Exercise 5.4 is the optional word-problem set that closes Chapter 5.
The textbook star-marks it with the note that these exercises are not from the examination point of view (NCERT, p. 23), but it is still worth a careful attempt: every question here forces you to carry the general-term formula and the sum formula into a real situation — the first negative term of a shrinking AP, a ladder’s rungs, a row of numbered houses, a concrete terrace.
This page gives the complete exercise 5.4 class 10 maths NCERT solutions, question by question, with the reasoning written out so you can attempt the next problem on your own.
Work through the five questions in the order they appear. Each solution opens with the idea behind the method, shows every substitution with its units, and ends with the common mistake to avoid. If a question asks for a final value, confirm it against the story — each solution shows you how.
Exercise 5.4 Class 10 Maths NCERT Solutions
What this exercise teaches. The five questions are application problems that pull together both the general-term formula and the sum formula in real-life contexts — hunting a negative term, a ladder’s rungs, house numbers, a concrete terrace.
The exercise is marked optional, but attempting it builds the one skill the board’s own AP word problems reward: turning a sentence into an AP before any formula is applied.
Every question solved on this page is reproduced word for word from the official NCERT Class 10 Mathematics textbook, so you can verify the exercise text and the formula derivations directly against the source document, page by page.
Concepts you need for these five questions. Only three tools appear here, and each question names which one it needs once you read the story.
- General term: \( a_n = a + (n-1)d \) gives the nth term of an AP with first term \( a \) and common difference \( d \) (NCERT, p. 58). The position \( n \) counts terms, so it must be a positive integer.
- Sum of first \( n \) terms: \( S = \frac{n}{2}[2a + (n-1)d] \) when you know \( a \) and \( d \); and the shorter form \( S = \frac{n}{2}(a + l) \) when the last term \( l \) is given and \( d \) is not (NCERT, p. 65).
- Recognise the AP first: before substituting anything, check that consecutive values differ by a fixed amount. Only then do the formulas apply.
Question 1: Which term of the AP: 121, 117, 113, …, is its first negative term? [Hint: Find n for aₙ < 0]
The trigger word is negative. “First negative term” asks for an inequality, not an equation: you want the smallest \( n \) for which \( a_n \lt 0 \). Setting \( a_n = 0 \) gives \( n = 31.25 \), which is not a term number — it tells you no term is exactly zero, and that the sign change happens between the 31st and 32nd terms.
Step 1: Read off \( a \) and \( d \).
The first term is \( a = 121 \), and the common difference is \( d = 117 – 121 = -4 \).
Step 2: Write the general term and simplify it.
\[ a_n = 121 + (n-1)(-4) = 125 – 4n \]
Step 3: Apply the inequality from the hint, \( a_n \lt 0 \).
\[ 125 – 4n \lt 0 \Rightarrow 4n \gt 125 \Rightarrow n \gt 31.25 \]
Step 4: Because \( n \) counts terms, the first positive integer greater than 31.25 is \( n = 32 \).
Step 5 (check): Substitute 31 and 32 to confirm the sign change.
\[ a_{31} = 125 – 4(31) = 1, \quad a_{32} = 125 – 4(32) = -3 \]
Final answer: The 32nd term, which is \( -3 \).
Question 2: The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
The two terms are a pair of numbers whose sum is 6 and product is 8 — exactly the data that fix the roots of the quadratic \( t^2 – 6t + 8 = 0 \), namely 2 and 4. The question never says which of them is \( a_3 \), so both orders must be tried, and each order gives its own AP.
Step 1: Write the two terms using the general term.
\( a_3 = a + 2d \) and \( a_7 = a + 6d \), and \( a_3 + a_7 = 6 \), \( a_3 \cdot a_7 = 8 \).
\[ t^2 – 6t + 8 = 0 \Rightarrow (t-2)(t-4) = 0 \Rightarrow t = 2, 4 \]
Step 2 (first AP): take \( a_3 = 2 \), \( a_7 = 4 \).
\[ a + 2d = 2, \quad a + 6d = 4 \Rightarrow 4d = 2 \Rightarrow d = \tfrac{1}{2}, \quad a = 1 \]
Step 3: sum the first sixteen terms of this AP.
