Exercise 5.2 — Arithmetic Progressions

1. 15 marksNCERT Cl-10 Maths, Ex 5.2, Q1 (five parts)

   Fill in the missing quantity in each row, where a is the first term, d the common difference and aₙ the nth term: (i) a = 7, d = 3, n = 8, find aₙ; (ii) a = −18, n = 10, aₙ = 0, find d; (iii) d = −3, n = 18, aₙ = −5, find a; (iv) a = −18.9, d = 2.5, aₙ = 3.6, find n; (v) a = 3.5, d = 0, n = 105, find aₙ.

   Hint. One formula does all five: aₙ = a + (n − 1)d. Substitute the three you know and solve for the fourth.

   One formula covers all five rows: aₙ = a + (n − 1)d. The (n − 1) is there because reaching the nth term takes one step fewer than the number of terms — the first term needs no steps at all.

   (i) Find a₈. a₈ = 7 + (8 − 1)(3) = 7 + 21 = 28, so seven steps of 3 are added to the start.

   ✦ a₈ = 28

   (ii) Find d. 0 = −18 + 9d, since reaching the 10th term takes 9 steps. That gives 9d = 18, so d = 2.

   ✦ d = 2

   (iii) Find a. −5 = a + 17(−3) = a − 51, which means a = 46. The first term is far above the 18th because the sequence is falling.

   ✦ a = 46

   (iv) Find n. 3.6 = −18.9 + (n − 1)(2.5). The total climb is 22.5, and each step is 2.5, so nine steps are needed and n = 10.

   ✦ n = 10

   (v) Find a₁₀₅. With d = 0 nothing is ever added, so every term equals the first.

   ✦ a₁₀₅ = 3.5

   Where students slip. Using n instead of (n − 1) in the formula. The first term needs no steps added, the second needs one — so reaching the nth term takes n − 1 steps, not n.

   Another way. In (iv) you can shortcut: the total climb is 3.6 − (−18.9) = 22.5, and each step is 2.5, so 9 steps are needed, putting you at the 10th term.

2. 24 marksNCERT Cl-10 Maths, Ex 5.2, Q2

   Choose the correct answer and justify it: (i) the 30th term of the AP 10, 7, 4, … ; (ii) the 11th term of the AP −3, −1/2, 2, ….

   Hint. Find d first, then apply the formula. Watch the sign in part (i).

   (i) AP 10, 7, 4, …

   Here a = 10 and d = 7 − 10 = −3. The terms are falling, so the 30th term must be well below 10 — worth noting before computing, because it rules out a positive answer.

   a₃₀ = 10 + 29(−3) = 10 − 87 = −77.

   ✦ a₃₀ = −77

   (ii) AP −3, −1/2, 2, …

   Here a = −3 and d = −1/2 − (−3) = −1/2 + 3 = 5/2, since subtracting −3 means adding 3. Check the next gap: 2 − (−1/2) = 5/2 ✓

   a₁₁ = −3 + 10(5/2) = −3 + 25 = 22.

   ✦ a₁₁ = 22

   Where students slip. In (ii), computing d as −1/2 − 3 = −7/2 by dropping the minus on the first term. The first term is −3, so subtracting it means adding 3.

   Another way. Sanity-check the sign of your answer before computing: in (i) the terms are falling, so the 30th term must be well below 10 — a positive answer would be wrong on sight.

3. 34 marksNCERT Cl-10 Maths, Ex 5.2, Q3

   Find the missing terms in each AP: (i) 2, __, 26; (ii) __, 13, __, 3; (iii) 5, __, __, 9½; (iv) −4, __, __, __, __, 6; (v) __, 38, __, __, __, −22.

   Hint. Count how many steps separate the two known terms — that tells you how many d's fit between them.

   (i) 2, __, 26. These are the 1st and 3rd terms, so two steps apart: 26 = 2 + 2d ⟹ d = 12. The middle term is 14.

