This page carries the complete exercise 5.1 class 10 maths ncert solutions for the chapter Arithmetic Progressions, with every question and sub-part solved step by step. Exercise 5.1 tests one core skill: deciding whether a given list of numbers is an arithmetic progression (AP), then reading off its first term and common difference.
All four questions are reproduced here exactly as printed in the official NCERT Class 10 Mathematics textbook, so you can follow every part without the book open.
Each answer opens with the idea behind it, then shows the working, then flags a mistake students actually make on that question. Below you will find the concepts you need, full solutions to Questions 1–4, a three-step method you can reuse on any list, and answers to the doubts that usually come up.
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Exercise 5.1 Class 10 Maths NCERT Solutions
This exercise trains you to recognise an arithmetic progression and to name its two building blocks: the first term and the common difference. Every question reduces to one idea — an AP is a list built by adding one fixed number each time. Once you can test whether a list moves by a constant addition, the whole exercise falls into place.
Only four ideas are needed for Questions 1–4:
- An arithmetic progression (AP) is a list of numbers in which each term, except the first, is obtained by adding a fixed number to the preceding term (NCERT, p. 52).
- That fixed number is the common difference \(d\), and it may be positive, negative or zero (NCERT, p. 52).
- The general form of an AP is \(a,\ a+d,\ a+2d,\ a+3d,\ \dots\), where \(a\) is the first term (NCERT, p. 53).
- The test: the differences \(a_2 – a_1,\ a_3 – a_2,\ a_4 – a_3,\ \dots\) must all be equal, that is \(d = a_{k+1} – a_k\) (NCERT, p. 54).
Subtraction order matters. The common difference is the later term minus the earlier term, never the reverse. For the list \(3, 1, -1, -3\), the difference is \(1 – 3 = -2\) — writing \(3 – 1 = 2\) flips the sign and can wrongly reject a valid AP.
Question 1: In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
- (i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
- (ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
- (iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
- (iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8 % per annum.
An AP needs each new term to be the previous term plus one fixed number. Before computing anything, ask: is each amount obtained by adding or by multiplying? If the list multiplies, it is not an AP.
Part (i): The first km costs ₹ 15 and every additional km adds ₹ 8, so the fares are 15, 23, 31, 39, … Each fare = previous fare + 8.
This is an AP with \(a = 15\) and \(d = 8\).
Part (ii): Removing \(\frac{1}{4}\) of the air leaves \(\frac{3}{4}\) of it.
The next removal again takes \(\frac{1}{4}\) of what remains, so the amount left each time is \(\frac{3}{4}\) of the previous amount — a multiplication, not an addition.
This is not an AP.
Part (iii): The first metre costs ₹ 150 and each subsequent metre adds ₹ 50, so the costs are 150, 200, 250, 300, … Each cost = previous cost + 50.
This is an AP with \(a = 150\) and \(d = 50\).
Part (iv): At 8% compound interest, the amount every year is the previous amount multiplied by 1.08.
The yearly increases grow (₹ 800, then ₹ 864, and so on), so no single common difference exists.
This is not an AP.
Common error: for the taxi or the well, students start the list at the per-km or per-metre addition (8, 16, 24 or 50, 100, 150). The first term is the starting charge — ₹ 15 for the first km and ₹ 150 for the first metre. Ask “add or multiply?” before you write any list down.
Question 2: Write first four terms of the AP, when the first term a and the common difference d are given as follows:
- (i) \(a = 10, \quad d = 10\)
- (ii) \(a = -2, \quad d = 0\)
- (iii) \(a = 4, \quad d = -3\)
- (iv) \(a = -1, \quad d = \frac{1}{2}\)
- (v) \(a = -1.25, \quad d = -0.25\)
Starting from the first term \(a\), add \(d\) each time to produce the next three terms. Watch the sign of \(d\) — a negative \(d\) makes the list fall, and a zero \(d\) keeps every term identical. Both are perfectly valid APs.
Part (i): \(a = 10,\ d = 10\): the terms are \(10, 20, 30, 40\).
Part (ii): \(a = -2,\ d = 0\): adding 0 each time gives \(-2, -2, -2, -2\).
Part (iii): \(a = 4,\ d = -3\): adding \(-3\) each time gives \(4, 1, -2, -5\).
Part (iv): \(a = -1,\ d = \frac{1}{2}\): adding \(\frac{1}{2}\) each time gives \(-1, -\frac{1}{2}, 0, \frac{1}{2}\).
Part (v): \(a = -1.25,\ d = -0.25\): adding \(-0.25\) each time gives \(-1.25, -1.50, -1.75, -2.00\).
Common error: mixing up the sign of \(d\), or stopping a decimal list at three terms. Check by subtracting consecutive terms — the difference you recover must equal the \(d\) you were given, so \(-2.00 – (-1.75) = -0.25\) confirms part (v).
