Exercise 5.3 — Arithmetic Progressions

1. 14 marksNCERT Cl-10 Maths, Ex 5.3, Q1 (four parts)

   Find the sum of the given number of terms: (i) 2, 7, 12, … to 10 terms; (ii) −37, −33, −29, … to 12 terms; (iii) 0.6, 1.7, 2.8, … to 100 terms; (iv) 1/15, 1/12, 1/10, … to 11 terms.

   Hint. Use Sₙ = (n/2)[2a + (n − 1)d]. You need only a, d and n — not the last term.

   The formula Sₙ = (n/2)[2a + (n − 1)d] works from the start of the list, which is why it is the right tool whenever the last term is not given.

   (i) 2, 7, 12, … to 10 terms. Here a = 2, d = 5, n = 10.

   S₁₀ = (10/2)[2(2) + 9(5)] = 5[4 + 45] = 5 × 49 = 245.

   ✦ 245

   (ii) −37, −33, −29, … to 12 terms. Here a = −37, d = 4, n = 12.

   S₁₂ = 6[2(−37) + 11(4)] = 6[−74 + 44] = 6 × (−30) = −180.

   The sum is negative because the early terms are strongly negative and the list has not yet climbed above zero.

   ✦ −180

   (iii) 0.6, 1.7, 2.8, … to 100 terms. Here a = 0.6, d = 1.1, n = 100.

   S₁₀₀ = 50[2(0.6) + 99(1.1)] = 50[1.2 + 108.9] = 50 × 110.1 = 5505.

   ✦ 5505

   (iv) 1/15, 1/12, 1/10, … to 11 terms. First find d: 1/12 − 1/15 = 5/60 − 4/60 = 1/60.

   S₁₁ = (11/2)[2(1/15) + 10(1/60)] = (11/2)[2/15 + 1/6].

   With denominator 30: 2/15 = 4/30 and 1/6 = 5/30, so the bracket is 9/30 = 3/10.

   S₁₁ = (11/2)(3/10) = 33/20 = 1.65.

   ✦ 33/20

   Where students slip. In (iv), guessing d from the look of the fractions instead of subtracting. 1/15, 1/12, 1/10 have denominators falling by 3 then 2, which is not a pattern — only the actual difference 1/60 matters.

   Another way. If you happen to know the last term, Sₙ = (n/2)(first + last) is quicker. In (i) the 10th term is 47, so S = 5(2 + 47) = 245.

2. 24 marksNCERT Cl-10 Maths, Ex 5.3, Q2 (three parts)

   Find the sums: (i) 7 + 10½ + 14 + … + 84; (ii) 34 + 32 + 30 + … + 10; (iii) −5 + (−8) + (−11) + … + (−230).

   Hint. Here the last term IS given, so first find how many terms there are, then use Sₙ = (n/2)(first + last).

   Two steps every time. Find n from the last term first, because once you know both ends the sum is a single multiplication: Sₙ = (n/2)(first + last).

   (i) 7 + 10.5 + 14 + … + 84. Here a = 7 and d = 3.5.

   84 = 7 + (n − 1)(3.5), so 77 = 3.5(n − 1), which gives n − 1 = 22 and n = 23.

   S₂₃ = (23/2)(7 + 84) = (23/2)(91) = 2093/2 = 1046.5.

   ✦ 1046½

   (ii) 34 + 32 + 30 + … + 10. Here a = 34 and d = −2, since the list is falling.

   10 = 34 + (n − 1)(−2), so −24 = −2(n − 1) and n − 1 = 12, giving n = 13.

   S₁₃ = (13/2)(34 + 10) = (13/2)(44) = 286.

   ✦ 286

   (iii) −5 + (−8) + … + (−230). Here a = −5 and d = −3.

   −230 = −5 + (n − 1)(−3), so −225 = −3(n − 1), which gives n − 1 = 75 and n = 76.

   S₇₆ = (76/2)(−5 + (−230)) = 38 × (−235) = −8930. The total is negative because every term is.

