: Side of $A_{n}$ = Diagonal of $A_{n + 1}$ .
For $n = 1$ : $s_{1} = d_{2}$ .

Diagonal Formula

For any square, diagonal $d = 2 \times side$ .
Therefore, $d_{2} = 2 s_{2}$ .

Relating $s_{n}$ and $s_{n + 1}$

Equating the two expressions: $s_{1} = 2 s_{2}$ .
Rearranging: $s_{2} = \frac{s _{1}}{2}$ .
In general: $s_{n + 1} = \frac{s _{n}}{2}$ .

Geometric Progression

The side lengths $s_{1}, s_{2}, s_{3}, \dots$ form a Geometric Progression (G.P.).
First term $a = s_{1} = 12$ .
Common ratio $r = \frac{1}{2}$ .

General Term $s_{n}$

General term of a G.P.: $s_{n} = a r^{n - 1}$ .
Substituting the values:
$s_{n} = 12 (\frac{1}{2})^{n - 1}$ .

Area of $A_{n}$

Area of a square is the square of its side: $A r e a_{n} = s_{n}^{2}$ .
$A r e a_{n} = [12 (\frac{1}{2})^{n - 1}]^{2}$ .

Simplifying the Expression

Squaring the terms: $1 2^{2} = 144$ .
$(\frac{1}{2})^{2 (n - 1)} = (\frac{1}{2})^{n - 1}$ .
$A r e a_{n} = \frac{144}{2 ^{n - 1}}$ .

Applying the Condition

We are given that the area must be strictly less than 1.
$A r e a_{n} < 1$ .
$\frac{144}{2 ^{n - 1}} < 1$ .

Solving for $n$

Since $2^{n - 1}$ is always positive, we can cross-multiply.
$144 < 2^{n - 1}$ .
Or, $2^{n - 1} > 144$ .

Checking Powers of 2

Let's evaluate powers of 2:
$2^{7} = 128$ (which is less than 144).
$2^{8} = 256$ (which is greater than 144).
Therefore, the smallest integer for $n - 1$ is $8$ .

Final Answer

$n - 1 = 8 ⟹ n = 9$ .
The smallest value of $n$ for which $A r e a_{n} < 1$ is 9.

00:00 / 00:00

The Sigma Insight: Geometric Progression (G.P.)

Solution Diagram

Analyzing the Setup

Imagine an infinite sequence of squares

A_{1}, A_{2}, A_{3}, \dots

where each square is nestled within the previous one. We are given that the side length of the first square,

A_{1}

, is

12

cm.

The fundamental relationship provided is that the side length of square

A_{n}

is exactly equal to the diagonal of the next square,

A_{n + 1}

. This geometric constraint governs the entire progression.

Decoding the Geometric Progression

Let

s_{n}

denote the side length of square

A_{n}

. From the geometry of a square, the diagonal

d

is related to the side

s

by the formula

d = s 2

Applying this to our sequence, the diagonal of the

(n + 1)

-th square is

d_{n + 1} = s_{n + 1} 2

. Given the condition

s_{n} = d_{n + 1}

, we substitute to obtain:

s_{n} = s_{n + 1} 2

Rearranging this yields the recurrence relation:

s_{n + 1} = \frac{s _{n}}{2}

This confirms that the side lengths form a Geometric Progression (G.P.) with the first term

a = s_{1} = 12

and a common ratio

r = \frac{1}{2}

. The general term for the side length is:

s_{n} = 12 (\frac{1}{2})^{n - 1}

The Transition to Area

The area of the

n

-th square is defined as

A r e a_{n} = s_{n}^{2}

. Substituting our expression for

s_{n}

, we get:

A r e a_{n} = [12 (\frac{1}{2})^{n - 1}]^{2}

Squaring the components, we find

1 2^{2} = 144

and

(\frac{1}{2})^{2 (n - 1)} = (\frac{1}{2})^{n - 1}

. Thus, the area simplifies to:

A r e a_{n} = \frac{144}{2 ^{n - 1}}

Solving the Final Inequality

We seek the smallest integer

n

such that the area is less than

1

^{2}

. This leads to the inequality:

\frac{144}{2 ^{n - 1}} < 1

Multiplying both sides by

2^{n - 1}

(which is always positive), we obtain:

2^{n - 1} > 144

We evaluate the powers of

2

to find the threshold: $2^{7} = 128$ (which is $< 144$ )

2^{8} = 256

(which is

> 144

)

To satisfy the inequality, we require

n - 1 \geq 8

. Therefore,

n \geq 9

The ninth square (

n = 9

) is the first square in the sequence to have an area smaller than

1

^{2}

Common Traps & FAQs

Wait, why is the diagonal of a square

2

times its side? I always forget this.

