Exercise 5.4 — Arithmetic Progressions

1. 14 marksNCERT Cl-10 Maths, Ex 5.4, Q1 (optional)

   Which term of the AP 121, 117, 113, … is its first negative term?

   Hint. You want the first n for which aₙ < 0 — an inequality, not an equation.

   Step 1 — Here a = 121 and d = 117 − 121 = −4, so the terms fall steadily.

   Step 2 — The general term: aₙ = 121 + (n − 1)(−4) = 121 − 4n + 4 = 125 − 4n.

   Step 3 — We want the first term below zero, so solve the inequality 125 − 4n < 0.

   Step 4 — Rearrange: 125 < 4n, so n > 31.25.

   Step 5 — n must be a whole number, and the smallest whole number greater than 31.25 is 32.

   Step 6 — Check both sides of the boundary. a₃₁ = 125 − 124 = 1, which is still positive; a₃₂ = 125 − 128 = −3, which is negative ✓

   ✦ The 32nd term is the first negative one, and its value is −3.

   Where students slip. Solving 125 − 4n = 0 and rounding. That gives n = 31.25, and rounding down to 31 lands on a term that is still positive. Because the condition is an inequality, you must round *up*.

   Another way. Ask how many steps of 4 fit into 121: since 121/4 = 30.25, thirty-one steps overshoot zero. Thirty-one steps past the first term is the 32nd term.

2. 24 marksNCERT Cl-10 Maths, Ex 5.4, Q2 (optional)

   The sum of the third and the seventh terms of an AP is 6, and their product is 8. Find the sum of the first sixteen terms of the AP.

   Hint. Two numbers with a known sum and product are the roots of a quadratic — build it rather than guessing.

   Step 1 — Write the two terms: a₃ = a + 2d and a₇ = a + 6d.

   Step 2 — Their sum is 6: (a + 2d) + (a + 6d) = 6, which gives 2a + 8d = 6, that is a + 4d = 3.

   Note that a + 4d is the fifth term, so this says a₅ = 3 — the middle term of the two, as you would expect.

   Step 3 — Their product is 8: (a + 2d)(a + 6d) = 8.

   Step 4 — Use Step 2 to write both factors around a₅. Since a + 2d = (a + 4d) − 2d = 3 − 2d and a + 6d = 3 + 2d, the product becomes

   (3 − 2d)(3 + 2d) = 9 − 4d² = 8.

   Step 5 — So 4d² = 1, giving d = ½ or d = −½. Both are genuine — the AP can rise or fall.

   Step 6 — Find a in each case from a = 3 − 4d.

   • If d = ½: a = 3 − 2 = 1. Then S₁₆ = (16/2)[2(1) + 15(½)] = 8[2 + 7.5] = 8 × 9.5 = 76.
   • If d = −½: a = 3 + 2 = 5. Then S₁₆ = 8[2(5) + 15(−½)] = 8[10 − 7.5] = 8 × 2.5 = 20.

   Step 7 — Check the first case: a₃ = 2 and a₇ = 4, summing to 6 and multiplying to 8 ✓

   ✦ S₁₆ = 76 when d = ½, or S₁₆ = 20 when d = −½. Both answers are valid.

   Where students slip. Reporting only one answer. The quadratic in d has two roots and neither is impossible here, so both APs satisfy the conditions and both sums must be given.

   Another way. Since the two terms have sum 6 and product 8, they are the roots of x² − 6x + 8 = 0, that is 2 and 4. Whether a₃ = 2 and a₇ = 4 or the other way round gives the two cases directly.

3. 34 marksNCERT Cl-10 Maths, Ex 5.4, Q3 (optional)

   A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. The distance between the top and bottom rungs is 2½ m. What length of wood is needed for the rungs?

   Hint. First find how many rungs there are from the spacing, then sum their lengths as an AP.

   Step 1 — Convert to a single unit. The gap between top and bottom rungs is 2½ m = 250 cm.

   Step 2 — Find the number of rungs. The rungs are 25 cm apart, so 250 cm spans 250/25 = 10 gaps. Ten gaps means eleven rungs — the classic fencepost count, one more than the number of gaps.