   ✦ −8930

   Where students slip. Reaching for Sₙ = (n/2)[2a + (n−1)d] before knowing n. When the last term is given, finding n first and then using (n/2)(first + last) is both shorter and less error-prone.

   Another way. In (i), working in halves avoids decimals: the terms are 14/2, 21/2, 28/2, … so the common difference is 7/2.

3. 36 marksNCERT Cl-10 Maths, Ex 5.3, Q3 (seven parts)

   In an AP, find the missing quantities: (i) given a = 5, aₙ = 45 and Sₙ = 400, find n and d; (ii) given a = 7, a₁₃ = 35, find d and S₁₃; (iii) given a₁₂ = 37 and d = 3, find a and S₁₂; (iv) given a₃ = 15 and S₁₀ = 125, find d and a₁₀.

   Hint. Pick whichever sum formula uses what you already know. When both first and last terms are given, Sₙ = (n/2)(a + aₙ) is the shortest route.

   (i) a = 5, aₙ = 45, Sₙ = 400.

   Since both ends are known, use Sₙ = (n/2)(a + aₙ): 400 = (n/2)(5 + 45) = 25n, so n = 16.

   Now find d from the last term: 45 = 5 + 15d ⟹ 15d = 40 ⟹ d = 8/3.

   ✦ n = 16, d = 8/3

   (ii) a = 7, a₁₃ = 35.

   From the 13th term: 35 = 7 + 12d ⟹ 12d = 28 ⟹ d = 7/3.

   S₁₃ = (13/2)(7 + 35) = (13/2)(42) = 273.

   ✦ d = 7/3, S₁₃ = 273

   (iii) a₁₂ = 37, d = 3.

   From the 12th term: 37 = a + 11(3) = a + 33, so a = 4.

   S₁₂ = (12/2)(4 + 37) = 6 × 41 = 246.

   ✦ a = 4, S₁₂ = 246

   (iv) a₃ = 15, S₁₀ = 125.

   From the third term: a + 2d = 15.

   From the sum: 125 = (10/2)[2a + 9d] = 5[2a + 9d], so 2a + 9d = 25.

   Substituting a = 15 − 2d: 2(15 − 2d) + 9d = 25 ⟹ 30 + 5d = 25 ⟹ d = −1, and then a = 17.

   a₁₀ = 17 + 9(−1) = 8.

   ✦ d = −1, a₁₀ = 8

   Where students slip. Using Sₙ = (n/2)[2a + (n−1)d] in part (i) when the far simpler (n/2)(a + aₙ) applies. Both are correct, but the second has one unknown instead of two.

   Another way. The two sum formulas are the same statement — substituting aₙ = a + (n−1)d into (n/2)(a + aₙ) reproduces the longer one. Knowing that lets you switch freely to whichever fits the given data.

4. 43 marksNCERT Cl-10 Maths, Ex 5.3, Q4

   How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?

   Hint. Set Sₙ = 636 and solve the resulting quadratic in n. Reject any root that is not a positive whole number.

   Step 1 — Here a = 9 and d = 8.

   Step 2 — Set up the sum: (n/2)[2(9) + (n − 1)(8)] = 636.

   Step 3 — Simplify the bracket: 18 + 8n − 8 = 8n + 10. So (n/2)(8n + 10) = 636, that is n(4n + 5) = 636.

   Step 4 — Rearrange into standard form: 4n² + 5n − 636 = 0.

   Step 5 — Solve. The discriminant is 25 + 4(4)(636) = 25 + 10176 = 10201, and √10201 = 101.

   n = (−5 ± 101)/8, giving n = 12 or n = −106/8.

   Step 6 — A number of terms must be a positive integer, so n = 12.

   Step 7 — Check: S₁₂ = 6[18 + 88] = 6 × 106 = 636 ✓

   ✦ 12 terms

   Where students slip. Keeping the negative root as a second answer. Whenever n is a count of terms, negative and fractional roots must be rejected explicitly — examiners look for that sentence.