\[ S_{16} = \frac{16}{2}[2(1) + 15(\tfrac{1}{2})] = 8(2 + 7.5) = 76 \]
Step 4 (second AP): take \( a_3 = 4 \), \( a_7 = 2 \).
\[ a + 2d = 4, \quad a + 6d = 2 \Rightarrow 4d = -2 \Rightarrow d = -\tfrac{1}{2}, \quad a = 5 \]
Step 5: sum the first sixteen terms of this second AP.
\[ S_{16} = \frac{16}{2}[2(5) + 15(-\tfrac{1}{2})] = 8(10 – 7.5) = 20 \]
Final answer: \( S_{16} = 76 \) for the AP 1, \( \tfrac{3}{2} \), 2, … and \( S_{16} = 20 \) for the AP 5, \( \tfrac{9}{2} \), 4, … . Both APs satisfy the given conditions, so both sums are valid answers.
Question 3: A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2½ m apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = 250/25 + 1]
Before any AP work, convert to one unit: \( 2\tfrac{1}{2} \) m is 250 cm. With rungs 25 cm apart, the 250 cm span is divided into \( 250/25 = 10 \) gaps, and the hint’s +1 adds the bottom rung, which starts the first gap. So there are 11 rungs, not 10. The rung lengths then form an AP.
The figure shows the ladder as a stack of evenly spaced rungs whose lengths shrink uniformly toward the top — exactly the 45, 43, 41, …, 25 pattern the question describes. Counting that stack needs the +1: the bottom rung is the starting point of the first 25 cm gap.
Step 1: Convert the separation and count the rungs.
\[ 2\tfrac{1}{2} \text{ m} = 250 \text{ cm}, \quad \text{Number of rungs} = \frac{250}{25} + 1 = 10 + 1 = 11 \]
Step 2: The rung lengths are \( 45, 43, 41, \ldots, 25 \), an AP with \( a = 45 \), \( d = -2 \), \( n = 11 \).
Check the last rung.
\[ a_{11} = 45 + (11-1)(-2) = 45 – 20 = 25 \]
Step 3: Total wood is the sum of the 11 rungs.
The last term is known, so use \( S = \frac{n}{2}(a + l) \).
\[ S_{11} = \frac{11}{2}(45 + 25) = \frac{11}{2} \times 70 = 385 \text{ cm} = 3.85 \text{ m} \]
Final answer: 385 cm of wood, i.e. 3.85 m.
Question 4: 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 the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint: Sₓ₋₁ = S₄₉ − Sₓ]
The hint turns a verbal symmetry statement into a single equation in \( x \). The houses before house \( x \) are 1, 2, …, \( x-1 \), and their sum must equal the sum of the houses after \( x \), which is the total \( S_{49} \) minus the sum up to \( x \). Because 1, 2, 3, … form the AP with \( a = 1 \), \( d = 1 \), every partial sum is \( \frac{n(n+1)}{2} \) (NCERT, p. 67).
Step 1: Write each sum in the hint using \( S_n = \frac{n(n+1)}{2} \).
\[ S_{x-1} = \frac{(x-1)x}{2}, \quad S_{49} = \frac{49 \times 50}{2} = 1225, \quad S_x = \frac{x(x+1)}{2} \]
Step 2: Substitute into \( S_{x-1} = S_{49} – S_x \) and solve.
\[ \frac{(x-1)x}{2} = 1225 – \frac{x(x+1)}{2} \]
\[ x^2 – x = 2450 – x^2 – x \Rightarrow 2x^2 = 2450 \Rightarrow x^2 = 1225 \Rightarrow x = \pm 35 \]
Step 3: Since \( x \) is a house number, it must be positive, so \( x = 35 \).
Step 4 (check): both halves must add to \( S_{49} = 1225 \).
\[ S_{34} = \frac{34 \times 35}{2} = 595, \quad S_{49} – S_{35} = 1225 – \frac{35 \times 36}{2} = 1225 – 630 = 595 \]
Final answer: \( x = 35 \); the sum of houses 1 to 34 equals the sum of houses 36 to 49, both 595.
Question 5: A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = 1/4 × 1/2 × 50 m³]
Each step is a rectangular prism of concrete: volume = length × rise × tread. The key is that every step sits on top of all the rises below it. Step 1 is \( \tfrac{1}{4} \) m high, step 2 is \( 2 \times \tfrac{1}{4} \) m high, so each step’s volume is 6.25 m³ more than the one below — the fifteen volumes form an AP.
