Continuing with Exercise 8.3, we will now apply the explicit GP formula \(t_n = a \cdot r^{n-1}\) to more complex scenarios.
We will analyze the shrinking geometry of bouncing balls and use the laws of exponents (indices) to find the exact number of terms hidden inside finite Geometric Progressions. Let's solve!
We will analyze the shrinking geometry of bouncing balls and use the laws of exponents (indices) to find the exact number of terms hidden inside finite Geometric Progressions. Let's solve!
1 Bouncing Ball Decay
2 Finding the Number of Terms
Select the correct option:
If a sequence is defined by the explicit geometric formula \(t_n = 5 \cdot 2^{n-1}\), what is the value of the 4th term?
Concept:
1
Use the formula \(t_n = a \cdot r^{n-1}\). According to BODMAS/PEMDAS, exponents must be evaluated before multiplication.
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Hint:
1
Substitute \(n=4\). The exponent is \(4-1 = 3\).
2
Calculate \(2^3\), then multiply that result by 5.
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Mistakes to Avoid:
!
Multiplying base before exponent: A fatal flaw is writing \(5 \cdot 2^3 = 10^3 = 1000\). The exponent ONLY applies to the 2!
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Solution:
\(t_4 = 5 \cdot 2^{4-1}\)
\(t_4 = 5 \cdot 2^3\)
\(t_4 = 5 \cdot 8 = 40\)
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Select the correct option:
What is the common ratio (\(r\)) of the Geometric Progression: \(-\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, \frac{1}{16} \dots\)?
Concept:
1
Calculate \(r = \frac{t_2}{t_1}\). When dividing fractions, multiply the top fraction by the reciprocal of the bottom fraction.
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Hint:
1
\(\frac{1}{4} \div -\frac{1}{2}\)
2
This becomes \(\frac{1}{4} \times -\frac{2}{1}\). Simplify it.
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Mistakes to Avoid:
!
Ignoring the alternating signs: The sequence goes \(-\), \(+\), \(-\), \(+\). This is the hallmark signature of a negative common ratio. If you choose a positive ratio, your answer is instantly wrong!
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Solution:
\(r = \frac{t_2}{t_1} = \frac{1/4}{-1/2}\)
\(r = \frac{1}{4} \times \frac{-2}{1} = \frac{-2}{4} = -\frac{1}{2}\)
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Select the correct option:
To find the number of terms (\(n\)) in the equation \(5 \cdot 3^{n-1} = 405\), what is the correct first algebraic step?
Concept:
1
When solving for an exponent variable in an equation like \(a \cdot r^{n-1} = x\), always isolate the base \(r\) first by dividing away \(a\).
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Hint:
1
We want \(3^{n-1}\) all by itself on the left side. What is currently attached to it? The 5.
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Mistakes to Avoid:
!
Multiplying base by coefficient: Writing \(15^{n-1} = 405\) breaks the fundamental rules of algebra. Exponents happen before multiplication!
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Solution:
\(5 \cdot 3^{n-1} = 405\)
Divide both sides by 5:
\(3^{n-1} = \frac{405}{5} = 81\)
Now solve: \(3^{n-1} = 3^4 \implies n=5\).
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Select the correct option:
If you determine that \(2^{n-1} = 64\), what is the value of \(n\)?
Concept:
1
Express the constant (64) as a base of 2. Then set the exponents equal to each other.
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Hint:
1
\(2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64\), which means \(2^6 = 64\).
2
So, \(n-1 = 6\). Solve for \(n\).
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Mistakes to Avoid:
!
Stopping at the exponent: The most common error in GP tests is finding the exponent (6) and circling Option A. The question asks for the position \(n\), which is always one higher than the exponent!
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Solution:
\(2^{n-1} = 64\)
\(2^{n-1} = 2^6\)
\(n - 1 = 6\)
\(n = 7\)
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Select the correct option:
A bouncing ball follows the explicit GP rule \(t_n = 54 \cdot (\frac{2}{3})^{n-1}\), where \(n\) is the bounce number. What does the \(54\) represent in reality?
Concept:
1
In \(t_n = a \cdot r^{n-1}\), the constant \(a\) is strictly the value of the sequence when \(n=1\).
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Hint:
1
If \(n\) represents the bounce number, what does the term at \(n=1\) represent physically?
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Mistakes to Avoid:
!
Confusing drop height with 1st bounce: The ball is dropped from a higher place. The sequence of "bounces" only begins AFTER the ball hits the floor the first time.
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Solution:
Substitute \(n=1\):
\(t_1 = 54 \cdot (\frac{2}{3})^0 = 54 \cdot 1 = 54\).
Since \(n\) is the bounce count, 54 represents the exact peak height of the very first bounce.
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