In these advanced problems, you won't just be finding numbers. You will be using the structure of the explicit formula \(t_n = a \cdot r^{n-1}\) to mathematically prove relationships between distant terms in a GP. Mastering these proofs is the ultimate test of your algebraic fluency with exponents!
As we progress through the End of Chapter exercises, we shift our focus from Arithmetic Progressions back to Geometric Progressions (GP).
In these advanced problems, you won't just be finding numbers. You will be using the structure of the explicit formula \(t_n = a \cdot r^{n-1}\) to mathematically prove relationships between distant terms in a GP. Mastering these proofs is the ultimate test of your algebraic fluency with exponents!
In these advanced problems, you won't just be finding numbers. You will be using the structure of the explicit formula \(t_n = a \cdot r^{n-1}\) to mathematically prove relationships between distant terms in a GP. Mastering these proofs is the ultimate test of your algebraic fluency with exponents!
1 The Geometric Mean Proof
2 Distant Term Proofs
★ Complex Ratio Proofs
Select the correct option:
If \(p, q,\) and \(r\) are three consecutive terms in a Geometric Progression, which of the following equations must be true?
Concept:
1
Because the sequence is a GP, the common ratio \(r\) must be identical: \(\frac{q}{p} = \frac{r}{q}\).
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Hint:
1
Cross multiply the fractions \(\frac{q}{p} = \frac{r}{q}\).
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Mistakes to Avoid:
!
Confusing AP and GP rules: Option A represents the defining property of an AP. Do not use it for a GP!
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Solution:
Cross-multiplying \(\frac{q}{p} = \frac{r}{q}\) yields \(q^2 = pr\).
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Select the correct option:
If you take an existing Geometric Progression and square every single term in it, what is the resulting sequence?
Concept:
1
If a sequence is \(a, ar, ar^2\), squaring them yields \(a^2, a^2r^2, a^2r^4\).
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Hint:
1
Test it: \(2, 4, 8\) has \(r=2\).
2
Square them: \(4, 16, 64\). What is the new ratio? \(\frac{16}{4} = 4\), which is \(2^2\).
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Mistakes to Avoid:
!
Assuming the ratio stays the same: Squaring the terms fundamentally changes the magnitude of the leaps between them, squaring the ratio in the process.
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Solution:
The new sequence is formed by \((a \cdot r^{n-1})^2 = a^2 \cdot (r^2)^{n-1}\).
This perfectly matches the GP formula, with a new first term \(a^2\) and new ratio \(r^2\).
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Select the correct option:
If \(x, y, z\) are in GP, what is the value of \(\frac{y}{x}\)?
Concept:
1
The fundamental definition of a GP is that the ratio of any term to its preceding term is constant across the entire sequence.
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Hint:
1
If \(\frac{y}{x} = r\), what does \(\frac{z}{y}\) equal?
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Mistakes to Avoid:
!
Skipping terms: \(\frac{z}{x}\) skips the middle term. This calculates \(r^2\), not \(r\).
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Solution:
By definition, \(\frac{t_2}{t_1} = \frac{t_3}{t_2}\).
Therefore, \(\frac{y}{x} = \frac{z}{y}\).
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Select the correct option:
When executing the algebraic proof \(y^2 = xz\) using standard variables (\(a, ar^3, ar^9\dots\)), what algebraic exponent rule is utilized to multiply \(x\) and \(z\)?
Concept:
1
GP proofs rely heavily on the Law of Exponents for Multiplication: When multiplying two terms with the same base, you add the exponents.
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Hint:
1
Think of \(x^2 \times x^3\). It is \((x \cdot x) \times (x \cdot x \cdot x) = x^5\). You added \(2+3\).
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Mistakes to Avoid:
!
Multiplying exponents: A very common error in GP proofs is evaluating \(r^3 \times r^{15}\) as \(r^{45}\). It must be \(r^{18}\).
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Solution:
The underlying mechanic that makes geometric sequences work is the product rule of exponents: \(r^m \times r^n = r^{m+n}\).
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Select the correct option:
If you have a Geometric Progression and you take the logarithm (log) of every single term, what happens to the sequence?
Concept:
1
The logarithm property states that \(\log(x^y) = y \cdot \log(x)\), and \(\log(xy) = \log(x) + \log(y)\).
2
Applying this to \(a \cdot r^{n-1}\) yields \(\log(a) + (n-1)\log(r)\).
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Hint:
1
Look at the transformed formula: \(\log(a) + (n-1)\log(r)\).
2
Does this look like \(A + (n-1)D\)? Yes!
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Mistakes to Avoid:
!
Assuming operations don't bridge: APs (addition) and GPs (multiplication) are intimately connected through exponents and logarithms.
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Solution:
Taking the log of a GP sequence \(a, ar, ar^2 \dots\) gives:
\(\log(a), \log(ar), \log(ar^2) \dots\)
\(\log(a), \log(a)+\log(r), \log(a)+2\log(r) \dots\)
This is an AP with first term \(\log(a)\) and common difference \(\log(r)\).
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