   ✦ 14

   (ii) __, 13, __, 3. The 2nd term is 13 and the 4th is 3, two steps apart: 3 = 13 + 2d ⟹ d = −5. So the 1st term is 13 − (−5) = 18 and the 3rd is 8.

   ✦ 18 and 8

   (iii) 5, __, __, 9½. The 1st is 5 and the 4th is 9.5, three steps apart: 9.5 = 5 + 3d ⟹ d = 1.5. Terms: 6.5 and 8.

   ✦ 6½ and 8

   (iv) −4, __, __, __, __, 6. The 1st is −4 and the 6th is 6, five steps apart: 6 = −4 + 5d ⟹ d = 2. Terms: −2, 0, 2, 4.

   ✦ −2, 0, 2, 4

   (v) __, 38, __, __, __, −22. The 2nd is 38 and the 6th is −22, four steps apart: −22 = 38 + 4d ⟹ d = −15. The 1st is 38 + 15 = 53, and the middle terms are 23, 8, −7.

   ✦ 53, 23, 8, −7

   Where students slip. Counting terms instead of gaps. Between the 1st and 6th terms there are five steps, not six — the number of d's is always one less than the number of terms spanned.

   Another way. For (i) there is a neat shortcut: the middle term of three consecutive AP terms is always the average of its neighbours, so it is (2 + 26)/2 = 14.

4. 43 marksNCERT Cl-10 Maths, Ex 5.2, Q4

   Which term of the AP 3, 8, 13, 18, … is 78?

   Hint. Set aₙ = 78 and solve for n. A whole-number answer means it is a term.

   Step 1 — Here a = 3 and d = 8 − 3 = 5.

   Step 2 — Set the nth term equal to 78: 3 + (n − 1)(5) = 78.

   Step 3 — Simplify: (n − 1)(5) = 75, so n − 1 = 15 and n = 16.

   Step 4 — n came out a positive whole number, so 78 genuinely is a term of this AP.

   ✦ 78 is the 16th term.

   Where students slip. Reporting n = 15 by stopping at n − 1 = 15. The equation solves for n − 1; you still have to add 1.

   Another way. Check by counting: from 3 to 78 is a rise of 75, which at 5 per step takes 15 steps — landing on the 16th term.

5. 53 marksNCERT Cl-10 Maths, Ex 5.2, Q5

   Find the number of terms in each AP: (i) 7, 13, 19, …, 205; (ii) 18, 15½, 13, …, −47.

   Hint. Put the last term into aₙ = a + (n − 1)d and solve for n.

   To count the terms, treat the last term as the nth and solve for n. That works because the last term is just another term — nothing about it is special except its position.

   (i) 7, 13, 19, …, 205. Here a = 7 and d = 6.

   205 = 7 + (n − 1)(6), so 198 = 6(n − 1), which gives n − 1 = 33 and n = 34.

   ✦ 34 terms

   (ii) 18, 15½, 13, …, −47. Here a = 18 and d = 15.5 − 18 = −2.5.

   −47 = 18 + (n − 1)(−2.5), so −65 = −2.5(n − 1). Both sides are negative, so the quotient is positive: n − 1 = 26 and n = 27.

   ✦ 27 terms

   Where students slip. Dividing by a negative d and losing the sign in (ii). Both sides are negative there, so the quotient is positive — if you get a negative n, a sign has gone astray.

   Another way. Rearranged, n = (last − first)/d + 1. It is the same calculation, but having it as a single formula reduces slips under time pressure.

6. 63 marksNCERT Cl-10 Maths, Ex 5.2, Q6

   Check whether −150 is a term of the AP 11, 8, 5, 2, ….

   Hint. Solve for n as usual — but this time pay attention to whether n turns out to be a whole number.

   Step 1 — Here a = 11 and d = 8 − 11 = −3.

   Step 2 — Suppose −150 is the nth term: 11 + (n − 1)(−3) = −150.

