This MCQ module is based on: 5.5 Applications of AP in Real Life
TOPIC 13 OF 35
5.5 Applications of AP in Real Life
🎓 Class 10
Mathematics
CBSE
Theory
Ch 5 — Arithmetic Progressions
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 5.5 Applications of AP in Real Life
Targeting Class 10 level in Algebra, with Intermediate difficulty.
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5.5 Applications of AP in Real Life
APs show up across science, engineering, finance, and everyday life. Here we compile a rich set of applications and then work the chapter's optional and miscellaneous exercises.
Where APs Appear
- Finance: Fixed monthly investments, simple interest growth, EMI ladders.
- Construction: Stair heights, seats in rows, log piles, pipe stacks.
- Physics: Distance fallen in successive seconds under gravity — \(u, u+a, u+2a, \ldots\)
- Sports: Tournament seeding, interval training loads.
- Nature: Growth rings, salary increments.
Example 20 — Ladder rungs (NCERT)
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. If the top and bottom rungs are \(2\tfrac12\) m apart, what is the length of the wood required for the rungs?
Distance between first and last rung \(= 250\) cm; rungs are 25 cm apart, so number of rungs \(n = \tfrac{250}{25}+1 = 11\).
Rungs form an AP: \(45, \ldots, 25\) with \(n=11,\ a=45,\ l=25\).
Sum \(= \tfrac{11}{2}(45+25) = \tfrac{11}{2}\cdot 70 = 385\) cm. 3.85 metres of wood.
Fig 5.3: Ladder with 11 rungs tapering from 45 cm to 25 cm — total wood = 385 cm
Example 21 — House numbers on a street (classic NCERT)
Houses on a street 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 \(x\).
Sum before \(x\): \(1+2+\cdots+(x-1) = \tfrac{(x-1)x}{2}\).
Sum after \(x\): \((x+1)+\cdots+49 = \tfrac{49\cdot 50}{2} - \tfrac{x(x+1)}{2} = 1225 - \tfrac{x(x+1)}{2}\).
Set equal: \(\tfrac{(x-1)x}{2} = 1225 - \tfrac{x(x+1)}{2}\Rightarrow x^2 = 1225 \Rightarrow x = 35\).
Check: sum 1–34 = \(\tfrac{34\cdot 35}{2}=595\); sum 36–49 = \(1225 - 1 - \cdots\) = 595 ✓.
Example 22 — Spiral of semicircles
A spiral is made with radii of successive semicircles \(r_1=0.5,\ r_2=1.0,\ r_3=1.5,\ r_4=2.0,\ \ldots\) cm. Find the total length of a spiral made of thirteen consecutive semicircles. (Use \(\pi = \tfrac{22}{7}\).)
Length of \(k\)th semicircle \(= \pi r_k\). Total \(= \pi(r_1+r_2+\cdots+r_{13})\). The radii form an AP: \(a=0.5,\ d=0.5,\ n=13\), so \(S_{13} = \tfrac{13}{2}[1 + 12(0.5)] = \tfrac{13}{2}\cdot 7 = 45.5\).
Total length \(= \tfrac{22}{7}\times 45.5 = 143\) cm.
Example 23 — Flower potted plants in an AP
200 logs stacked: 20 in the bottom row, 19 in the next, 18 in the next. In how many rows are 200 logs placed and how many logs in the top row? (Done earlier — answer: 16 rows, 5 logs in top.)
Example 24 — Potato race
In a potato race, 10 potatoes are placed in a straight line, with the first potato 5 m from the bucket, and each subsequent potato 3 m further. The competitor runs from the bucket, picks up one potato, returns it, and repeats. What is the total distance run?
Distance for potato \(k\): \(2(5+3(k-1)) = 10 + 6(k-1)\). So distances form an AP: \(10, 16, 22, \ldots\) with 10 terms.
\(S_{10} = \tfrac{10}{2}[20 + 9(6)] = 5(20+54) = 5\cdot 74 = 370\) m.
Summary of Formulas
\(a_n = a + (n-1)d\)\(S_n = \tfrac{n}{2}[2a+(n-1)d] = \tfrac{n}{2}(a+l)\)
\(a_n = S_n - S_{n-1}\)
For three numbers in AP, take them as \((a-d),\ a,\ (a+d)\). For five, take \((a-2d),(a-d),a,(a+d),(a+2d)\).
Historical Note
The study of APs appears in ancient Indian mathematics in works by Āryabhaṭa (5th century CE) and Brahmagupta (7th century CE). Brahmagupta explicitly gave the formulas for the \(n\)th term and the sum of an AP, centuries before Gauss's famous pairing trick. The Gauss story endures because it captures the formula's elegance through a single insight.
Activity: Measure a Real Staircase
L3 ApplyMaterials: Measuring tape, notebook, a home or school staircase.
Predict: Do successive step-heights (from the floor) form an AP?
- Measure the height of step 1 from the floor, then step 2, step 3, and so on for at least 8 steps.
- Compute differences \(h_2-h_1,\ h_3-h_2, \ldots\) and record them.
- If all differences are (approximately) equal, the staircase is an AP. Compute \(a\) and \(d\).
- Use the formula to predict the height of the 20th step and check with the builder's plan if available.
Staircases are designed using the "rise-and-run" rule, which ensures a constant rise per step — exactly an AP. Indian building codes typically require a rise of 150–175 mm, so \(d\) is in that range.
Exercise 5.4 — Optional & Miscellaneous
Q1. Which term of the AP 121, 117, 113, ... is its first negative term?
