Exercise 6.4 — Geometric Progression

Introduction of geometric progression.

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Exercise 6.4 — Geometric Progressions (Class 10 Maths)

Exercise 6.4 is part of Chapter 6, Progressions, of Class 10 Mathematics for CBSE, Telangana, and Andhra Pradesh board students. While earlier exercises in this chapter deal with Arithmetic Progressions (AP), Exercise 6.4 introduces a new and equally important type of sequence — the Geometric Progression (GP). Every term in a GP is obtained by multiplying the previous term by a fixed number called the common ratio.

This exercise tests three skills: recognising whether a given situation produces a GP, generating GP terms from given values of a and r, and identifying whether a listed sequence is a GP and continuing it. The final question introduces an algebraic approach to finding an unknown term. Together, these problems prepare students for the GP sum formula and real-world compound-growth applications in higher classes.

Geometric Progressions Common Ratio Test Finding Next Terms Real-World GP Problems Algebraic GP Equation

What Is a Geometric Progression?

Definition — Geometric Progression (GP)
A sequence of numbers in which each term is obtained by multiplying the preceding term by a fixed non-zero number (except the first term) is called a Geometric Progression. The fixed multiplier is called the common ratio, denoted by r.

If the first term of a GP is a and the common ratio is r, then the sequence takes the form:

a, ar, ar², ar³, ar⁴, ……
General (nth) term: aₙ = a · r^(n-1)
Common ratio: r = a₂/a₁ = a₃/a₂ = a₄/a₃ = ……

The key test for a GP is simple: the ratio of any term to the term immediately before it must be the same throughout the sequence. If consecutive ratios differ even once, the sequence is NOT a GP.

Four Classic Examples of GP

SequenceFirst Term (a)Common Ratio (r)Ratios Equal?
3, 6, 12, 24, 48, … 32 ✔ Yes — GP
2, 10, 50, 250, 1250, … 25 ✔ Yes — GP
½, ⅛, 1/32, 1/128, … ½¼ ✔ Yes — GP
0.2, 0.04, 0.008, 0.0016, … 0.20.2 ✔ Yes — GP
📌 Key distinction from AP: In an Arithmetic Progression (AP), you add a fixed number (common difference) to get the next term. In a GP, you multiply by a fixed number (common ratio). The two concepts are completely different and students often confuse them.

Question 1 — Identifying GP in Real-World Situations

This question presents three practical scenarios and asks whether the numbers involved form a GP. The approach is always the same: generate the first few terms, then check whether the ratio between consecutive terms remains constant.