The figure shows the terrace cut open as a stepped block: each step is 50 m long, with a \( \tfrac{1}{4} \) m rise and a \( \tfrac{1}{2} \) m tread, and each new step adds one more rise on top of the ones below it.
Step 1: Volume of the first step (the hint’s calculation).
\[ V_1 = \frac{1}{4} \times \frac{1}{2} \times 50 = \frac{50}{8} = 6.25 \text{ m}^3 \]
- Step 1: The volumes climb by a fixed amount: 6.25, 12.5, 18.75, …, so \( a = 6.25 \), \( d = 6.25 \), \( n = 15 \).
- Step 2: Total concrete is the sum of the first 15 terms.
\[ S_{15} = \frac{15}{2}[2(6.25) + 14(6.25)] = \frac{15}{2}[12.5 + 87.5] = \frac{15}{2} \times 100 = 750 \text{ m}^3 \]
Final answer: 750 m³ of concrete.
How to Solve AP Word Problems: Method Recap
All five questions above follow the same path from story to answer. Keep this order and you can attempt any of them alone.
- Decide whether the data form an AP. Check that consecutive values differ by a constant amount. Q3’s rungs drop by 2 cm each; Q5’s step volumes grow by 6.25 m³ each. If the differences are not constant, the sum formulas do not apply.
- Name what you are given. List \( a \), \( d \), \( n \), \( l \) and \( S \) from the story. Q1 gives \( a \) and \( d \) and asks for \( n \); Q4 gives \( a = 1 \), \( d = 1 \) and asks for the position \( x \).
- Pick the right formula. Use \( a_n = a + (n-1)d \) when you need one particular term (Q1). Use \( S = \frac{n}{2}[2a + (n-1)d] \) when you know \( a \) and \( d \) (Q2, Q5). Use \( S = \frac{n}{2}(a + l) \) when the last term is known and \( d \) is not (Q3). The chapter summary lists all three (NCERT, p. 24).
- Substitute, then respect the domain. \( n \) must be a positive integer (Q1: \( n = 32 \), not 31.25), and reject roots the story cannot take (Q4: \( x = -35 \)).
- Check against the story. In Q1 substitute 31 and 32; in Q4 confirm the two halves sum to 1225. A result that contradicts the situation marks a slip.
Frequently Asked Questions on Exercise 5.4
Why is Exercise 5.4 marked optional in the Class 10 Maths NCERT textbook, and should I still solve it?
The star on Exercise 5.4 carries the note “These exercises are not from the examination point of view” (NCERT, p. 23), meaning this question set itself will not be asked directly in a board paper.
Still solve it: the general-term and sum formulas it trains are exactly the tools the board’s AP word problems need, and the exercise shows how to translate a story into an AP — the skill the marks actually reward.
How do you find the first negative term of an AP such as 121, 117, 113, …?
Write the general term \( a_n = a + (n-1)d \), then set it less than zero and solve the inequality. Here \( a_n = 125 – 4n \), so \( 125 – 4n \lt 0 \) gives \( n \gt 31.25 \), and the first valid term number is 32. The equality \( a_n = 0 \) only locates where the sign changes; the inequality names the first term on the negative side.
When should I use S = n/2[2a + (n−1)d] instead of S = n/2(a + l)?
Use \( S = \frac{n}{2}[2a + (n-1)d] \) when you know the first term \( a \) and the common difference \( d \) (Q2, Q5). Use \( S = \frac{n}{2}(a + l) \) when the last term \( l \) is given and \( d \) is not — that is Q3, where the 11th rung is 25 cm.
Both formulas give the same sum; the second is simply shorter when \( l \) is known (NCERT, p. 65).
How do I count the rungs of a ladder when the gap between consecutive rungs is given?
Divide the total distance between the top and bottom rungs by the gap, then add 1. The quotient counts the gaps; the +1 counts the starting rung, because rungs = gaps + 1. For the ladder in Q3: \( 250/25 + 1 = 11 \) rungs.
Need the rest of the chapter’s material in one place? Browse the Class 10 Maths notes index, or work back through the Class 10 notes hub and the wider CBSE notes library for the topics this exercise builds on.
Reference: NCERT Class 10 Mathematics textbook, chapter 5 (Arithmetic Progressions).
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