JEE Advanced 1999

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: 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?

Q: Wait, why do we reject the negative root when relating the side length and the diagonal?

That is a great observation! In pure algebra, $d^2 = 2a^2$ implies $d = \pm a\sqrt{2}$. However, we are dealing with physical lengths of squares. A length must always be a positive quantity. Therefore, we discard the negative root because a square with a negative side length has no physical meaning in this context.

Q: Why can't we just use the ratio of areas directly instead of finding the side length first?

You absolutely can! If the side of $S_n$ is the diagonal of $S_{n+1}$, then $a_n = \sqrt{2} a_{n+1}$. Squaring both sides gives $a_n^2 = 2 a_{n+1}^2$, which means $A_n = 2 A_{n+1}$. This tells us that each area is exactly half of the previous one. This is a geometric progression with a common ratio of $1/2$. Both methods are perfectly valid and lead to the same result!

Select Answer:

* Multiple Correct

Visualized Solution

Visualizing $S_{1}$

Let $a_{n}$ be the side length of square $S_{n}$ .
For the first square $S_{1}$ , the side length is given as $a_{1} = 10$ cm.

The Diagonal Condition

The problem states a unique relationship between consecutive squares.
Side of $S_{n}$ = Diagonal of $S_{n + 1}$ .
Visually, $S_{2}$ can be formed by connecting the midpoints of $S_{1}$ .

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

For any square with side $a$ , its diagonal is $2 a$ .
Applying the condition: $a_{n} = Diagonal of S_{n + 1}$ .
Therefore, $a_{n} = 2 a_{n + 1}$ .

Geometric Progression of Sides

Rearranging the equation: $a_{n + 1} = \frac{a _{n}}{2}$ .
Each subsequent side is obtained by multiplying the previous side by $\frac{1}{2}$ .
This forms a Geometric Progression (G.P.) with common ratio $r = \frac{1}{2}$ .

General Term $a_{n}$

The $n$ -th term of a G.P. is given by $a_{n} = a_{1} r^{n - 1}$ .
Substituting $a_{1} = 10$ and $r = \frac{1}{2}$ .
$a_{n} = 10 (\frac{1}{2})^{n - 1}$ .

Finding the Area $A_{n}$

The area of a square is the square of its side length: $A_{n} = a_{n}^{2}$ .
Substitute the expression for $a_{n}$ : $A_{n} = [10 (\frac{1}{2})^{n - 1}]^{2}$ .

Simplifying $A_{n}$

Squaring the terms: $1 0^{2} = 100$ and $(\frac{1}{2})^{2} = \frac{1}{2}$ .
$A_{n} = 100 (\frac{1}{2})^{n - 1}$ .
$A_{n} = \frac{100}{2 ^{n - 1}}$ .

Applying the Condition $A_{n} < 1$

We need to find $n$ such that the area is strictly less than 1 sq. cm.
$\frac{100}{2 ^{n - 1}} < 1$ .

Rearranging the Inequality

Since $2^{n - 1}$ is always positive, we can safely multiply both sides by it.
$100 < 2^{n - 1}$ .
Or, $2^{n - 1} > 100$ .

Finding the Critical Value

Let's check the powers of 2.
$2^{6} = 64$ (which is not greater than 100).
$2^{7} = 128$ (which is greater than 100).
Therefore, the exponent $n - 1$ must be at least 7.

Final Values of $n$

$n - 1 \geq 7 ⟹ n \geq 8$ .
The possible values for $n$ are $8, 9, 10, 11, \dots$
Looking at the given options (7, 8, 9, 10), the correct values are 8, 9, and 10.

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

Imagine you are standing before a sequence of squares, each one perfectly nested within the last. We start with a square

S_{1}

that has a side length of

10

cm. This serves as our anchor and starting point for the sequence.

The Geometric Bridge

The core of this problem lies in the relationship between consecutive squares. We are told that the side of

S_{n}

is equal to the diagonal of

S_{n + 1}

a_{n}

is the side of

S_{n}

and

a_{n + 1}

is the side of

S_{n + 1}

, the condition given is:

a_{n} = a_{n + 1} 2

By rearranging this, we find the relationship for the side lengths:

a_{n + 1} = \frac{a _{n}}{2}

This reveals that the side lengths form a geometric progression with a common ratio of

r = \frac{1}{2}

The Transition to Area

The question asks us about the area,

A_{n}

, of the

n

-th square. Since

A_{n} = a_{n}^{2}

, we square our expression for the side length.