Question 1 · Part (i)
Sharmila's salary starts at ₹5,00,000 and increases by 10% every year. Is the list of yearly salaries a GP?
Year 1: a₁ = 5,00,000 Year 2: a₂ = 5,00,000 × (110/100) = 5,50,000 Year 3: a₃ = 5,50,000 × (110/100) = 6,05,000 Year 4: a₄ = 6,05,000 × (110/100) = 6,65,500 Check ratios: a₂/a₁ = 5,50,000 / 5,00,000 = 11/10 a₃/a₂ = 6,05,000 / 5,50,000 = 11/10 a₄/a₃ = 6,65,500 / 6,05,000 = 11/10 Common ratio r = 11/10 = 1.1 throughout. ∴ The salaries form a G.P.
₹5,00,000
×1.1→
₹5,50,000
×1.1→
₹6,05,000
×1.1→
₹6,65,500
…
✅ Real-life insight: Any salary, investment, or quantity that grows by a fixed percentage each period automatically forms a GP. The common ratio equals (1 + rate/100). A 10% annual increase gives r = 1.1; a 20% increase gives r = 1.2; and so on. This is the mathematical foundation of compound interest.
Question 1 · Part (ii)
A staircase has 30 steps. The bottom step needs 100 bricks and each step needs 2 bricks less than the one below. Is the brick count per step a GP?
Step 1 (bottom): a₁ = 100 Step 2: a₂ = 100 − 2 = 98 Step 3: a₃ = 98 − 2 = 96 Step 4: a₄ = 96 − 2 = 94 Check ratios: a₂/a₁ = 98/100 = 49/50 a₃/a₂ = 96/98 = 48/49 ← different from 49/50 a₄/a₃ = 94/96 = 47/48 ← different again Common ratio is NOT constant. ∴ The brick counts do NOT form a G.P.
💡 Why not? Subtracting a fixed number each time produces an Arithmetic Progression (AP), not a GP. The brick sequence 100, 98, 96, 94, … is an AP with common difference −2. For a GP, you need to multiply, not add or subtract, by a fixed number.
Question 1 · Part (iii)
Midpoints of the sides of an equilateral triangle with side 24 cm are joined to form smaller triangles, again and again. Is the sequence of perimeters a GP?
Outer triangle side = 24 cm → a₁ = 3 × 24 = 72 cm Midpoint theorem: each inner side = half the outer side = 12 cm Inner triangle 1 → a₂ = 3 × 12 = 36 cm Inner triangle 2 → a₃ = 3 × 6 = 18 cm Inner triangle 3 → a₄ = 3 × 3 = 9 cm Check ratios: a₂/a₁ = 36/72 = 1/2 a₃/a₂ = 18/36 = 1/2 a₄/a₃ = 9/18 = 1/2 Common ratio r = 1/2 throughout. ∴ The perimeters form a G.P.
12 cm 12 cm 12 cm 24 cm 24 cm 24 cm P₁ = 72 P₂ = 36 P₃ = 18 P₄ = 9
Nested equilateral triangles
Each inner perimeter = ½ of previous → GP with r = ½
📌 Midpoint Theorem connection: The line joining midpoints of two sides of a triangle is parallel to the third side and exactly half its length. This is why each successive triangle's side is exactly half the previous — making the perimeter sequence a perfect GP with r = ½.
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Question 2 — Write the Next Three Terms of a GP Given a and r

When you know the first term a and the common ratio r, generating terms is straightforward: multiply repeatedly by r. The formula for the nth term is aₙ = a · r^(n−1).

Question 2 · Part (i)
a = 4, r = 3 — Find the next three terms
1st term: a₁ = a = 4 2nd term: a₂ = ar = 4 × 3 = 12 3rd term: a₃ = ar² = 4 × 3² = 4 × 9 = 36 4th term: a₄ = ar³ = 4 × 3³ = 4 × 27 = 108 Next three terms: 12, 36, 108
4
×3→
12
×3→
36
×3→
108
Question 2 · Part (ii)
a = √5, r = 1/√5 — Find the next three terms
1st term: a₁ = √5 2nd term: a₂ = √5 × (1/√5) = 1 Actually: a₂ = ar = √5 × (1/√5) = 1/√5 × √5·... a₂ = √5 · (1/√5) = 1 wait — let us use ar = 5^(1/2) · 5^(-1/2) = 5^0 = 1 Correct: a₂ = √5 × (1/√5) = 1 3rd term: a₃ = ar² = √5 × (1/√5)² = √5 × (1/5) = 1/√5 4th term: a₄ = ar³ = √5 × (1/√5)³ = √5 × (1/5√5) = 1/(5) Next three terms: 1, 1/√5, 1/5
📌 Note: The textbook presents the same working with a = √5, which gives 2nd term = 1 (not listed as a term to find), so the next three terms after √5 are 1, 1/√5, and 1/(5√5) — always double-check which term you start listing from.
Question 2 · Part (iii)
a = 81, r = −1/3 — Find the next three terms
1st term: a₁ = 81 2nd term: a₂ = 81 × (−1/3) = −27 3rd term: a₃ = 81 × (−1/3)² = 81 × (1/9) = 9 4th term: a₄ = 81 × (−1/3)³ = 81 × (−1/27) = −3 Next three terms: −27, 9, −3
81
×(−⅓)→
−27
×(−⅓)→
9
×(−⅓)→
−3
💡 Negative common ratio: When r is negative, the signs alternate — positive, negative, positive, negative, … This is a typical pattern in GP questions that catches students off guard. The magnitude still forms a GP; only the sign flips each time.
Question 2 · Part (iv)
a = 1/64, r = 2 — Find the next three terms
1st term: a₁ = 1/64 2nd term: a₂ = (1/64) × 2 = 1/32 3rd term: a₃ = (1/64) × 2² = (1/64) × 4 = 1/16 4th term: a₄ = (1/64) × 2³ = (1/64) × 8 = 1/8 Next three terms: 1/32, 1/16, 1/8
1/64
×2→
1/32
×2→
1/16
×2→
1/8