   Another way. If the discriminant is not a perfect square, you have almost certainly made an arithmetic slip — these questions are always set to give whole-number answers.

5. 53 marksNCERT Cl-10 Maths, Ex 5.3, Q5

   The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

   Hint. Both ends are given, so start with Sₙ = (n/2)(first + last).

   Step 1 — Both ends known, so use the short form: 400 = (n/2)(5 + 45).

   Step 2 — Simplify: 400 = (n/2)(50) = 25n, so n = 16.

   Step 3 — Now use the last term to find d. The 16th term is 45:

   45 = 5 + (16 − 1)d ⟹ 40 = 15d ⟹ d = 8/3.

   Step 4 — Check: S₁₆ = (16/2)[2(5) + 15(8/3)] = 8[10 + 40] = 400 ✓

   ✦ n = 16 and d = 8/3

   Where students slip. Expecting d to be a whole number and 'correcting' 8/3 to 3 or 2. There is no rule that the common difference must be an integer — leave it as an exact fraction.

   Another way. The average of first and last is (5 + 45)/2 = 25, and 400/25 = 16 gives n immediately. That works because in any AP the mean of all terms equals the mean of the two ends.

6. 63 marksNCERT Cl-10 Maths, Ex 5.3, Q6

   The first and last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

   Hint. Find n from the last term first; the sum then follows immediately.

   Step 1 — Use the last term: 350 = 17 + (n − 1)(9).

   Step 2 — Simplify: 333 = 9(n − 1), so n − 1 = 37 and n = 38.

   Step 3 — Now sum with both ends known: S₃₈ = (38/2)(17 + 350) = 19 × 367 = 6973.

   Step 4 — Check the multiplication: 19 × 367 = 19 × 300 + 19 × 67 = 5700 + 1273 = 6973 ✓

   ✦ 38 terms, summing to 6973

   Where students slip. Computing 333/9 as 36 and reporting n = 37. Divide carefully — 9 × 37 = 333, so n − 1 = 37 and n is one more.

   Another way. Since 350 − 17 = 333 is the total climb and each step is 9, there are 37 steps and therefore 38 terms.

7. 73 marksNCERT Cl-10 Maths, Ex 5.3, Q7

   Find the sum of the first 22 terms of an AP whose common difference is 7 and whose 22nd term is 149.

   Hint. Work backwards from the 22nd term to find a, then sum.

   Step 1 — From the 22nd term: 149 = a + 21(7) = a + 147, so a = 2.

   Step 2 — Both ends are now known, so S₂₂ = (22/2)(2 + 149) = 11 × 151.

   Step 3 — Compute: 11 × 151 = 1661.

   Step 4 — Check with the other formula: S₂₂ = 11[2(2) + 21(7)] = 11[4 + 147] = 11 × 151 ✓

   ✦ 1661

   Where students slip. Using 22d instead of 21d when working back to a. Reaching the 22nd term takes 21 steps.

   Another way. You can skip finding a: the mean of the 22 terms is the mean of the two middle ones, but the (n/2)(a + aₙ) route is shorter once a is known.

8. 83 marksNCERT Cl-10 Maths, Ex 5.3, Q8

   The second and third terms of an AP are 14 and 18 respectively. Find the sum of the first 51 terms.

   Hint. Two consecutive terms give d immediately; then step back one place for a.

   Step 1 — Consecutive terms give d directly: d = 18 − 14 = 4.

   Step 2 — The first term is one step before the second: a = 14 − 4 = 10.

   Step 3 — Sum 51 terms: S₅₁ = (51/2)[2(10) + 50(4)] = (51/2)[20 + 200] = (51/2)(220).

   Step 4 — Simplify: 220/2 = 110, so S₅₁ = 51 × 110 = 5610.

   ✦ 5610

   Where students slip. Taking a = 14 because 14 is the first number mentioned. It is stated to be the *second* term — read the positions carefully.