Starting with

a_{1} = 10

, the general term for the side length is

a_{n} = 10 (\frac{1}{2})^{n - 1}

. Squaring this, we obtain:

A_{n} = [10 (\frac{1}{2})^{n - 1}]^{2}

Simplifying this expression, where

1 0^{2} = 100

and

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

, we arrive at the general formula for the area:

A_{n} = \frac{100}{2 ^{n - 1}}

The Inequality Challenge

We now find when the area

A_{n}

drops below

1

sq. cm by setting up the following inequality:

\frac{100}{2 ^{n - 1}} < 1

Multiplying both sides by

2^{n - 1}

, we obtain:

100 < 2^{n - 1}

We test the powers of

2

to find the smallest integer

n

that satisfies this condition:

2^{6} = 64

(which is less than

100

)

2^{7} = 128

(which is greater than

100

)

Since

128 > 100

, the condition

2^{n - 1} > 100

is satisfied when

n - 1 \geq 7

. This implies

n \geq 8

Conclusion: Starting from the $8$ -th square, the area will be less than

1

sq. cm.

Common Traps & FAQs

Wait, why do we reject the negative root when relating the side length and the diagonal?

That is a great observation! In pure algebra,

d^{2} = 2 a^{2}

implies

d = \pm a 2

. However, we are dealing with physical lengths of squares. A length must always be a positive quantity. Therefore, we discard the negative root because a square with a negative side length has no physical meaning in this context.

Why can't we just use the ratio of areas directly instead of finding the side length first?

You absolutely can! If the side of

S_{n}

is the diagonal of

S_{n + 1}

, then

a_{n} = 2 a_{n + 1}

. Squaring both sides gives

a_{n}^{2} = 2 a_{n + 1}^{2}

, which means

A_{n} = 2 A_{n + 1}

. This tells us that each area is exactly half of the previous one. This is a geometric progression with a common ratio of

1/2

. Both methods are perfectly valid and lead to the same result!

How do I know if the inequality sign flips when I solve

100/ 2^{n - 1} < 1

The inequality sign only flips if you multiply or divide by a negative number. Since

2^{n - 1}

is an exponential function, it is always positive for any real

n

. Therefore, when you multiply both sides by

2^{n - 1}

, the inequality sign remains exactly as it is. You are safe to proceed!

Select Answer:

Visualized Solution

Visualizing $S_{1}$

The Diagonal Condition

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

Geometric Progression of Sides

General Term $a_{n}$

Finding the Area $A_{n}$

Simplifying $A_{n}$

Applying the Condition $A_{n} < 1$

Rearranging the Inequality

Finding the Critical Value

Final Values of $n$

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

Analyzing the Setup

The Geometric Bridge

The Transition to Area

The Inequality Challenge

Similar Questions

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

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 ......... .

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 $A B C$ 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 $α$ and $β$ be the roots of the quadratic equation $x^{2} sin θ - x (sin θ cos θ + 1) + cos θ = 0$ ( $0 < θ < 4 5^{\circ}$ ), and $α < β$ . Then $\sum_{n = 0}^{\infty} (α^{n} + \frac{( - 1 ) ^{n}}{β ^{n}})$ 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 :-

If $2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + .... + 2 \cdot 3^{9} + 3^{10} = S - 2^{11}$ then $S$ is equal to :

Select Answer:

Visualized Solution

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

The Diagonal Condition

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

Geometric Progression of Sides

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

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

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

Applying the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An​<1

Rearranging the Inequality

Finding the Critical Value

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

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

Analyzing the Setup

The Geometric Bridge

The Transition to Area

The Inequality Challenge

Similar Questions

.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; } 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 α and β be the roots of the quadratic equation x2sinθ−x(sinθcosθ+1)+cosθ=0 (0<θ<45∘), and α<β. Then ∑n=0∞​(αn+βn(−1)n​) 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 :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 210+29⋅31+28⋅32+....+2⋅39+310=S−211 then S is equal to :

Visualizing $S_{1}$

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

General Term $a_{n}$

Finding the Area $A_{n}$

Simplifying $A_{n}$

Applying the Condition $A_{n} < 1$

Final Values of $n$

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 ......... .

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 $α$ and $β$ be the roots of the quadratic equation $x^{2} sin θ - x (sin θ cos θ + 1) + cos θ = 0$ ( $0 < θ < 4 5^{\circ}$ ), and $α < β$ . Then $\sum_{n = 0}^{\infty} (α^{n} + \frac{( - 1 ) ^{n}}{β ^{n}})$ 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 :-

If $2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + .... + 2 \cdot 3^{9} + 3^{10} = S - 2^{11}$ then $S$ is equal to :