Problem 49 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(S_{1}, S_{2}, \ldots\) 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 \(10 \mathrm{~cm}\), then for which of the following values of \(n\) is the area of \(S_{s}\) less than 1 square \(\mathrm{cm} ?\) (A) 7 (B) 8 (C) 9 (D) 10

Step-by-Step Solution

Answer

(B) 8

Step 1: Understand the relationship between side and diagonal

For a square with side length \( a \), the diagonal length \( d \) is given by \( d = a\sqrt{2} \). In this sequence of squares, the side of square \( S_{n} \) is the diagonal of square \( S_{n+1} \).

Step 2: Express recursive side length relation

We have \( a_{n} \) as the side length of square \( S_{n} \). Therefore, \( a_{n+1} = \frac{a_{n}}{\sqrt{2}} \) because the diagonal \( a_{n} \) of \( S_{n+1} \) is the side length of \( S_{n} \).

Step 3: Formulate side length for general n

Starting from \( a_1 = 10 \text{ cm} \), use the recursive relation: \[ a_{n} = \frac{10}{(\sqrt{2})^{n-1}} \] to express side lengths for general \( n \).

Step 4: Calculate area of square S_n

The area of square \( S_{n} \) is \( (a_{n})^2 \). Substitute \( a_{n} \): \[ (a_{n})^2 = \left( \frac{10}{(\sqrt{2})^{n-1}} \right)^2 = \frac{100}{2^{n-1}} \].

Step 5: Determine n where area is less than 1

We need \( \frac{100}{2^{n-1}} < 1 \). Solving \( 100 < 2^{n-1} \), we take log base 2: \[ \log_{2}(100) < n - 1 \]. We calculate \( \log_{2}(100) \approx 6.644 \), so \( n - 1 > 6.644 \), thus \( n > 7.644 \).

Step 6: Choose the smallest integer value for n

To satisfy \( n > 7.644 \), the smallest integer \( n \) is 8.

Key Concepts

Recursive Sequences

Mathematical problem-solving often involves patterns and sequences, among which recursive sequences are particularly interesting. A recursive sequence is defined where each term depends on preceding terms. In essence, the next value is found from the previous one.

This exercise involves squares, where each square's side length becomes the diagonal of the next one. This creates a sequence where each term, denoted as \(a_n\), provides crucial information for the next term, represented mathematically as \(a_{n+1} = \frac{a_{n}}{\sqrt{2}}\). The sequence continues recursively, following this pattern.

Benefits of using recursive sequences include:

• They simplify complex relationships into understandable steps.
• They are ideal for computations where each step builds on the last, leading to more complex yet organized solutions.

By mastering the concept of recursive sequences, one gains the ability to solve intricate problems by breaking them down into iterative processes.

Geometry

Geometry, the study of shapes and spaces, allows us to understand spatial relationships and properties of objects. In this exercise, we're focusing on squares, which are fundamental geometric shapes with equal sides and angles.

Each square in our sequence presents a specific geometric characteristic — the relationship between its side and its diagonal. Given a square with side length \(a\), its diagonal is \(d = a\sqrt{2}\). This geometrical property underpins the problem as it sets up the relationship between consecutive squares.

Understanding this geometrical principle is crucial because:

• It helps visualize how each square transitions into the next.
• It emphasizes fundamental geometric properties, enhancing overall spatial reasoning skills.

Geometry, therefore, serves as the foundational element that leads to understanding the recursive relationship in the sequence of squares.

Square Properties

Squares are unique among geometric shapes because of their defining properties—they have equal sides and equal angles of 90 degrees. In this exercise, square properties are key to understanding the sequence of calculations needed to find the terms and areas involved.

For a square with side \(a\), the area is simply \(a^2\). Employing this property, the exercise requires calculating when the area of a square becomes less than 1 square cm. Using the recursive sequence, the area formula can be expressed in terms of \(n\) as \(\left( \frac{10}{(\sqrt{2})^{n-1}} \right)^2 = \frac{100}{2^{n-1}}\).

Key properties of squares to remember include:

• Equilateral sides aid in straightforward computations.
• The relationship between side and diagonal facilitates understanding of internal and external shape transformations.

By using these square properties, calculating the areas and following the transitions of squares in our sequence becomes a manageable task.