Chapter 6.2 — Exercise 6.1 — Recognising AP

Four Real Situations, Tested

Not every "increases by a rule" situation is actually an AP in disguise — some rules act on percentages or fractions of the current amount rather than fixed amounts, and that single distinction changes everything about the resulting sequence.

• Taxi fare: ₹20 charged for the first km, then ₹8 for each additional km beyond that — fares run 20, 28, 36, 44, … Each term simply adds a fixed ₹8. This is an AP, d = 8.
• Air in a cylinder: a vacuum pump removes exactly ¼ of the remaining air each time it runs. Starting from 1024 litres: 1024, 768, 576, 432, … The differences are −256, −192, −144 — shrinking each time, not constant. Not an AP, because removing a quarter of an already shrinking amount removes less in absolute terms every round.
• Well-digging cost: ₹150 charged for the first metre, rising by a flat ₹50 for every metre after that — costs run 150, 200, 250, 300, … This is an AP, with d = 50.
• Compound interest: ₹10,000 invested at 8% per annum grows to 10800, then 11664, then 12597.12, and so on year after year. The differences are 864 and 933.12 — not equal, because 8% of a growing balance is itself growing. Not an AP.

The pattern across all four: a fixed amount added or subtracted each step (₹8 per km, ₹50 per metre) always produces an AP; a fixed percentage or fraction of the current value (¼ of the remaining air, 8% interest on the current balance) never does, because the actual amount added or removed changes as the underlying total changes. It's worth being able to tell which kind a new situation is before doing any arithmetic at all: if the rule can be phrased as "add/subtract [a specific number]," expect an AP; if it can only be phrased as "add/subtract [a percentage or fraction] of whatever the current value is," expect something else entirely — a pattern this chapter revisits directly once geometric progressions are introduced.

Building an AP From a and d

Given just the first term and the common difference, the first four terms follow directly from the general form a, a+d, a+2d, a+3d:

The second row is worth pausing on: a common difference of exactly 0 is perfectly valid, and produces a sequence where every term is identical. It might not look like it's "progressing" anywhere, but it satisfies the AP definition just as completely as any of the others — the definition only ever requires a fixed number to be added each step, and 0 is as fixed a number as any other.

The fifth row is worth a second glance too, purely as an arithmetic-accuracy check: subtracting a negative common difference (−0.25) from an already-negative starting value (−1.25) means every term keeps getting more negative, moving further from zero rather than back toward it. Tracking the sign of both a and d separately, rather than trying to judge "increasing or decreasing" by eye, avoids the sign-flip errors that decimal APs like this one make easy to commit.

Working Backward: Finding a and d

Given only the sequence itself, a is simply the first term, and d is the difference between any two consecutive terms — though it's good practice to check the difference between a second pair too, confirming it matches, rather than trusting just one subtraction:

• 3, 1, −1, −3, … → a = 3, d = 1−3 = −2
• −5, −1, 3, 7, … → a = −5, d = −1−(−5) = 4
• 1/3, 5/3, 9/3, 13/3, … → a = 1/3, d = 5/3−1/3 = 4/3
• 0.6, 1.7, 2.8, 3.9, … → a = 0.6, d = 1.7−0.6 = 1.1

Thirteen Sequences, Classified

The exercise's largest single question tests the consecutive-difference rule against thirteen separate sequences — some obviously arithmetic at a glance, others deliberately constructed to look arithmetic without actually being one, right up until every difference is checked.

Five sequences deserve individual attention beyond the summary table above, since each one illustrates a genuinely distinct idea the exercise is deliberately testing for.

• 3, 3+√2, 3+2√2, 3+3√2, … — the common difference is √2, an irrational number. The AP test doesn't require d to be rational, only constant: (3+√2)−3=√2, (3+2√2)−(3+√2)=√2, and so on. This is an AP, and the next three terms are 3+4√2, 3+5√2, 3+6√2.
• a, 2a, 3a, 4a, … — a sequence built from a variable instead of numbers. The differences are still constant: 2a−a=a, 3a−2a=a, 4a−3a=a. This is an AP, d=a, with next terms 5a, 6a, 7a.
• a, a², a³, a⁴, … — a close cousin of the sequence above, but built from powers instead of multiples. Here a²−a is not the same as a³−a² in general (try a=3: 9−3=6, but 27−9=18). Not an AP — a useful contrast with the previous sequence, since both "look algebraic" but only one actually has a constant difference.
• √2, √8, √18, √32, … — simplifying each term first reveals the pattern: √8=2√2, √18=3√2, √32=4√2, so the sequence is really √2, 2√2, 3√2, 4√2, … This is an AP, d=√2, with next terms 5√2, 6√2, 7√2.
• √3, √6, √9, √12, … — these look similar to the previous sequence, but √6−√3 is not equal to √9−√6 (√9=3 simplifies away its root entirely, breaking the pattern the others share). Not an AP — the closest possible contrast to the previous, genuinely-AP square-root sequence, and a reminder to actually compute every difference rather than trust a visual pattern.

From Recognising to Calculating

Every AP identified in this exercise stopped at "here are the next three terms" — genuinely useful for confirming a pattern by hand, but far too slow a method for finding, say, the fiftieth term of any of them. Exercise 6.2 replaces that term-by-term counting with a direct formula, aₙ = a + (n−1)d, that jumps straight to any term without listing the ones before it. For the AP definition and general form this exercise relies on throughout, revisit the chapter introduction, which introduces the exact notation this exercise's answers are all expressed in.