Question 3 — Test Each Sequence: GP or Not? Then Find Next Three Terms

For each sequence, divide consecutive terms to check for a constant ratio. If all ratios match, it is a GP and you can extend it. If even one ratio differs, it is not a GP.

PartSequencea₂/a₁a₃/a₂GP?rNext Three Terms
(i) 4, 8, 16, … 22 ✔ Yes 2 32, 64, 128
(ii) 1/3, −1/6, 1/12, … −1/2−1/2 ✔ Yes −1/2 −1/24, 1/48, −1/96
(iii) 5, 55, 555, … 11111/11 ✘ No — —
(iv) −2, −6, −18, … 33 ✔ Yes 3 −54, −162, −486
(v) 1/2, 1/4, 1/6, … 1/22/3 ✘ No — —
(vi) 3, −3², 3³, … −3−3 ✔ Yes −3 −3⁴, 3⁵, −3⁶
(vii) x, 1, 1/x, … 1/x1/x ✔ Yes 1/x 1/x², 1/x³, 1/x⁴
(viii) 1/√2, −2, 8/√2, … −2√2−2√2 ✔ Yes −2√2 −16, 32√2, −128
(ix) 0.4, 0.04, 0.004, … 0.10.1 ✔ Yes 0.1 0.0004, 0.00004, 0.000004

Now let's look at the detailed working for each part:

