Class 9 - Ganita Manjari Chapter 8 - Part 22
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Class 9 Ganita Manjari Chapter 8: Sequences and Progressions
Part 22
THE LOGIC OF MULTIPLICATION End of Chapter - GP Proofs Three consecutive GP terms: \(a, \quad a \cdot r, \quad a \cdot r^2\) Proof: \((ar)^2 = (a)(ar^2)\) The square of the middle is the product of the extremes.
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!
1 The Geometric Mean Proof
Ganita Manjari - Pg 195
End of Chapter Exercise: If \(a, b,\) and \(c\) are three consecutive terms of a Geometric Progression, prove that \(b^2 = ac\).
Concept:
1
By definition, the ratio between consecutive terms in a GP is always equal. Therefore, the ratio of the second to the first term must equal the ratio of the third to the second term.
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Hint:
1
Set up the ratio equation: \(\frac{b}{a} = \frac{c}{b}\).
2
Cross-multiply to solve.
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Mistakes to Avoid:
!
Using AP logic: Do not write \(b - a = c - b \implies 2b = a + c\). That is the Arithmetic Mean! We are proving the Geometric Mean, which relies purely on multiplication and division.
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Solution:
Since \(a, b,\) and \(c\) are in GP, they must share a common ratio \(r\).
\(\frac{b}{a} = r\)
\(\frac{c}{b} = r\)

Equating the ratios:
\(\frac{b}{a} = \frac{c}{b}\)

Cross-multiplying:
\(b \cdot b = a \cdot c\)
\(b^2 = ac\). Hence Proved.
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2 Distant Term Proofs
Ganita Manjari - Pg 195
End of Chapter Exercise: The 4th, 10th, and 16th terms of a Geometric Progression are \(x, y,\) and \(z\) respectively. Prove that \(x, y,\) and \(z\) are themselves in GP.
Concept:
1
Express \(x, y,\) and \(z\) using the standard explicit formula \(t_n = a \cdot r^{n-1}\).
2
To prove that three numbers are in GP, you must prove that the square of the middle term equals the product of the outer terms (i.e., prove \(y^2 = xz\)).
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Hint:
1
\(x = ar^3\), \(y = ar^9\), \(z = ar^{15}\).
2
Calculate \(y^2 = (ar^9)^2\).
3
Calculate \(x \cdot z = (ar^3)(ar^{15})\). Do they match?
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Mistakes to Avoid:
!
Adding exponents incorrectly: When multiplying \(r^3 \times r^{15}\), you MUST add the exponents (\(3+15=18\)). Do not multiply them to get \(r^{45}\)! Likewise, \((r^9)^2 = r^{18}\), not \(r^{81}\).
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Solution:
1. Define the terms:
\(x = t_4 = a \cdot r^3\)
\(y = t_{10} = a \cdot r^9\)
\(z = t_{16} = a \cdot r^{15}\)

2. Calculate \(y^2\) (LHS):
\(y^2 = (a \cdot r^9)^2 = a^2 \cdot r^{18}\)

3. Calculate \(x \cdot z\) (RHS):
\(x \cdot z = (a \cdot r^3)(a \cdot r^{15}) = a^2 \cdot r^{3+15} = a^2 \cdot r^{18}\)

Conclusion:
Since \(y^2 = xz\), the terms \(x, y, z\) form a Geometric Progression. Hence Proved.
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★ Complex Ratio Proofs
Ganita Manjari - Pg 195
End of Chapter Exercise Q20 (*): If \(a, b,\) and \(c\) are in Geometric Progression, prove that:

\(\frac{a+b}{b+c} = \frac{a}{b}\)
Concept:
1
This is a HOTS algebraic proof. When given a GP, express every variable in terms of the first term (\(a\)) and the common ratio (\(r\)).
2
Substitute these definitions into the Left Hand Side (LHS) and Right Hand Side (RHS) of the equation independently, and show that they simplify to the exact same expression.
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Hint:
1
Let \(a = a\), \(b = ar\), and \(c = ar^2\).
2
Substitute into LHS: \(\frac{a + ar}{ar + ar^2}\). Factor out common terms from the numerator and denominator.
3
Substitute into RHS: \(\frac{a}{ar}\). Simplify.
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Mistakes to Avoid:
!
Illegal canceling: In the fraction \(\frac{a + ar}{ar + ar^2}\), you cannot just cross out the \(ar\) terms on the top and bottom! You must factor them out first: \(\frac{a(1+r)}{ar(1+r)}\).
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Solution:
Since \(a, b, c\) are in GP, let:
\(a = a\)
\(b = ar\)
\(c = ar^2\)

Evaluating LHS:
\(\frac{a+b}{b+c} = \frac{a + ar}{ar + ar^2}\)
\(= \frac{a(1+r)}{ar(1+r)}\)
The \((1+r)\) and \(a\) terms cancel out, leaving: \(\frac{1}{r}\).

Evaluating RHS:
\(\frac{a}{b} = \frac{a}{ar} = \frac{1}{r}\).

Conclusion:
Since LHS = RHS = \(\frac{1}{r}\), the equation holds true. Hence Proved.
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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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Chapter 8: Sequences & Progressions

Part 1: Intro to Sequences & Patterns Ganita Manjari Pgs 174-175 Part 2: Explicit Rules & Predicting Terms Ganita Manjari Pgs 176-177 Part 3: Recursive Rules & Fibonacci Ganita Manjari Pgs 178-179 Part 4: Exercise 8.1 - Mastery (Q1-Q3) Ganita Manjari Pg 179 Part 5: Exercise 8.1 - Recursive (Q4-Q6) Ganita Manjari Pg 179-180 Part 6: Intro to AP & Common Difference Ganita Manjari Pgs 180-182 Part 7: AP Formulas & Taxi Fare Math Ganita Manjari Pgs 182-183 Part 8: Sum of First n Natural Numbers Ganita Manjari Pgs 183-185 Part 9: Exercise 8.2 - AP Basics (Q1-Q3) Ganita Manjari Pg 185 Part 10: Exercise 8.2 - Missing Terms (Q4-Q5) Ganita Manjari Pgs 185-186 Part 11: Exercise 8.2 - AP Word Problems Ganita Manjari Pg 186 Part 12: Intro to GP & Common Ratios Ganita Manjari Pgs 186-188 Part 13: Fun with Fractals: SierpiΕ„ski Ganita Manjari Pgs 188-190 Part 14: GP Graphs & Bouncing Balls Ganita Manjari Pgs 190-193 Part 15: Exercise 8.3 - GP Basics Ganita Manjari Pg 193 Part 16: Exercise 8.3 - Rebounds & Roots Ganita Manjari Pgs 193-194 Part 17: Exercise 8.3 - SierpiΕ„ski Carpet Ganita Manjari Pg 194 Part 18: End of Chapter - AP Logic Ganita Manjari Pgs 194-195 Part 19: End of Chapter - Divisibility Ganita Manjari Pg 195 Part 20: End of Chapter - Consecutive Sums Ganita Manjari Pg 195 Part 21: End of Chapter - Dual Equations Ganita Manjari Pg 195 Part 22: End of Chapter - GP Proofs Ganita Manjari Pg 195 Part 23: End of Chapter - Complex Logic Ganita Manjari Pg 195 Part 24: End of Chapter - Recursive Chains Ganita Manjari Pg 195 Part 25: Chapter Summary & Finale Ganita Manjari Pg 196
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