Exercise 6.1 — Recognising AP

Recognition of arithmetic progression.

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Exercise 6.1 — Progressions (Arithmetic Progressions)

Exercise 6.1 is the opening exercise of Chapter 6, Progressions, in Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). It builds the foundation for the entire chapter by teaching you how to recognise an Arithmetic Progression (AP) — both in real-life situations and in plain number sequences — and how to work out the two values that completely describe any AP: the first term (a) and the common difference (d).

This exercise has four question types: checking whether everyday situations (taxi fares, well-digging costs, vacuum pumps, compound interest) form an AP; writing out the first four terms when a and d are given; working backwards to find a and d from a list of numbers; and identifying which of thirteen given sequences are genuine APs, then extending the genuine ones by three more terms.

Real-Life AP Situations Writing AP Terms Finding a & d Identify & Extend

💡 Foundation fact: In an Arithmetic Progression, every term (after the first) is obtained by adding a fixed number d, called the common difference, to the term before it. If the difference between consecutive terms keeps changing, the sequence is not an AP — no matter how "regular" it looks at first glance.

What Is an Arithmetic Progression (AP)?

A list of numbers in which each term after the first is found by adding a constant value to the previous term is called an Arithmetic Progression. That constant value is the common difference, written as d. The first number in the list is called the first term, written as a.

General form of an AP: a, a+d, a+2d, a+3d, ...

Example AP with a = 10 and d = 10 — each term is exactly 10 more than the one before it.

To test whether any list of numbers is an AP, calculate the difference between consecutive terms throughout the list:

d = a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = ...

If every difference comes out the same — the list is an AP, and that constant value is d.
If even one difference is different — the list is not an AP, even if the first one or two differences matched.
d can be positive (an increasing AP), negative (a decreasing AP), or zero (a constant sequence where every term is the same).

📌 Important distinction: Many "tricky" sequences in this exercise are not APs because they grow by a constant ratio rather than a constant difference — these are Geometric Progressions (GP) instead. Recognising the difference is the single most useful skill for this exercise.

Arithmetic Progression (AP)

Same number is added each time.

2, 4, 6, 8, ... (d = 2)

Taxi fare per extra km
Simple interest year on year
Cost of digging that rises by a fixed amount

Geometric Progression (GP)

Same number is multiplied each time.

2, 4, 8, 16, ... (ratio = 2)

Air removed in fixed fractions by a vacuum pump
Compound interest year on year
Population growing at a fixed percentage rate

Question 1 — Do These Real-Life Situations Form an AP?

This question takes four everyday situations and asks you to write out the resulting list of numbers, then decide whether that list is an Arithmetic Progression. The trick in each part is to actually compute two or three terms and check the differences — never assume from the wording alone.

Part (i) — AP ✓

Taxi fare: Rs. 20 for the first km, then Rs. 8 for every extra km

The fare for the first km is fixed at Rs. 20. After that, every additional km adds exactly Rs. 8, so the fares for 1 km, 2 km, 3 km, 4 km... are:

Fares: 20, 28, 36, 44, ... a₂ − a₁ = 28 − 20 = 8 a₃ − a₂ = 36 − 28 = 8 a₄ − a₃ = 44 − 36 = 8 (same difference every time)

✅ Conclusion: The differences are constant (= 8), so this list forms an AP with a = 20 and d = 8.

Part (ii) — Not an AP ✗

Vacuum pump removing ¼ of the remaining air each time

Assume the cylinder starts with 1024 litres of air. Each time the pump runs, it removes ¼ of whatever air remains — not a fixed amount, but a fixed fraction of a shrinking quantity.