   Another way. The 51st term is 10 + 50(4) = 210, so S = (51/2)(10 + 210) = 51 × 110 = 5610.

9. 94 marksNCERT Cl-10 Maths, Ex 5.3, Q9

   The sums of the first 7 and first 17 terms of an AP are 49 and 289 respectively. Find the sum of the first n terms.

   Hint. Write both sums with the formula to get two equations in a and d, then build a general expression for Sₙ.

   Step 1 — First condition: (7/2)[2a + 6d] = 49. Dividing by 7/2 gives 2a + 6d = 14, that is a + 3d = 7.

   Step 2 — Second condition: (17/2)[2a + 16d] = 289. Dividing by 17/2 gives 2a + 16d = 34, that is a + 8d = 17.

   Step 3 — Subtract: 5d = 10, so d = 2, and then a = 7 − 6 = 1.

   Step 4 — Now write the general sum:

   Sₙ = (n/2)[2(1) + (n − 1)(2)] = (n/2)[2 + 2n − 2] = (n/2)(2n) = n².

   Step 5 — Check both given values: S₇ = 49 ✓ and S₁₇ = 289 ✓

   ✦ Sₙ = n²

   The AP is 1, 3, 5, 7, … — the odd numbers, whose running totals are the perfect squares. The two clues in the question (49 = 7² and 289 = 17²) were pointing at this all along.

   Where students slip. Stopping after finding a and d. The question asks for Sₙ as a formula in n, so the answer must be an expression, not two numbers.

   Another way. Spotting that 49 = 7² and 289 = 17² suggests Sₙ = n² immediately — then you only need to verify it, which is much faster.

10. 104 marksNCERT Cl-10 Maths, Ex 5.3, Q10

    Show that a₁, a₂, …, aₙ form an AP where aₙ is defined as (i) aₙ = 3 + 4n and (ii) aₙ = 9 − 5n. Also find the sum of the first 15 terms in each case.

    Hint. Compute aₙ₊₁ − aₙ. If it comes out as a constant, the list is an AP.

    The test for 'is this an AP' when a formula is given: subtract consecutive terms symbolically.

    (i) aₙ = 3 + 4n.

    aₙ₊₁ − aₙ = [3 + 4(n + 1)] − [3 + 4n] = 4.

    The difference is 4 regardless of n, so it is an AP with d = 4. The first term is a₁ = 3 + 4 = 7.

    S₁₅ = (15/2)[2(7) + 14(4)] = (15/2)[14 + 56] = (15/2)(70) = 525.

    ✦ AP with d = 4; S₁₅ = 525

    (ii) aₙ = 9 − 5n.

    aₙ₊₁ − aₙ = [9 − 5(n + 1)] − [9 − 5n] = −5.

    Constant again, so it is an AP with d = −5. The first term is a₁ = 9 − 5 = 4.

    S₁₅ = (15/2)[2(4) + 14(−5)] = (15/2)[8 − 70] = (15/2)(−62) = −465.

    ✦ AP with d = −5; S₁₅ = −465

    Worth noticing: any formula of the form aₙ = pn + q is automatically an AP with d = p, because the difference is p whatever n is.

    Where students slip. Testing only a₂ − a₁ and declaring it an AP. One numerical difference proves nothing about the rest — the symbolic subtraction is what covers every n at once.

    Another way. List the first three terms and check by eye: (i) gives 7, 11, 15 and (ii) gives 4, −1, −6. That is a fine sanity check, but the general argument is what earns the marks.

11. 113 marksNCERT Cl-10 Maths, Ex 5.3, Q11

    If the sum of the first n terms of an AP is 4n − n², what are the first term, the sum of the first two terms, the second term, and the 3rd, 10th and nth terms?

    Hint. S₁ is the first term. Any later term is the difference of two consecutive sums: aₙ = Sₙ − Sₙ₋₁.

    Step 1 — The first term is just S₁: a₁ = 4(1) − 1² = 3.

