Exercise 5.1 — Arithmetic Progressions

1. 14 marksNCERT Cl-10 Maths, Ex 5.1, Q1 (four situations)

   In which of these situations does the list of numbers involved form an AP, and why? (i) A taxi charges ₹15 for the first km and ₹8 for each additional km. (ii) The air in a cylinder is removed by a vacuum pump that takes out 1/4 of the air remaining at each stroke. (iii) The cost of digging a well rises by ₹50 for each successive metre, starting at ₹150 for the first metre. (iv) An amount of ₹10,000 is deposited at 8% compound interest per year.

   Hint. A list is an AP only if you get from each term to the next by ADDING a fixed number. Multiplying by a fixed factor is not the same thing.

   The whole question turns on one distinction: adding a constant gives an AP, multiplying by a constant does not.

   (i) The taxi. Fares are 15, 23, 31, 39, … Each km after the first adds ₹8, a fixed amount.

   ✦ AP, with a = 15 and d = 8.

   (ii) The vacuum pump. Each stroke removes a quarter of what is left, so 3/4 of the air remains. Starting from a volume V the amounts are V, (3/4)V, (9/16)V, …

   The differences are not equal — each term is 3/4 of the one before, which is multiplication, not addition.

   ✦ Not an AP.

   (iii) The well. Costs are 150, 200, 250, 300, … with a fixed ₹50 added each metre.

   ✦ AP, with a = 150 and d = 50.

   (iv) Compound interest. Each year the amount is multiplied by 1.08, giving 10000, 10800, 11664, … The differences are 800, then 864 — not equal.

   This is the standard trap of the question: simple interest would give an AP, compound interest never does.

   ✦ Not an AP.

   Where students slip. Calling (ii) and (iv) APs because the *rule* is constant. The rule being constant is not enough — it is the constant *difference* that defines an AP, and a fixed percentage change produces a constant ratio instead.

   Another way. A quick test for any list: compute two consecutive differences. If they disagree, stop — it is not an AP, and you do not need to check further.

2. 24 marksNCERT Cl-10 Maths, Ex 5.1, Q2 (four parts)

   Write the first four terms of the AP when the first term a and the common difference d are given: (i) a = 10, d = 10; (ii) a = −2, d = 0; (iii) a = 4, d = −3; (iv) a = −1, d = 1/2.

   Hint. Just keep adding d. The only thing to watch is the sign of d.

   Each term is the previous one plus d.

   (i) a = 10, d = 10. 10, then 20, 30, 40.

   ✦ 10, 20, 30, 40

   (ii) a = −2, d = 0. Adding zero changes nothing, so every term repeats.

   ✦ −2, −2, −2, −2

   This is a legitimate AP. A common difference of zero is allowed — the list is constant, but the difference between consecutive terms is still fixed.

   (iii) a = 4, d = −3. A negative d means the terms decrease: 4, 1, −2, −5.

   ✦ 4, 1, −2, −5

   (iv) a = −1, d = 1/2. −1, then −1/2, 0, 1/2.

   ✦ −1, −1/2, 0, 1/2

   Where students slip. In (iii), adding 3 instead of −3 after the first step. Write the operation as '+ (−3)' rather than 'subtract 3' and the signs take care of themselves.

   Another way. You can also use aₙ = a + (n−1)d directly for each n from 1 to 4, which is worth doing once to see the formula agreeing with simple repeated addition.

3. 33 marksNCERT Cl-10 Maths, Ex 5.1, Q3 (four parts)

   For each AP below, write down the first term and the common difference: (i) 3, 1, −1, −3, …; (ii) −5, −1, 3, 7, …; (iii) 1/3, 5/3, 9/3, 13/3, …; (iv) 0.6, 1.7, 2.8, 3.9, …

   Hint. The first term is simply the first number. Get d by subtracting any term from the one after it.

   The first term needs no work — it is just the number written first. For d, subtract any term from the one that follows it, and then confirm with a second pair, because a single difference cannot tell you whether the list is really arithmetic.