Start: 1024 L Removed 1st time: ¼ × 1024 = 256 L → remaining = 1024 − 256 = 768 L Removed 2nd time: ¼ × 768 = 192 L → remaining = 768 − 192 = 576 L Removed 3rd time: ¼ × 576 = 144 L → remaining = 576 − 144 = 432 L Air left after each step: 1024, 768, 576, 432, ... a₂ − a₁ = 768 − 1024 = −256 a₃ − a₂ = 576 − 768 = −192 a₄ − a₃ = 432 − 576 = −144 (differences keep changing)

⛔ Conclusion: The difference between successive terms is not constant (−256, then −192, then −144), so this list does not form an AP. Because each step removes a constant fraction of the remaining air, the remaining-air values actually form a Geometric Progression instead.

Part (iii) — AP ✓

Cost of digging a well: Rs. 150 for the first metre, rising by Rs. 50 each metre after

The cost for the first metre is Rs. 150, and every metre after that adds a fixed Rs. 50:

Costs: 150, 200, 250, 300, ... a₂ − a₁ = 200 − 150 = 50 a₃ − a₂ = 250 − 200 = 50 a₄ − a₃ = 300 − 250 = 50 (same difference every time)

✅ Conclusion: The costs form an AP with a = 150 and d = 50.

Part (iv) — Not an AP ✗

Rs. 10,000 deposited at 8% compound interest per year

Because the interest is compound, each year's interest is calculated on the previous year's total amount (not on the original Rs. 10,000), so the amount grows by a fixed multiplying factor of 1.08 each year — not by a fixed rupee amount.

Amount after year 1: 10000 × 108/100 = 10800 Amount after year 2: 10800 × 108/100 = 11664 Amount after year 3: 11664 × 108/100 = 12597.12 Amounts: 10800, 11664, 12597.12, ... a₂ − a₁ = 11664 − 10800 = 864 a₃ − a₂ = 12597.12 − 11664 = 933.12 (differences keep changing)

⛔ Conclusion: 864 ≠ 933.12, so the difference is not constant — the amounts do not form an AP. Compound interest always produces a Geometric Progression, never an Arithmetic one.

Situation	Resulting List	Is it an AP?
(i) Taxi fare	20, 28, 36, 44, ...	Yes
(ii) Vacuum pump (¼ removed)	1024, 768, 576, 432, ...	No
(iii) Well-digging cost	150, 200, 250, 300, ...	Yes
(iv) Compound interest @ 8%	10800, 11664, 12597.12, ...	No

💡 Pattern to remember: Situations involving a fixed addition (extra km charge, extra metre charge, simple interest) always give an AP. Situations involving a fixed percentage or fraction (compound interest, repeated fractional removal, population growth rate) always give a GP, never an AP.

Question 2 — Writing the First Four Terms of an AP

Once you know the first term a and the common difference d, every other term of the AP can be generated using the general nth-term rule:

aₙ = a + (n − 1)d

So the first four terms are simply a, a + d, a + 2d, and a + 3d. The table below works out all five parts of this question:

Part	a	d	First Four Terms (a, a+d, a+2d, a+3d)
(i)	10	10	10, 20, 30, 40
(ii)	−2	0	−2, −2, −2, −2
(iii)	4	−3	4, 1, −2, −5
(iv)	−1	1/2	−1, −1/2, 0, 1/2
(v)	−1.25	−0.25	−1.25, −1.50, −1.75, −2.00

Worked Example — Part (iii)

a = 4, d = −3

1st term = a = 4 2nd term = a + d = 4 + (−3) = 1 3rd term = a + 2d = 4 + 2(−3) = 4 − 6 = −2 4th term = a + 3d = 4 + 3(−3) = 4 − 9 = −5

1st

2nd

3rd

−2

4th

−5

📌 Watch the sign of d: When d is negative (parts iii, v), every term gets smaller — the AP is decreasing. When d = 0 (part ii), every term is identical — a special "constant" AP. When d is a fraction (part iv), keep terms as fractions or convert to decimals consistently; don't mix the two in the same answer.