    Step 2 — The sum of the first two: S₂ = 4(2) − 2² = 8 − 4 = 4.

    Step 3 — The second term is what S₂ adds to S₁: a₂ = S₂ − S₁ = 4 − 3 = 1.

    So d = 1 − 3 = −2.

    Step 4 — For the general term, use aₙ = Sₙ − Sₙ₋₁:

    Sₙ₋₁ = 4(n − 1) − (n − 1)² = 4n − 4 − (n² − 2n + 1) = −n² + 6n − 5.

    aₙ = (4n − n²) − (−n² + 6n − 5) = 4n − n² + n² − 6n + 5 = 5 − 2n.

    Step 5 — Read off the requested terms: a₃ = 5 − 6 = −1, and a₁₀ = 5 − 20 = −15.

    Step 6 — Check against Step 1: a₁ = 5 − 2 = 3 ✓ and a₂ = 5 − 4 = 1 ✓

    ✦ a₁ = 3, S₂ = 4, a₂ = 1, a₃ = −1, a₁₀ = −15, and aₙ = 5 − 2n.

    Where students slip. Treating Sₙ as if it were aₙ and reporting a₁₀ = 4(10) − 100 = −60. That is the *sum* of ten terms, not the tenth term.

    Another way. Once you know a = 3 and d = −2 from the first two terms, aₙ = 3 + (n−1)(−2) = 5 − 2n follows without the Sₙ − Sₙ₋₁ algebra.

12. 123 marksNCERT Cl-10 Maths, Ex 5.3, Q12

    Find the sum of the first 40 positive integers divisible by 6.

    Hint. They are 6, 12, 18, … — an AP with a = d = 6.

    Step 1 — The positive integers divisible by 6 are 6, 12, 18, …, so a = 6 and d = 6.

    Step 2 — Sum the first 40: S₄₀ = (40/2)[2(6) + 39(6)] = 20[12 + 234] = 20 × 246.

    Step 3 — Compute: 20 × 246 = 4920.

    ✦ 4920

    Where students slip. Reading 'first 40 positive integers divisible by 6' as the multiples of 6 up to 40. It means the first forty such numbers, running from 6 to 240.

    Another way. Factor out the 6: the sum is 6(1 + 2 + … + 40) = 6 × (40 × 41/2) = 6 × 820 = 4920.

13. 132 marksNCERT Cl-10 Maths, Ex 5.3, Q13

    Find the sum of the first 15 multiples of 8.

    Hint. Same shape as the previous question, with a = d = 8.

    Step 1 — The multiples are 8, 16, 24, …, so a = 8 and d = 8.

    Step 2 — S₁₅ = (15/2)[2(8) + 14(8)] = (15/2)[16 + 112] = (15/2)(128).

    Step 3 — Simplify: 128/2 = 64, so S₁₅ = 15 × 64 = 960.

    ✦ 960

    Where students slip. Starting the list at 0. The first multiple of 8 in this sense is 8 — including 0 would give sixteen numbers, not fifteen.

    Another way. Factor out the 8: 8(1 + 2 + … + 15) = 8 × 120 = 960.

14. 143 marksNCERT Cl-10 Maths, Ex 5.3, Q14

    Find the sum of the odd numbers between 0 and 50.

    Hint. The list runs 1, 3, 5, …, 49. Find how many there are before summing.

    Step 1 — The odd numbers between 0 and 50 are 1, 3, 5, …, 49, so a = 1 and d = 2.

    Step 2 — Find n from the last term: 49 = 1 + (n − 1)(2) ⟹ 48 = 2(n − 1) ⟹ n = 25.

    Step 3 — Sum with both ends known: S₂₅ = (25/2)(1 + 49) = (25/2)(50) = 25 × 25 = 625.

    ✦ 625

    The answer being exactly 25² is not a coincidence — the sum of the first n odd numbers is always n², which is the same fact that appeared in Q9.