   Step 3 — Simplify: (n − 1)(−3) = −161, so n − 1 = 161/3.

   Step 4 — That gives n = 161/3 + 1 = 164/3 ≈ 54.67.

   Step 5 — Term numbers must be positive integers. Since 164/3 is not an integer, no term of this AP equals −150. The value falls between the 54th and 55th terms.

   ✦ No — −150 is not a term of this AP.

   Where students slip. Rounding 54.67 to 55 and answering 'yes, the 55th term'. Check it: the 55th term is 11 + 54(−3) = −151, not −150. A non-integer n is the answer, not something to tidy up.

   Another way. Divisibility gives it away faster: every term is 11 − 3k, so terms leave remainder 2 when divided by 3. Since −150 is exactly divisible by 3, it cannot be one.

7. 73 marksNCERT Cl-10 Maths, Ex 5.2, Q7

   Find the 31st term of an AP whose 11th term is 38 and whose 16th term is 73.

   Hint. Write both given terms with the formula. Subtracting the two equations eliminates a and leaves d.

   Step 1 — Write both terms: a + 10d = 38 and a + 15d = 73.

   Step 2 — Subtract the first from the second. The a cancels: 5d = 35, so d = 7.

   Step 3 — Substitute back: a + 70 = 38, so a = −32.

   Step 4 — Now the 31st term: a₃₁ = −32 + 30(7) = −32 + 210 = 178.

   Step 5 — Check against a given value: a₁₆ = −32 + 15(7) = −32 + 105 = 73 ✓

   ✦ a₃₁ = 178

   Where students slip. Writing a₁₁ as a + 11d. It is a + 10d — the 11th term is ten steps past the first.

   Another way. Faster route: from the 16th to the 31st is 15 steps, so a₃₁ = 73 + 15(7) = 178, no need to find a at all.

8. 83 marksNCERT Cl-10 Maths, Ex 5.2, Q8

   An AP has 50 terms; its third term is 12 and its last term is 106. Find the 29th term.

   Hint. The last term is the 50th, so write a₃ and a₅₀ and eliminate a.

   Step 1 — a₃ = a + 2d = 12 and a₅₀ = a + 49d = 106.

   Step 2 — Subtract: 47d = 94, so d = 2.

   Step 3 — Then a + 4 = 12, giving a = 8.

   Step 4 — a₂₉ = 8 + 28(2) = 8 + 56 = 64.

   Step 5 — Check: a₅₀ = 8 + 49(2) = 106 ✓

   ✦ a₂₉ = 64

   Where students slip. Treating '50 terms' as if the last term were a₅₁. If a list has 50 terms, the last one is the 50th.

   Another way. From the 3rd to the 29th is 26 steps, so a₂₉ = 12 + 26(2) = 64 directly.

9. 93 marksNCERT Cl-10 Maths, Ex 5.2, Q9

   The 3rd and 9th terms of an AP are 4 and −8 respectively. Which term of this AP is zero?

   Hint. Find a and d first, then set aₙ = 0 and solve for n.

   Step 1 — a + 2d = 4 and a + 8d = −8.

   Step 2 — Subtract: 6d = −12, so d = −2.

   Step 3 — Then a − 4 = 4, giving a = 8.

   Step 4 — Set the nth term to zero: 8 + (n − 1)(−2) = 0 ⟹ (n − 1)(−2) = −8 ⟹ n − 1 = 4 ⟹ n = 5.

   Step 5 — Check: the AP runs 8, 6, 4, 2, 0 — and 0 is indeed the fifth term ✓

   ✦ The 5th term is zero.

   Where students slip. Concluding that a term of zero is impossible because d is negative. A decreasing AP passes through zero on its way to negative values — the only question is whether it lands exactly on it.

   Another way. Since a₃ = 4 and each step drops 2, zero arrives two steps later — at the 5th term.

10. 103 marksNCERT Cl-10 Maths, Ex 5.2, Q10

    The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

    Hint. You do not need a at all — the difference between two terms depends only on d.