   Step 3 — The rung lengths form an AP from 25 cm to 45 cm with 11 terms. (Their common difference is 20/10 = 2 cm, though we will not need it.)

   Step 4 — Sum with both ends known: S₁₁ = (11/2)(25 + 45) = (11/2)(70) = 11 × 35 = 385.

   Step 5 — Sanity check: the average rung is 35 cm and there are 11 of them, giving 385 cm ✓

   ✦ 385 cm of wood is needed (that is 3.85 m).

   Where students slip. Counting 10 rungs from the 10 gaps. Rungs sit at both ends of the span, so the count is gaps + 1 — the same reason a fence with 10 panels needs 11 posts.

   Another way. You can find d = 2 cm and use the long sum formula, but with both first and last lengths given, (n/2)(first + last) is a single line.

4. 44 marksNCERT Cl-10 Maths, Ex 5.4, Q4 (optional)

   The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding house x equals the sum of the numbers of the houses following it. Find this value of x.

   Hint. The sum 1 + 2 + … + n is n(n+1)/2. Write both sides with that and solve.

   Step 1 — Houses before x are numbered 1 to x − 1, so their sum is

   S_before = (x − 1)x/2.

   Step 2 — Houses after x run from x + 1 to 49. Their sum is everything up to 49 minus everything up to x:

   S_after = 49(50)/2 − x(x + 1)/2 = 1225 − x(x + 1)/2.

   Step 3 — Set them equal: (x − 1)x/2 = 1225 − x(x + 1)/2.

   Step 4 — Multiply through by 2: x² − x = 2450 − x² − x.

   Step 5 — The −x cancels from both sides, leaving 2x² = 2450, so x² = 1225.

   Step 6 — Therefore x = ±35, and a house number must be positive, so x = 35.

   Step 7 — Check: houses 1–34 sum to 34(35)/2 = 595, and houses 36–49 sum to 1225 − 35(36)/2 = 1225 − 630 = 595 ✓ Equal, as required.

   ✦ x = 35

   Where students slip. Including house x itself on one side. The question says houses *preceding* and *following* x, so house 35 belongs to neither sum — which is exactly why the −x terms cancel.

   Another way. The tidy cancellation at Step 5 is worth noticing: it means x² is exactly half of 49 × 50 / 2 × 2, so x = √1225 = 35 falls out without solving a full quadratic.

5. 55 marksNCERT Cl-10 Maths, Ex 5.4, Q5 (optional)

   A small terrace at a football ground is made of 15 steps, each 50 m long and built of solid concrete. Each step has a rise of ¼ m and a tread of ½ m. Calculate the total volume of concrete required to build the terrace.

   Hint. Each step sits on top of those below it, so its height is a multiple of the rise. Volume = height × tread × length.

   Step 1 — Understand the shape. The bottom step is one rise high. The second step must be built up from the ground too, so it is two rises high, the third is three rises high, and so on. This is the part of the question that does the work — the steps are not all the same height.

   Step 2 — Volume of the nth step = height × tread × length:

   Vₙ = (n × ¼) × ½ × 50 = n × 6.25 m³.

   Step 3 — So the volumes are 6.25, 12.5, 18.75, … an AP with a = 6.25 and d = 6.25.

   Step 4 — Total for 15 steps:

   S₁₅ = (15/2)[2(6.25) + 14(6.25)] = (15/2)[12.5 + 87.5] = (15/2)(100) = 750.

   Step 5 — Check another way: the total is 6.25 × (1 + 2 + … + 15) = 6.25 × 120 = 750 ✓

   ✦ 750 m³ of concrete is required.

   Where students slip. Treating every step as a single ¼ m × ½ m × 50 m block and multiplying by 15, giving 93.75 m³. That would be true only if the steps floated — each one actually rests on all the concrete beneath it.

   Another way. Factor the 6.25 out immediately and the problem reduces to summing 1 to 15, which is 120. Keeping the common factor outside avoids decimals entirely.

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.