   (i) 3, 1, −1, −3. a = 3, and d = 1 − 3 = −2. Check: −1 − 1 = −2 ✓

   ✦ a = 3, d = −2

   (ii) −5, −1, 3, 7. a = −5, and d = −1 − (−5) = 4, since subtracting a negative adds. Check: 3 − (−1) = 4 ✓

   ✦ a = −5, d = 4

   (iii) 1/3, 5/3, 9/3, 13/3. a = 1/3, and d = 5/3 − 1/3 = 4/3. The denominators are already common, so only the numerators move — they climb by 4 each time. Check: 9/3 − 5/3 = 4/3 ✓

   ✦ a = 1/3, d = 4/3

   (iv) 0.6, 1.7, 2.8, 3.9. a = 0.6, and d = 1.7 − 0.6 = 1.1. Check: 2.8 − 1.7 = 1.1 ✓

   ✦ a = 0.6, d = 1.1

   Where students slip. Computing d backwards as (first) − (second). In (ii) that would give −4 instead of 4, and every later answer built on it would be wrong. Always later term minus earlier term.

   Another way. In (iii) the terms are 1/3, 5/3, 9/3, 13/3 — leaving the numerators over a common denominator makes the step of 4/3 visible without any subtraction.

4. 45 marksNCERT Cl-10 Maths, Ex 5.1, Q4 (eleven lists)

   Which of the following lists form an AP? For those that do, state the common difference and write three more terms. Lists include: 2, 4, 8, 16, …; 2, 5/2, 3, 7/2, …; −1.2, −3.2, −5.2, −7.2, …; −10, −6, −2, 2, …; 3, 3+√2, 3+2√2, 3+3√2, …; 0.2, 0.22, 0.222, …; 0, −4, −8, −12, …; 1, 3, 9, 27, …; and a, a², a³, ….

   Hint. Compute at least two differences for each list. Equal differences means AP; anything else means stop.

   2, 4, 8, 16, … Differences 2, 4, 8 — each term doubles.

   ✦ Not an AP.

   2, 5/2, 3, 7/2, … Differences all 1/2.

   ✦ AP, d = 1/2. Next three: 4, 9/2, 5.

   −1.2, −3.2, −5.2, −7.2, … Differences all −2.

   ✦ AP, d = −2. Next three: −9.2, −11.2, −13.2.

   −10, −6, −2, 2, … Differences all 4.

   ✦ AP, d = 4. Next three: 6, 10, 14.

   3, 3+√2, 3+2√2, 3+3√2, … Each step adds √2 — the terms are irrational but the difference is constant, which is all that matters.

   ✦ AP, d = √2. Next three: 3+4√2, 3+5√2, 3+6√2.

   0.2, 0.22, 0.222, … Differences 0.02 then 0.002.

   ✦ Not an AP. (Tempting, because the pattern looks regular — but regular is not the same as arithmetic.)

   0, −4, −8, −12, … Differences all −4.

   ✦ AP, d = −4. Next three: −16, −20, −24.

   1, 3, 9, 27, … Each term is tripled.

   ✦ Not an AP.

   a, a², a³, … Each term is multiplied by a.

   ✦ Not an AP in general. (If a = 1 every term is 1, which is an AP with d = 0 — but as a general list it is not.)

   Where students slip. Judging 0.2, 0.22, 0.222 as an AP because the digits follow an obvious pattern. Compute the differences: 0.02 and 0.002 are not equal, so it fails.

   Another way. For lists like 1, 3, 9, 27 check the ratio instead — a constant ratio identifies a geometric progression, which is a useful thing to be able to name even though it is not on this syllabus.

Solutions written by the tuition.in editorial team and checked against the rationalised NCERT Class 10 Mathematics textbook and the CBSE 2026-27 syllabus (nothing was removed from this chapter; Exercise 5.4 is the optional exercise and is still present). Questions are referenced from the NCERT textbook for identification.