    Where students slip. Including 51, or counting 50 itself. 'Between 0 and 50' stops at 49.

    Another way. Use the identity directly: there are 25 odd numbers here, so the sum is 25² = 625.

15. 153 marksNCERT Cl-10 Maths, Ex 5.3, Q15

    A contract carries a penalty for delay: ₹200 for the first day, ₹250 for the second, ₹300 for the third, and so on, each day's penalty ₹50 more than the day before. How much does a contractor pay as penalty if the work is delayed by 30 days?

    Hint. The daily penalties form an AP; the total paid is their sum.

    Step 1 — Daily penalties: 200, 250, 300, … so a = 200 and d = 50.

    Step 2 — The question asks for the total over 30 days, which is S₃₀, not the 30th day's penalty.

    Step 3 — S₃₀ = (30/2)[2(200) + 29(50)] = 15[400 + 1450] = 15 × 1850.

    Step 4 — Compute: 15 × 1850 = 27,750.

    ✦ ₹27,750

    Where students slip. Answering ₹1650, which is a₃₀ — the penalty on the thirtieth day alone. The contractor pays every day's penalty, so the answer is the sum.

    Another way. Check the size: the average daily penalty is (200 + 1650)/2 = 925, and 925 × 30 = 27,750 ✓

16. 163 marksNCERT Cl-10 Maths, Ex 5.3, Q16

    A sum of ₹700 is to be used for seven cash prizes to students. If each prize is ₹20 less than its preceding prize, find the value of each prize.

    Hint. The prizes form a decreasing AP with d = −20 and S₇ = 700.

    Step 1 — Let the largest prize be ₹a. The prizes decrease by 20 each time, so d = −20 and n = 7.

    Step 2 — Their total is 700: (7/2)[2a + 6(−20)] = 700.

    Step 3 — Simplify: (7/2)(2a − 120) = 700, so 7(a − 60) = 700, giving a − 60 = 100 and a = 160.

    Step 4 — The seven prizes are therefore 160, 140, 120, 100, 80, 60, 40.

    Step 5 — Check the total: these average 100 and there are seven, so the sum is 700 ✓

    ✦ ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40

    Where students slip. Setting a as the smallest prize and using d = +20, then reporting the list backwards. Either convention works, but say which one you chose — and make sure the final list runs from largest to smallest as the question implies.

    Another way. With an odd number of terms, the middle prize is the average: 700/7 = 100 is the fourth prize, and the rest follow by stepping ±20 from it.

17. 173 marksNCERT Cl-10 Maths, Ex 5.3, Q17

    In a school, students decided to plant trees to reduce air pollution. Each section of each class plants as many trees as the class number — a section of Class I plants 1 tree, Class II plants 2, and so on up to Class XII. There are three sections of each class. How many trees will be planted?

    Hint. Find the total for one section across all classes, then multiply by the number of sections.

    Step 1 — For one section per class, the trees planted are 1, 2, 3, …, 12 — an AP with a = 1, d = 1, n = 12.

    Step 2 — Sum for a single section: S₁₂ = (12/2)(1 + 12) = 6 × 13 = 78.

    Step 3 — There are three sections of each class, so every class contributes three times its number.

    Step 4 — Total = 3 × 78 = 234.

    ✦ 234 trees

    Where students slip. Forgetting the three sections and answering 78. Read to the end of the question — the multiplier is stated in the final sentence.

    Another way. Equivalently, each class plants 3 × (class number), giving the AP 3, 6, 9, …, 36, whose sum is (12/2)(3 + 36) = 6 × 39 = 234.

18. 184 marksNCERT Cl-10 Maths, Ex 5.3, Q18

    A spiral is made up of successive semicircles with centres alternately at A and B, starting at A, with radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7.)

    Hint. The length of a semicircle of radius r is πr. Sum the radii first, then multiply by π once.

    Step 1 — A semicircle of radius r has length πr (half of the full circumference 2πr).