Question 3: For the following APs, write the first term and the common difference:
- (i) \(3, 1, -1, -3, \dots\)
- (ii) \(-5, -1, 3, 7, \dots\)
- (iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots\)
- (iv) \(0.6, 1.7, 2.8, 3.9, \dots\)
The first term \(a\) is simply the first number printed. The common difference is the second term minus the first — later term minus earlier term, even when the later term is smaller.
Part (i): \(a = 3\), and \(d = 1 – 3 = -2\).
Part (ii): \(a = -5\), and \(d = -1 – (-5) = -1 + 5 = 4\).
Part (iii): \(a = \frac{1}{3}\), and \(d = \frac{5}{3} – \frac{1}{3} = \frac{4}{3}\).
Part (iv): \(a = 0.6\), and \(d = 1.7 – 0.6 = 1.1\).
Common error: in part (i), \(3 – 1 = 2\) looks tempting, but the difference must be later minus earlier, so \(1 – 3 = -2\). Reversing the order flips the sign and rejects a perfectly good AP. Regenerate the list from your \(a\) and \(d\) to verify: \(3,\ 3 + (-2),\ 3 + 2(-2),\ \dots\) matches the printed terms.
Question 4: Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
- (i) \(2, 4, 8, 16, \dots\)
- (ii) \(2, \frac{5}{2}, 3, \frac{7}{2}, \dots\)
- (iii) \(-1.2, -3.2, -5.2, -7.2, \dots\)
- (iv) \(-10, -6, -2, 2, \dots\)
- (v) \(3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots\)
- (vi) \(0.2, 0.22, 0.222, 0.2222, \dots\)
- (vii) \(0, -4, -8, -12, \dots\)
- (viii) \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots\)
- (ix) \(1, 3, 9, 27, \dots\)
- (x) \(a, 2a, 3a, 4a, \dots\)
- (xi) \(a, a^2, a^3, a^4, \dots\)
- (xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\)
- (xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots\)
- (xiv) \(1^2, 3^2, 5^2, 7^2, \dots\)
- (xv) \(1^2, 5^2, 7^2, 73, \dots\)
For each list, find consecutive differences as later term minus earlier term. If they are all equal, the list is an AP and that equal value is \(d\); you then extend it by adding \(d\) three times.
A list that follows any other rule — multiplying, squaring, growing differences — is not an AP, because an AP requires a constant added difference.
Part (i): \(4-2=2,\ 8-4=4,\ 16-8=8\) — differences are not equal (the list multiplies by 2).
Not an AP.
Part (ii): \(\frac{5}{2}-2 = \frac{1}{2},\ 3-\frac{5}{2} = \frac{1}{2},\ \frac{7}{2}-3 = \frac{1}{2}\) — constant.
AP, \(d = \frac{1}{2}\); next three terms: \(4, \frac{9}{2}, 5\).
Part (iii): \(-3.2-(-1.2) = -2,\ -5.2-(-3.2) = -2,\ -7.2-(-5.2) = -2\) — constant.
AP, \(d = -2\); next three terms: \(-9.2, -11.2, -13.2\).
Part (iv): \(-6-(-10) = 4,\ -2-(-6) = 4,\ 2-(-2) = 4\) — constant.
AP, \(d = 4\); next three terms: \(6, 10, 14\).
Part (v): \((3+\sqrt{2})-3 = \sqrt{2},\ (3+2\sqrt{2})-(3+\sqrt{2}) = \sqrt{2},\ (3+3\sqrt{2})-(3+2\sqrt{2}) = \sqrt{2}\) — constant.
AP, \(d = \sqrt{2}\); next three terms: \(3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}\).
Part (vi): \(0.22-0.2 = 0.02,\ 0.222-0.22 = 0.002,\ 0.2222-0.222 = 0.0002\) — the differences shrink, they are not fixed.
Not an AP.
Part (vii): \(-4-0 = -4,\ -8-(-4) = -4,\ -12-(-8) = -4\) — constant.
AP, \(d = -4\); next three terms: \(-16, -20, -24\).
Part (viii): every difference is \(-\frac{1}{2}-(-\frac{1}{2}) = 0\) — constant.
AP, \(d = 0\); next three terms: \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\).
Part (ix): \(3-1=2,\ 9-3=6,\ 27-9=18\) — differences grow (the list multiplies by 3).
Not an AP.
Part (x): \(2a-a=a,\ 3a-2a=a,\ 4a-3a=a\) — constant.
AP, \(d = a\); next three terms: \(5a, 6a, 7a\).
Part (xi): \(a^2-a,\ a^3-a^2,\ a^4-a^3\) — not equal (the terms multiply by \(a\)).
