Square S_1 has side length 12 centimeters. For each integer n geq 1, square S_n+1 has a side length equal...
GMAT Advanced Math : (Adv_Math) Questions
Square \(\mathrm{S_1}\) has side length 12 centimeters. For each integer \(\mathrm{n \geq 1}\), square \(\mathrm{S_{n+1}}\) has a side length equal to one-half the side length of square \(\mathrm{S_n}\). If \(\mathrm{A(n)}\) represents the area, in square centimeters, of square \(\mathrm{S_n}\), which equation gives \(\mathrm{A(n)}\) in terms of n?
\(\mathrm{A(n) = 12(1/2)^{(n-1)}}\)
\(\mathrm{A(n) = 144(1/2)^{(n-1)}}\)
\(\mathrm{A(n) = 144(1/4)^n}\)
\(\mathrm{A(n) = 144(1/4)^{(n-1)}}\)
\(\mathrm{A(n) = 12^n(1/2)^{(n-1)}}\)
1. TRANSLATE the problem information
- Given information:
- Square \(\mathrm{S_1}\) has side length 12 cm
- Each subsequent square: side length = (1/2) × previous side length
- \(\mathrm{A(n)}\) = area of square \(\mathrm{S_n}\)
- Need to find \(\mathrm{A(n)}\) in terms of n
2. INFER the pattern by calculating first few terms
- Let's find the first few side lengths and areas:
- \(\mathrm{S_1}\): side = 12 cm → \(\mathrm{A(1) = 12^2 = 144}\) cm²
- \(\mathrm{S_2}\): side = 6 cm → \(\mathrm{A(2) = 6^2 = 36}\) cm²
- \(\mathrm{S_3}\): side = 3 cm → \(\mathrm{A(3) = 3^2 = 9}\) cm²
- Key insight: Side lengths form sequence 12, 6, 3, ... but we need the AREA sequence!
3. INFER the geometric sequence relationships
- Side lengths: 12, 6, 3, 1.5, ...
This is geometric with first term 12 and common ratio 1/2 - Areas: 144, 36, 9, 2.25, ...
Check ratios: \(\mathrm{36/144 = 1/4}\), \(\mathrm{9/36 = 1/4}\)
This is geometric with first term 144 and common ratio 1/4 - Why 1/4? Because area = \(\mathrm{(side)^2}\), so area ratio = \(\mathrm{(side\ ratio)^2 = (1/2)^2 = 1/4}\)
4. SIMPLIFY using the geometric sequence formula
- For geometric sequence: \(\mathrm{a_n = a_1 \times r^{(n-1)}}\)
- For areas: \(\mathrm{A(n) = 144 \times (1/4)^{(n-1)}}\)
Answer: D
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER reasoning: Students recognize the side length pattern but fail to connect it properly to the area pattern. They see that side lengths are halved each time and incorrectly assume areas are also halved each time.
This leads them to think \(\mathrm{A(n)}\) follows the same ratio as side lengths (1/2 instead of 1/4), causing them to select Choice A (\(\mathrm{A(n) = 12(1/2)^{(n-1)}}\)) or Choice B (\(\mathrm{A(n) = 144(1/2)^{(n-1)}}\)).
Second Most Common Error:
Inadequate SIMPLIFY execution: Students correctly identify that areas form a geometric sequence but make algebraic errors when determining the common ratio. They might recognize that area involves squaring but fail to apply the exponent rule \(\mathrm{(1/2)^2 = 1/4}\) correctly.
This confusion about the ratio can lead them to select Choice C (\(\mathrm{A(n) = 144(1/4)^n}\)) by using the wrong exponent form.
The Bottom Line:
This problem requires students to think in two layers: recognizing the geometric pattern in side lengths, then translating that understanding to the geometric pattern in areas. The key insight is that when linear dimensions change by a factor, areas change by the square of that factor.
\(\mathrm{A(n) = 12(1/2)^{(n-1)}}\)
\(\mathrm{A(n) = 144(1/2)^{(n-1)}}\)
\(\mathrm{A(n) = 144(1/4)^n}\)
\(\mathrm{A(n) = 144(1/4)^{(n-1)}}\)
\(\mathrm{A(n) = 12^n(1/2)^{(n-1)}}\)