Q3 · Part (i)
Sequence: 4, 8, 16, …
a₂/a₁ = 8/4 = 2 a₃/a₂ = 16/8 = 2 → r = 2 (constant) → GP a₄ = 4 × 2³ = 32 a₅ = 4 × 2⁴ = 64 a₆ = 4 × 2⁵ = 128 Next three terms: 32, 64, 128
Q3 · Part (ii)
Sequence: 1/3, −1/6, 1/12, …
a₂/a₁ = (−1/6) ÷ (1/3) = −1/6 × 3 = −1/2 a₃/a₂ = (1/12) ÷ (−1/6) = 1/12 × (−6) = −1/2 → r = −1/2 → GP a₄ = (1/3) × (−1/2)³ = (1/3) × (−1/8) = −1/24 a₅ = (1/3) × (−1/2)⁴ = (1/3) × (1/16) = 1/48 a₆ = (1/3) × (−1/2)⁵ = (1/3) × (−1/32) = −1/96 Next three terms: −1/24, 1/48, −1/96
Q3 · Part (iii)
Sequence: 5, 55, 555, … — Is this a GP?
a₂/a₁ = 55/5 = 11 a₃/a₂ = 555/55 = 111/11 ≈ 10.09 ← not equal to 11 Ratios are different. ∴ 5, 55, 555 is NOT a GP.
⛔ This is a common trick question. The sequence looks like it might have a pattern, but the ratios are 11, 111/11, 1111/111, … which are all different. Don't be misled by the visual pattern of repeating digits.
Q3 · Part (iv)
Sequence: −2, −6, −18, …
a₂/a₁ = −6/−2 = 3 a₃/a₂ = −18/−6 = 3 → r = 3 → GP a₄ = −2 × 3³ = −2 × 27 = −54 a₅ = −2 × 3⁴ = −2 × 81 = −162 a₆ = −2 × 3⁵ = −2 × 243 = −486 Next three terms: −54, −162, −486
Q3 · Part (v)
Sequence: 1/2, 1/4, 1/6, … — Is this a GP?
a₂/a₁ = (1/4) ÷ (1/2) = 1/4 × 2 = 1/2 a₃/a₂ = (1/6) ÷ (1/4) = 1/6 × 4 = 2/3 ← not equal to 1/2 Ratios are different. ∴ 1/2, 1/4, 1/6 is NOT a GP.
📌 Students often confuse this with the GP 1/2, 1/4, 1/8, 1/16 … (which has r = 1/2). The sequence 1/2, 1/4, 1/6, 1/8 … has denominators in AP (2, 4, 6, 8 …), not a GP of denominators.
Q3 · Part (vi)
Sequence: 3, −3², 3³, … (written as powers of 3 with alternating sign)
a₁ = 3, a₂ = −9, a₃ = 27 a₂/a₁ = −9/3 = −3 a₃/a₂ = 27/−9 = −3 → r = −3 → GP a₄ = 3 × (−3)³ = 3 × (−27) = −81 = −3⁴ a₅ = 3 × (−3)⁴ = 3 × 81 = 243 = 3⁵ a₆ = 3 × (−3)⁵ = 3 × (−243) = −729 = −3⁶ Next three terms: −3⁴, 3⁵, −3⁶ i.e., −81, 243, −729
Q3 · Part (vii)
Sequence: x, 1, 1/x, … (x ≠ 0)
a₂/a₁ = 1/x a₃/a₂ = (1/x) ÷ 1 = 1/x → r = 1/x → GP a₄ = x · (1/x)³ = 1/x² a₅ = x · (1/x)⁴ = 1/x³ a₆ = x · (1/x)⁵ = 1/x⁴ Next three terms: 1/x², 1/x³, 1/x⁴
💡 Algebraic GP: This is a GP expressed in terms of a variable x. The approach is identical — compute the ratio, confirm it's constant, then apply aₙ = a · rⁿ⁻¹. Works the same whether numbers or algebraic expressions are involved.
Q3 · Part (viii)
Sequence: 1/√2, −2, 8/√2, …
a₂/a₁ = (−2) ÷ (1/√2) = −2√2 a₃/a₂ = (8/√2) ÷ (−2) = −4/√2 = −4·√2/2 = −2√2 → r = −2√2 → GP a₄ = (1/√2)(−2√2)³ = (1/√2)(−16√2) = −16 a₅ = (1/√2)(−2√2)⁴ = (1/√2)(64) = 32√2 a₆ = (1/√2)(−2√2)⁵ = (1/√2)(−128√2) = −128 Next three terms: −16, 32√2, −128
Q3 · Part (ix)
Sequence: 0.4, 0.04, 0.004, …
a₂/a₁ = 0.04/0.4 = 0.1 a₃/a₂ = 0.004/0.04 = 0.1 → r = 0.1 → GP a₄ = 0.4 × (0.1)³ = 0.0004 a₅ = 0.4 × (0.1)⁴ = 0.00004 a₆ = 0.4 × (0.1)⁵ = 0.000004 Next three terms: 0.0004, 0.00004, 0.000004
📌 Each term has one extra zero after the decimal point. In general, aₙ = 4 × 10^(−n). This type of GP appears in questions about geometric series and converging sums in higher classes.

Question 4 — Find x so that x, x+2, x+6 are Consecutive Terms of a GP

This algebraic question uses the defining property of a GP: the ratio between consecutive terms must be constant. Equivalently, the square of the middle term equals the product of the first and third terms. This leads to a simple linear equation.

Question 4 · Solution
x, x+2, x+6 are in GP. Find x.
Set up the GP condition: For three terms to be in GP, the ratio of 2nd to 1st must equal the ratio of 3rd to 2nd.
a₂/a₁ = a₃/a₂ → (x+2)/x = (x+6)/(x+2)
Cross-multiply:
(x+2)² = x(x+6)
Expand both sides:
x² + 4x + 4 = x² + 6x
Simplify:
x² + 4 − x² = 6x − 4x
4 = 2x
x = 2
Verify: With x = 2, the three terms are 2, 4, 8 — which is a GP with r = 2. ✔
(x+2)/x = (x+6)/(x+2) ⟹ (x+2)² = x(x+6) ⟹ x² + 4x + 4 = x² + 6x ⟹ 4 = 2x ⟹ x = 2 Verification: 2, 4, 8 → r = 4/2 = 8/4 = 2 ✔
✅ Alternative method (middle-term property): In any GP, the square of the middle term = product of the first and third terms. So (x+2)² = x · (x+6). This is the same equation and gives the same result. This property is useful when the GP condition is given in a different form.
GP Middle-Term Property: (b)² = a × c
i.e. if a, b, c are in GP then b² = ac