Not an AP.
Part (xii): Simplify first: \(\sqrt{8} = 2\sqrt{2},\ \sqrt{18} = 3\sqrt{2},\ \sqrt{32} = 4\sqrt{2}\), so the list is \(\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}\).
Differences are \(\sqrt{2}\) each.
AP, \(d = \sqrt{2}\); next three terms: \(5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2}\).
Part (xiii): \(\sqrt{9} = 3\) breaks the pattern: differences \(\sqrt{6}-\sqrt{3},\ 3-\sqrt{6},\ \sqrt{12}-3\) are not equal.
Not an AP.
Part (xiv): The squares are \(1, 9, 25, 49\); differences \(9-1=8,\ 25-9=16,\ 49-25=24\) are not equal.
Not an AP.
Part (xv): The list is \(1, 25, 49, 73\); differences \(25-1=24,\ 49-25=24,\ 73-49=24\) are constant.
AP, \(d = 24\); next three terms: \(97, 121, 145\).
Common error: calling a list an AP just because it has a pattern. \(1, 3, 9, 27\) has a clean pattern (multiply by 3), yet it is not an AP — only a constant added difference counts. Before declaring either way, test at least two differences; for surds, fractions and decimals, simplify each term first.
Three Steps to Test Any List for an AP
Every question in this exercise runs on the same routine. You can reuse it on any list of numbers.
- Write consecutive differences as later term minus earlier term — second minus first, third minus second, fourth minus third.
- Check whether they are all equal. If they are, the list is an AP and that equal value is the common difference \(d\). If any two differences differ, it is not an AP.
- To extend an AP, keep adding \(d\) to the last known term.
Simplify surds, fractions and decimals before testing. The most common trap is the difference between adding and multiplying:
| Adds a fixed number → AP | Multiplies or varies → not an AP |
|---|---|
| \(2, \frac{5}{2}, 3, \frac{7}{2}\) (adds \(\frac{1}{2}\)) | \(2, 4, 8, 16\) (multiplies by 2) |
| \(3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}\) (adds \(\sqrt{2}\)) | \(1, 3, 9, 27\) (multiplies by 3) |
| \(0, -4, -8, -12\) (adds \(-4\)) | \(0.2, 0.22, 0.222\) (differences shrink) |
The first term and common difference you extract here feed directly into the \(n\)th-term formula and the sum formula, which arrive in Exercises 5.2 and 5.3. If you want to revise the definitions first, our Class 10 Maths notes gather the chapter’s groundwork, and the quadratic equations solutions build the same habit of testing a pattern before trusting it.
The habit of asking “constant or changing?” carries into later chapters — in the triangles chapter you test constant ratios the same way you test a constant difference here. With every chapter mapped in one place, the Class 10 index is the fastest route around.
FAQs: Common Mistakes in Exercise 5.1
Why does the list 2, 4, 8, 16 not form an AP even though it follows a pattern?
An AP requires each term to be the previous term plus a fixed number, not times a fixed number. Here each term is the previous one multiplied by 2, so the differences are 2, 4, 8 — they grow instead of staying equal, and the list fails the test \(d = a_{k+1} – a_k\).
A pattern alone never makes an AP; only a constant added difference does.
Can the common difference of an AP be zero or negative?
Yes, both are valid. A zero \(d\) makes every term the same, as in \(-2, -2, -2, -2\) from Question 2(ii), and a negative \(d\) makes the list decrease, as in \(3, 1, -1, -3\) with \(d = -2\). The definition only requires a fixed number to be added each time — it never says that number must be positive.
What is the fastest way to check whether a given list is an AP?
Test two consecutive differences: second term minus first, and third term minus second. If they match, compute one more (fourth minus third) to be sure. For lists with surds, fractions or decimals, simplify every term first — the list \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}\) only reveals \(d = \sqrt{2}\) once you rewrite it as \(\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}\).
Why is compound interest not an arithmetic progression in Question 1?
At 8% compound interest, each year’s amount is the previous amount multiplied by 1.08, not increased by a fixed amount. The yearly increases themselves keep growing (₹ 800, then ₹ 864, then more), so no single common difference exists. Simple interest would form an AP, because it adds the same amount every year.
Do I need to simplify surds before checking whether a list is an AP?
Yes. Only after rewriting \(\sqrt{8} = 2\sqrt{2}\), \(\sqrt{18} = 3\sqrt{2}\) and \(\sqrt{32} = 4\sqrt{2}\) do the differences become visible as a constant \(\sqrt{2}\). Working with unsimplified surds hides the pattern and makes the common difference hard to spot — simplifying first is the cleaner and safer route.
Reference: NCERT Class 10 Mathematics textbook, chapter 5 Arithmetic Progressions.
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