    Step 1 — Write both terms: a₁₇ = a + 16d and a₁₀ = a + 9d.

    Step 2 — The condition says a₁₇ − a₁₀ = 7.

    Step 3 — Substitute: (a + 16d) − (a + 9d) = 7. The a cancels, leaving 7d = 7.

    Step 4 — So d = 1.

    ✦ d = 1

    The structure worth noticing: the gap between the mth and nth terms is always (m − n)d. Here 17 − 10 = 7 steps, so 7d = 7.

    Where students slip. Trying to find a as well and getting stuck because the question does not determine it. Infinitely many APs satisfy this condition — only d is fixed, and only d was asked for.

    Another way. Apply the general fact directly: aₘ − aₙ = (m − n)d, so d = 7/(17 − 10) = 1.

11. 113 marksNCERT Cl-10 Maths, Ex 5.2, Q11

    Which term of the AP 3, 15, 27, 39, … will be 132 more than its 54th term?

    Hint. Find the 54th term, add 132, then work out which term that value is.

    Step 1 — Here a = 3 and d = 12.

    Step 2 — The 54th term: a₅₄ = 3 + 53(12) = 3 + 636 = 639.

    Step 3 — The target is 132 more than that value, so it is 639 + 132 = 771. Note this is a difference of values, not of positions.

    Step 4 — Solve 3 + (n − 1)(12) = 771, which gives (n − 1)(12) = 768, so n − 1 = 64 and n = 65.

    ✦ The 65th term.

    A neater way to see it: since each step adds 12, being 132 higher means climbing 132/12 = 11 extra steps, and 54 + 11 = 65.

    Where students slip. Adding 132 to the term *number* rather than the term *value* — answering 54 + 132 = 186. The 132 is a difference between values.

    Another way. The shortcut in the last line generalises: to be k more than the mth term, go k/d steps further along, provided k/d is a whole number.

12. 123 marksNCERT Cl-10 Maths, Ex 5.2, Q12

    Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?

    Hint. Write the nth terms of both APs and subtract. See what survives.

    Step 1 — Let the first terms be a and b, with the same common difference d for both.

    Step 2 — Their 100th terms are a + 99d and b + 99d. Subtracting:

    (a + 99d) − (b + 99d) = a − b.

    The d terms cancel completely. So a − b = 100.

    Step 3 — Their 1000th terms are a + 999d and b + 999d. Subtracting gives a − b again.

    Step 4 — So the difference is the same at every position — it is always a − b = 100.

    ✦ 100

    Geometrically: two APs with equal common difference are parallel sequences. They never converge or diverge, so the gap between them is fixed from the start.

    Where students slip. Scaling the answer to 1000 by proportion, giving 1000. Nothing here grows with n — the d terms cancel identically, whatever n is.

    Another way. Test it with numbers: take d = 1 with first terms 100 and 0. The 100th terms are 199 and 99 (difference 100); the 1000th are 1099 and 999 (difference 100).

13. 133 marksNCERT Cl-10 Maths, Ex 5.2, Q13

    How many three-digit numbers are divisible by 7?

    Hint. The multiples of 7 form an AP with d = 7. Find the first and last three-digit ones, then count the terms.

    Step 1 — The smallest three-digit multiple of 7: 100 ÷ 7 ≈ 14.3, so the first is 7 × 15 = 105.

    Step 2 — The largest: 999 ÷ 7 ≈ 142.7, so the last is 7 × 142 = 994.

    Step 3 — These form an AP: 105, 112, …, 994 with a = 105 and d = 7.

    Step 4 — Solve 994 = 105 + (n − 1)(7) ⟹ 889 = 7(n − 1) ⟹ n − 1 = 127 ⟹ n = 128.

    ✦ There are 128 three-digit numbers divisible by 7.

    Where students slip. Starting at 98 or ending at 1001 — both are multiples of 7 but not three-digit numbers. Check that your endpoints are inside the range before counting.