Think of a square with side

s

. If you draw a diagonal, you create a right-angled triangle with two sides of length

s

and the hypotenuse as the diagonal

d

. By the Pythagorean theorem,

d^{2} = s^{2} + s^{2} = 2 s^{2}

. Taking the square root of both sides gives

d = 2 s

. It is a fundamental geometric property that will serve you well in many problems!

Why is the common ratio

1/ 2

and not

2

? I keep getting them flipped.

Great question! The problem states that the side of

A_{n}

equals the diagonal of

A_{n + 1}

. So,

s_{n} = 2 s_{n + 1}

. To find the ratio for the sequence

s_{1}, s_{2}, s_{3}, \dots

, we need

s_{n + 1}

in terms of

s_{n}

. Dividing both sides by

2

gives

s_{n + 1} = s_{n} / 2

. Since we are multiplying by

1/ 2

to get the next term, that is our common ratio

r

Why do we square the side length to get the area? Is there a shortcut?

The area of a square is defined as

s i d e \times s i d e

, or

s^{2}

. There is no shortcut because the area is fundamentally the square of the side length. When you have an expression like

s_{n} = 12 (1/ 2)^{n - 1}

, squaring it is the most direct way to find the area. Just remember to square both the coefficient (12) and the geometric part

(1/ 2)^{n - 1}

Enter Numerical Value:

Visualized Solution

Sequence of Squares

Side and Diagonal Relation

Diagonal Formula

Relating $s_{n}$ and $s_{n + 1}$

Geometric Progression

General Term $s_{n}$

Area of $A_{n}$

Simplifying the Expression

Applying the Condition

Solving for $n$

Checking Powers of 2

Final Answer

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

Decoding the Geometric Progression

The Transition to Area

Solving the Final Inequality

Similar Questions

Let $S_{1}, S_{2}, \dots$ be squares such that for each $n \geq 1$ , the length of a side of $S_{n}$ equals the length of a diagonal of $S_{n + 1}$ . If the length of a side of $S_{1}$ is 10 cm, then for which of the following values of $n$ is the area of $S_{n}$ less than 1 sq. cm?

If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to ......... .

Let, $A_{n} = (\frac{3}{4}) - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} - \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $B_{n} = 1 - A_{n}$ . Then, the least odd natural number p, so that $B_{n} > A_{n}$ for all $n \geq p$ , is :-

Let $A_{1}, A_{2}, A_{3}, \dots$ be an increasing geometric progression of positive real numbers. If $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$ and $A_{2} + A_{4} = \frac{7}{36}$ , then, the value of $A_{6} + A_{8} + A_{10}$ is equal to

Let $A B C$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $A B C$ and the same process is repeated infinitely many times. If $P$ is the sum of perimeters and $Q$ is the sum of areas of all the triangles formed in this process, then :

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

Enter Numerical Value:

Visualized Solution

Sequence of Squares

Side and Diagonal Relation

Diagonal Formula

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sn​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sn+1​

Geometric Progression

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

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

Simplifying the Expression

Applying the Condition

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

Checking Powers of 2

Final Answer

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

Decoding the Geometric Progression

The Transition to Area

Solving the Final Inequality

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an​=43​−(43​)2+(43​)3+⋯+(−1)n−1(43​)n and bn​=1−an​, then find the least natural number n0​ such that bn​>an​ for all n≥n0​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​=1 and for n≥1,an+1​=21​an​+n2(n+1)2n2−2n−1​. Then ∑n=1∞​(an​−n22​) is equal to ......... .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let, An​=(43​)−(43​)2+(43​)3−⋯+(−1)n−1(43​)n and Bn​=1−An​. Then, the least odd natural number p, so that Bn​>An​ for all n≥p, is :-

Relating $s_{n}$ and $s_{n + 1}$

General Term $s_{n}$

Area of $A_{n}$

Solving for $n$

If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to ......... .

Let, $A_{n} = (\frac{3}{4}) - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} - \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $B_{n} = 1 - A_{n}$ . Then, the least odd natural number p, so that $B_{n} > A_{n}$ for all $n \geq p$ , is :-