Question 3 — Finding the First Term and Common Difference

This question works in the opposite direction to Question 2: instead of being given a and d, you are given the AP itself, and must read off a (simply the first listed number) and calculate d using:

d = a₂ − a₁

Part	Given AP	First Term (a)	Common Difference (d)
(i)	3, 1, −1, −3, ...	3	−2
(ii)	−5, −1, 3, 7, ...	−5	4
(iii)	1/3, 5/3, 9/3, 13/3, ...	1/3	4/3
(iv)	0.6, 1.7, 2.8, 3.9, ...	0.6	1.1

Worked Example — Part (iii)

AP: 1/3, 5/3, 9/3, 13/3, ...

First term: a = 1/3 d = a₂ − a₁ = 5/3 − 1/3 = (5−1)/3 = 4/3 Check: a₃ − a₂ = 9/3 − 5/3 = 4/3 ✓ matches Check: a₄ − a₃ = 13/3 − 9/3 = 4/3 ✓ matches

⚠️ Common slip in part (ii): Watch the signs carefully — the list is −5, −1, 3, 7, so d = (−1) − (−5) = −1 + 5 = 4, a positive common difference, even though the list begins with negative numbers.

Question 4 — Identify the AP and Write the Next Three Terms

This is the longest and most important part of Exercise 6.1 — thirteen sequences to test, one by one, using the same method every time: compute at least two consecutive differences (a₂ − a₁ and a₃ − a₂); if they match, it's an AP, write d and use aₙ₊₁ = aₙ + d to extend it by three more terms; if they don't match, it's not an AP and no common difference exists.

#	Sequence	a₂−a₁	a₃−a₂	AP?	d	Next 3 Terms
(i)	2, 4, 8, 16, ...	2	4	No	—	— (this is a GP)
(ii)	2, 5/2, 3, 7/2, ...	1/2	1/2	Yes	1/2	4, 9/2, 5
(iii)	−1.2, −3.2, −5.2, −7.2, ...	−2	−2	Yes	−2	−9.2, −11.2, −13.2
(iv)	−10, −6, −2, 2, ...	4	4	Yes	4	6, 10, 14
(v)	3, 3+√2, 3+2√2, 3+3√2, ...	√2	√2	Yes	√2	3+4√2, 3+5√2, 3+6√2
(vi)	0.2, 0.22, 0.222, 0.2222, ...	0.02	0.002	No	—	—
(vii)	0, −4, −8, −12, ...	−4	−4	Yes	−4	−16, −20, −24
(viii)	−1/2, −1/2, −1/2, −1/2, ...	0	0	Yes	0	−1/2, −1/2, −1/2
(ix)	1, 3, 9, 27, ...	2	6	No	—	— (this is a GP)
(x)	a, 2a, 3a, 4a, ...	a	a	Yes	a	5a, 6a, 7a
(xi)	a, a², a³, a⁴, ...	a²−a	a³−a²	No	—	—
(xii)	√2, √8, √18, √32, ...	√2	√2	Yes	√2	5√2, 6√2, 7√2
(xiii)	√3, √6, √9, √12, ...	≈0.72	≈0.55	No	—	—

Worked Example — Part (v): Surds in an AP

3, 3+√2, 3+2√2, 3+3√2, ...

a₂ − a₁ = (3+√2) − 3 = √2 a₃ − a₂ = (3+2√2) − (3+√2) = √2 a₄ − a₃ = (3+3√2) − (3+2√2) = √2 ✓ constant

✅ This is an AP with d = √2. Next terms: a₅ = 3+3√2+√2 = 3+4√2, a₆ = 3+5√2, a₇ = 3+6√2.