    Another way. Count directly with the multiplier: the multiples run from 7 × 15 to 7 × 142, so the count is 142 − 15 + 1 = 128.

14. 143 marksNCERT Cl-10 Maths, Ex 5.2, Q14

    How many multiples of 4 lie between 10 and 250?

    Hint. Find the first multiple above 10 and the last below 250, then count the terms of that AP.

    Step 1 — The first multiple of 4 greater than 10 is 12.

    Step 2 — The largest multiple of 4 below 250: 250 ÷ 4 = 62.5, so it is 4 × 62 = 248.

    Step 3 — The AP is 12, 16, …, 248 with a = 12 and d = 4.

    Step 4 — Solve 248 = 12 + (n − 1)(4) ⟹ 236 = 4(n − 1) ⟹ n − 1 = 59 ⟹ n = 60.

    ✦ 60 multiples of 4 lie between 10 and 250.

    Where students slip. Including 8 or 252 by taking 'between 10 and 250' too loosely. Both endpoints must be strictly inside the stated range.

    Another way. By multiplier: the multiples run from 4 × 3 to 4 × 62, so the count is 62 − 3 + 1 = 60.

15. 153 marksNCERT Cl-10 Maths, Ex 5.2, Q15

    For what value of n are the nth terms of the APs 63, 65, 67, … and 3, 10, 17, … equal?

    Hint. Write both nth terms with the formula, set them equal, and solve the resulting linear equation.

    Step 1 — First AP: a = 63, d = 2, so its nth term is 63 + (n − 1)(2) = 61 + 2n.

    Step 2 — Second AP: a = 3, d = 7, so its nth term is 3 + (n − 1)(7) = 7n − 4.

    Step 3 — Set them equal: 61 + 2n = 7n − 4.

    Step 4 — Solve: 65 = 5n, so n = 13.

    Step 5 — Check both: first AP gives 61 + 26 = 87; second gives 91 − 4 = 87 ✓

    ✦ n = 13, and the common value is 87.

    Where students slip. Setting the two *first* terms equal, or comparing terms at different positions. The question asks for the same n in both sequences.

    Another way. The second AP gains 5 per step on the first, and starts 60 behind. Closing a 60 gap at 5 per step takes 12 steps, landing at the 13th term.

16. 163 marksNCERT Cl-10 Maths, Ex 5.2, Q16

    Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

    Hint. The second condition involves only d — use it first, then the third term gives a.

    Step 1 — The second condition: a₇ − a₅ = 12. Since these are two steps apart, that is 2d = 12, so d = 6.

    Step 2 — The third term: a + 2d = 16, so a + 12 = 16 and a = 4.

    Step 3 — Build the AP: 4, 10, 16, 22, 28, …

    Step 4 — Check both conditions: the third term is 16 ✓, and a₇ − a₅ = 40 − 28 = 12 ✓

    ✦ The AP is 4, 10, 16, 22, …

    Where students slip. Writing a₇ − a₅ = 12d or 7d − 5d = 12 without care. It is (a + 6d) − (a + 4d) = 2d — two steps, not twelve.

    Another way. Take the conditions in the other order and you have two simultaneous equations in a and d, which works too but takes longer than spotting that the second condition isolates d.

17. 173 marksNCERT Cl-10 Maths, Ex 5.2, Q17

    Find the 20th term from the last term of the AP 3, 8, 13, …, 253.

    Hint. Counting from the end is the same as counting forwards in the reversed AP — which has common difference −d.

    Step 1 — The original AP has a = 3 and d = 5, ending at 253.

    Step 2 — Reverse it. Read backwards, the list starts at 253 and decreases by 5 each time, so it is an AP with first term 253 and common difference −5.

    Step 3 — The 20th term from the last is the 20th term of this reversed AP:

    253 + (20 − 1)(−5) = 253 − 95 = 158.