Common Mistakes to Avoid in Exercise 6.4

  • Confusing AP and GP: The most common error. In AP you add/subtract a constant; in GP you multiply/divide by a constant. The brick staircase problem (Q1-ii) is designed to catch students who automatically assume a decreasing sequence is a GP.
  • Not checking ALL consecutive ratios: Check at least two consecutive ratios — a₂/a₁ and a₃/a₂. Checking only one ratio is not enough to confirm a GP. In Q3(v), the first ratio looks like 1/2 but the second is 2/3.
  • Sign errors with negative common ratio: When r is negative (Q2-iii, Q3-ii, Q3-vi), be careful about powers: (−r)² is positive, (−r)³ is negative. Odd powers keep the negative; even powers remove it.
  • In Q4, forgetting to verify: After solving for x algebraically, substitute back and confirm the three values actually form a GP. Board examiners deduct marks if verification is missing.
  • Wrong formula for nth term: The nth term of a GP is aₙ = a·r^(n−1), not a·rⁿ. The exponent is (n−1), not n. This off-by-one error produces wrong answers for every term.
⛔ High-risk exam trap (Q3-v): The sequence 1/2, 1/4, 1/6, 1/8 … looks like a GP because the terms are halving, but they're not — the denominators increase by 2 (an AP of denominators). The actual GP would be 1/2, 1/4, 1/8, 1/16, … (denominators doubling). Always compute ratios; never guess from appearance.

Quick Reference — Key Formulas for GP

Formula / PropertyExpressionUsed In
General form of GP a, ar, ar², ar³, ar⁴, … All problems
nth term aₙ = a · r^(n−1) Q2, Q3 — finding next terms
Common ratio r = a₂/a₁ = a₃/a₂ = … Q1, Q3 — testing for GP
Middle-term property If a, b, c are in GP: b² = ac Q4 — algebraic GP
GP condition a₂/a₁ = a₃/a₂ = constant Q1, Q3 — testing sequences
Q1(i)GP — r = 11/10
Q1(ii)Not GP (AP)
Q1(iii)GP — r = 1/2
Q3(i)GP — 32, 64, 128
Q3(iii)Not GP
Q3(v)Not GP
Q4x = 2 → 2, 4, 8

What Exercise 6.4 Prepares You For

This exercise builds the foundation for Exercise 6.5, which introduces the formula for the sum of n terms of a GP (Sₙ = a(rⁿ − 1)/(r − 1)) and the sum of an infinite GP when |r| < 1. The common-ratio identification skill you practise here is essential for applying that formula correctly.

The compound-interest connection (Q1-i) is revisited in Comparing Quantities, where GP-based growth models appear in financial maths. The nested-triangle problem (Q1-iii) also resurfaces in the context of similarity of triangles and fractal-like geometric patterns.

For students appearing in Telangana and AP board exams, GP questions frequently appear as 2-mark or 4-mark problems. The most common question types are: (a) given a and r, find the nth term; (b) given first few terms, test for GP and extend; and (c) find an unknown using the GP condition (as in Q4). All three types are fully covered by this exercise.

📐 Board Exam Tip (Telangana & AP): Always show the ratio check explicitly — write out a₂/a₁ = … and a₃/a₂ = … before declaring a sequence to be a GP. Examiners award process marks for showing the ratio computation, not just for stating the conclusion. Similarly, in Q4-type problems, always verify your answer by substituting back and showing the three resulting values form a GP.
🎯 Revision checklist before your exam:
  • Can you state the definition of GP and identify the common ratio from any sequence?
  • Do you know the nth term formula aₙ = a·r^(n−1) and can apply it quickly?
  • Can you distinguish a GP from an AP when both appear in the same problem?
  • Do you know the middle-term property b² = ac and when to use it?
  • Have you practised Q1(iii) — the nested triangles problem — as it combines midpoint theorem with GP?
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