Worked Example — Part (xii) vs (xiii): The Surd Trap

√2, √8, √18, √32 (an AP) vs. √3, √6, √9, √12 (not an AP)

These two parts look almost identical at first glance — both are lists of square roots that appear to "grow steadily." But simplifying each surd reveals the real pattern:

√2 = √2, √8 = √(4×2) = 2√2, √18 = √(9×2) = 3√2, √32 = √(16×2) = 4√2 Simplified: √2, 2√2, 3√2, 4√2 → differences are all √2 → this IS an AP √3 ≈ 1.73, √6 ≈ 2.45, √9 = 3, √12 ≈ 3.46 Differences: 2.45−1.73=0.72, 3−2.45=0.55, 3.46−3=0.46 → NOT constant → NOT an AP

⛔ Exam trap: Never judge a surd sequence "by eye." Always simplify each term under the root first (look for a common factor you can pull out), then test the differences. Two sequences that look the same shape can have completely different answers.

📌 On part (xi): Whether a, a², a³, a⁴ forms an AP actually depends on the specific value of a — for example if a = 1, every term equals 1 and it trivially is an AP with d = 0. But for a general value of a (the standard assumption in this exercise), the differences a²−a and a³−a² are not equal, so it is treated as not an AP.

Common Mistakes to Avoid

Checking only one difference: Computing just a₂ − a₁ and stopping there can fool you — always check at least one more difference (a₃ − a₂) before declaring something an AP.
Confusing AP with GP: Sequences like 1, 3, 9, 27 or 2, 4, 8, 16 "look regular" but grow by multiplication, not addition — they are Geometric Progressions, not Arithmetic ones.
Not simplifying surds first: As seen in parts (xii) and (xiii), you must simplify expressions like √8 or √18 into the form k√2 before you can correctly test for a common difference.
Sign errors with negative numbers: When subtracting negatives (e.g. −1 − (−5)), remember this becomes −1 + 5 = 4, not −6. Double negatives are a frequent source of error in this exercise.
Forgetting that d = 0 is still valid: A constant sequence like −1/2, −1/2, −1/2, ... is technically a valid AP with common difference zero — don't dismiss it as "not an AP" just because nothing changes.
Mixing decimals and fractions inconsistently: Pick one form (fraction or decimal) for a given part and stay consistent throughout your working and final answer.

⛔ High-risk exam trap: Real-life situations involving compound growth (compound interest, percentage-based depreciation, repeated fractional removal) are a favourite board-exam trick to test whether students truly understand that AP requires a constant difference, not just "growing in a pattern."

Quick Reference — All Answers at a Glance

Question	Topic	Key Result
Q1(i)	Taxi fare	AP, a=20, d=8
Q1(ii)	Vacuum pump (air removal)	Not AP (it's a GP)
Q1(iii)	Well-digging cost	AP, a=150, d=50
Q1(iv)	Compound interest	Not AP (it's a GP)
Q2(i)–(v)	First four terms from a, d	See terms table above
Q3(i)–(iv)	Finding a and d from an AP	See a/d table above
Q4(i),(vi),(ix),(xi),(xiii)	Sequences that are NOT APs	5 of the 13 parts
Q4(ii),(iii),(iv),(v),(vii),(viii),(x),(xii)	Sequences that ARE APs	8 of the 13 parts

What This Exercise Prepares You For

Exercise 6.1 is the gateway to the rest of Chapter 6. Once you are confident at spotting an AP and extracting a and d, the next exercises build directly on this skill — deriving and applying the nth-term formula aₙ = a + (n−1)d to solve for unknown terms, and later, finding the sum of the first n terms of an AP using Sₙ = n/2 [2a + (n−1)d].

The surd-simplification skill practised in parts (v), (xii), and (xiii) of Question 4 connects directly back to the Real Numbers chapter, where simplifying expressions under a square root is taught in detail. The careful sign-handling needed throughout Question 3 also reinforces concepts from Rational Numbers.

📐 Board Exam Tip (CBSE, Telangana & AP): Question 4 of Exercise 6.1 is extremely popular in board exams as a standalone 2–3 mark question, often picking just two or three of the thirteen parts. Practise the full set so that recognising AP vs. non-AP patterns becomes instant, especially the surd-simplification cases and the AP-vs-GP real-life situations from Question 1.