    Step 4 — Sanity check: 158 should be a term of the original AP. Solving 3 + (n − 1)(5) = 158 gives n = 32, a whole number ✓

    ✦ The 20th term from the last is 158.

    Where students slip. Finding the 20th term from the beginning (98) instead. Read the phrase carefully — 'from the last term' reverses the direction of counting.

    Another way. You can also find the total number of terms first: 253 = 3 + (n−1)5 gives n = 51. The 20th from the last is then the (51 − 20 + 1) = 32nd from the start, which is 3 + 31(5) = 158.

18. 183 marksNCERT Cl-10 Maths, Ex 5.2, Q18

    The sum of the 4th and 8th terms of an AP is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

    Hint. Write each condition in terms of a and d, then solve the pair simultaneously.

    Step 1 — First condition: (a + 3d) + (a + 7d) = 24, which simplifies to 2a + 10d = 24, that is a + 5d = 12.

    Step 2 — Second condition: (a + 5d) + (a + 9d) = 44, which simplifies to 2a + 14d = 44, that is a + 7d = 22.

    Step 3 — Subtract the first from the second: 2d = 10, so d = 5.

    Step 4 — Substitute back: a + 25 = 12, so a = −13.

    Step 5 — First three terms: −13, −8, −3.

    Step 6 — Check: a₄ + a₈ = 2 + 22 = 24 ✓ and a₆ + a₁₀ = 12 + 32 = 44 ✓

    ✦ −13, −8, −3

    Where students slip. Halving 24 and 44 to guess individual terms. The two terms in each sum are not equal, so you cannot split them — form the equations and solve.

    Another way. Notice that a + 5d is the 6th term, so the first condition says a₆ = 12 directly, and the second says a₆ + a₁₀ = 44, giving a₁₀ = 32 and hence 4d = 20.

19. 193 marksNCERT Cl-10 Maths, Ex 5.2, Q19

    Subba Rao started work in 1995 at an annual salary of ₹5000, receiving an increment of ₹200 each year. In which year did his income reach ₹7000?

    Hint. The salaries form an AP. Find which term equals 7000, then convert that term number back into a year.

    Step 1 — Salaries form an AP with a = 5000 and d = 200.

    Step 2 — Find n with 5000 + (n − 1)(200) = 7000.

    Step 3 — Simplify: (n − 1)(200) = 2000, so n − 1 = 10 and n = 11.

    Step 4 — Convert to a year. The 1st term is 1995, so the 11th term is 1995 + 10 = 2005. This is the step most answers get wrong — you add n − 1 years, not n.

    ✦ His income reached ₹7000 in 2005.

    Where students slip. Answering 2006 by adding 11 to 1995. The first salary is already in 1995, so ten increments land you in 2005.

    Another way. Think in increments: reaching ₹7000 needs ₹2000 more, which is 10 increments of ₹200 — so ten years after 1995.

20. 203 marksNCERT Cl-10 Maths, Ex 5.2, Q20

    Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75 each week. In which week were her weekly savings ₹20.75?

    Hint. Weekly savings form an AP with a = 5 and d = 1.75.

    Step 1 — a = 5 and d = 1.75.

    Step 2 — Solve 5 + (n − 1)(1.75) = 20.75.

    Step 3 — Subtract: (n − 1)(1.75) = 15.75.

    Step 4 — Divide: n − 1 = 15.75 ÷ 1.75 = 9, so n = 10.

    Step 5 — Check: 5 + 9(1.75) = 5 + 15.75 = 20.75 ✓

    ✦ In the 10th week.

    Note that here the term number is the week number, since the first week is the first term — unlike the previous question, where a base year had to be added.

    Where students slip. Dividing 20.75 by 1.75 and working from there. The ₹5 starting amount is not part of the increments — subtract it first.

    Another way. Working in quarters avoids decimals: ₹1.75 is 7 quarter-rupees and the gap of ₹15.75 is 63 of them, so 63 ÷ 7 = 9 steps.

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.