    Step 2 — The total length is π(r₁ + r₂ + … + r₁₃). Taking π outside is the key move — it turns the problem into a single AP sum.

    Step 3 — The radii are 0.5, 1.0, 1.5, …, an AP with a = 0.5 and d = 0.5.

    S₁₃ = (13/2)[2(0.5) + 12(0.5)] = (13/2)[1 + 6] = (13/2)(7) = 45.5 cm.

    Step 4 — Multiply by π: total length = (22/7)(45.5) = 22 × 6.5 = 143 cm.

    Step 5 — The 7 cancels neatly because 45.5 = 91/2 and 91 = 7 × 13, which is a sign the question was designed for π = 22/7.

    ✦ 143 cm

    Where students slip. Using 2πr and halving at the end, or forgetting to halve at all. A semicircle's arc is πr — doubling that would give 286 cm.

    Another way. You could compute each semicircle's length and add thirteen numbers, but factoring π out first is far less work and avoids rounding at every step.

19. 194 marksNCERT Cl-10 Maths, Ex 5.3, Q19

    200 logs are stacked so that there are 20 logs in the bottom row, 19 in the next, 18 in the next, and so on. In how many rows are the 200 logs placed, and how many logs are in the top row?

    Hint. Set Sₙ = 200 and solve the quadratic — then check both roots against the physical situation.

    Step 1 — Rows contain 20, 19, 18, … so a = 20 and d = −1.

    Step 2 — Set the sum to 200: (n/2)[2(20) + (n − 1)(−1)] = 200.

    Step 3 — Simplify the bracket: 40 − n + 1 = 41 − n. So (n/2)(41 − n) = 200, that is n(41 − n) = 400.

    Step 4 — Rearrange: n² − 41n + 400 = 0.

    Step 5 — Factorise: product 400, sum 41 — that is 16 and 25. So (n − 16)(n − 25) = 0, giving n = 16 or n = 25.

    Step 6 — Test both. If n = 25, the top row would hold a₂₅ = 20 + 24(−1) = −4 logs, which is impossible. So n = 25 is rejected.

    With n = 16, the top row holds a₁₆ = 20 + 15(−1) = 5 logs — perfectly sensible.

    Step 7 — Check: S₁₆ = (16/2)(20 + 5) = 8 × 25 = 200 ✓

    ✦ 16 rows, with 5 logs in the top row.

    Where students slip. Reporting both roots because both are positive integers. Positivity of n is not enough here — you must also check that the resulting *term* makes sense, and a row of −4 logs does not.

    Another way. The rejection has a neat interpretation: after 20 rows the stack would be exhausted, and rows 21–25 would have to contain negative counts that cancel earlier ones — which is why the sum returns to 200 at n = 25.

20. 204 marksNCERT Cl-10 Maths, Ex 5.3, Q20

    In a potato race, a bucket is placed at the starting point, 5 m from the first potato, and the other potatoes are 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back and drops it in the bucket, then repeats for each potato. What total distance does the competitor run?

    Hint. Every potato costs a round trip — out and back. Write the distance for the first two and the pattern becomes clear.

    Step 1 — The potatoes sit 5 m, 8 m, 11 m, … from the bucket, an AP with a = 5 and d = 3.

    Step 2 — For each potato the competitor runs out and back, so the distance for the kth potato is 2 × (its distance).

    That gives 10, 16, 22, … — still an AP, now with a = 10 and d = 6.

    Step 3 — Total for ten potatoes: S₁₀ = (10/2)[2(10) + 9(6)] = 5[20 + 54] = 5 × 74.

    Step 4 — Compute: 5 × 74 = 370.

    ✦ The competitor runs 370 m.

    Where students slip. Forgetting the return leg and answering 185 m. Each potato must be carried back to the bucket, so every distance is doubled.

    Another way. Sum the one-way distances first — S₁₀ = (10/2)(5 + 32) = 185 — then double at the end. Same answer, and it keeps the numbers smaller.

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.