\(a=121,\ d=-4\). Need \(a_n < 0\Rightarrow 121-4(n-1)<0\Rightarrow n-1 > 30.25\Rightarrow n > 31.25\). So the 32nd term, \(a_{32}=121-124=-3\), is the first negative term.
Q2. The sum of the third and seventh terms of an AP is 6 and their product is 8. Find the sum of the first 16 terms.
\(a_3+a_7=(a+2d)+(a+6d)=2a+8d=6\Rightarrow a+4d=3\).
Product: \((a+2d)(a+6d)=8\). Substitute \(a=3-4d\): \((3-2d)(3+2d)=9-4d^2=8\Rightarrow d^2=\tfrac14\Rightarrow d=\pm\tfrac12\).
If \(d=\tfrac12\): \(a=1\); \(S_{16}=8(2+15\cdot\tfrac12)=8\cdot 9.5=76\).
If \(d=-\tfrac12\): \(a=5\); \(S_{16}=8(10-15/2)=8\cdot 2.5=20\).
Product: \((a+2d)(a+6d)=8\). Substitute \(a=3-4d\): \((3-2d)(3+2d)=9-4d^2=8\Rightarrow d^2=\tfrac14\Rightarrow d=\pm\tfrac12\).
If \(d=\tfrac12\): \(a=1\); \(S_{16}=8(2+15\cdot\tfrac12)=8\cdot 9.5=76\).
If \(d=-\tfrac12\): \(a=5\); \(S_{16}=8(10-15/2)=8\cdot 2.5=20\).
Q3. A ladder has rungs 25 cm apart and tapers from 45 cm to 25 cm over 2.5 m height. Find the total length of wood needed.
As in Example 20: 11 rungs; \(S=385\) cm = 3.85 m.
Q4. Houses are numbered 1 to 49. Find \(x\) such that the sum of numbers preceding \(x\) equals the sum following \(x\).
As in Example 21: \(x^2=1225\Rightarrow x=35\).
Q5. A small terrace at a football ground has 15 steps. Each step is 50 m long, 0.25 m high, and 0.5 m wide. Volume of concrete for the first step is \(50\times 0.25\times 0.5 = 6.25\) m³. Find the total volume of concrete required.
Step \(k\) requires height \(= k\cdot 0.25\) m (each step rises). Volume of step \(k\) = \(50\times 0.5\times 0.25 k = 6.25k\). Total \(=6.25(1+2+\cdots+15) = 6.25\cdot \tfrac{15\cdot 16}{2} = 6.25\cdot 120 = 750\) m³.
Q6. Find the sum of all three-digit numbers divisible by 7.
First \(= 105\), last \(=994\), \(d=7\). \(994=105+(n-1)\cdot 7\Rightarrow n=128\). \(S=\tfrac{128}{2}(105+994)=64\cdot 1099=70336\).
Competency-Based Questions
Scenario: An athlete trains for a marathon by running 3 km on day 1, and increasing her distance by 500 m every day.
Q1. How far will she run on day 10?
L3 Apply(c) 7.5 km. \(a_{10}=3+9(0.5)=7.5\).
Q2. Analyse the total distance she will run over 20 days.
L4 Analyse\(S_{20}=\tfrac{20}{2}[6+19\cdot 0.5]=10(6+9.5)=155\) km. The cumulative distance grows quadratically in \(n\), so later days contribute disproportionately.
Q3. Evaluate whether she can complete a 150 km training block in exactly 15 days with the same plan.
L5 Evaluate\(S_{15}=\tfrac{15}{2}[6+14\cdot 0.5]=7.5\cdot 13=97.5\) km — far short of 150. She cannot; she needs a larger \(d\). Solve \(150=\tfrac{15}{2}(6+14d)\Rightarrow 20=6+14d\Rightarrow d=1\) km/day.
Q4. Design a 21-day taper where the first day is 15 km and total distance is 210 km. Find \(d\). Is \(d\) positive or negative, and why?
L6 Create\(210=\tfrac{21}{2}[30+20d]\Rightarrow 20 = 30+20d\Rightarrow d=-0.5\) km/day. Negative, because a taper reduces daily load before the race.
Assertion–Reason Questions
Assertion (A): The sum of all two-digit numbers divisible by 5 is 945.
Reason (R): The numbers form an AP 10, 15, ..., 95 with 18 terms.
Reason (R): The numbers form an AP 10, 15, ..., 95 with 18 terms.
Answer: (a). \(S=\tfrac{18}{2}(10+95)=9(105)=945\) ✓.
Assertion (A): The sum of first 100 positive integers is 5050.
Reason (R): \(1+2+\cdots+n = \tfrac{n(n+1)}{2}\).
Reason (R): \(1+2+\cdots+n = \tfrac{n(n+1)}{2}\).
Answer: (a).
Assertion (A): If three numbers are in AP with sum 15 and product 80, they are 2, 5, 8.
Reason (R): Taking the numbers as \(a-d, a, a+d\) makes the sum \(3a\).
Reason (R): Taking the numbers as \(a-d, a, a+d\) makes the sum \(3a\).
Answer: (a). \(3a=15\Rightarrow a=5\); \((5-d)(5)(5+d)=80\Rightarrow 25-d^2=16\Rightarrow d=\pm 3\). Numbers: 2, 5, 8 (or 8, 5, 2).
Frequently Asked Questions — Arithmetic Progressions
What is Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool in NCERT Class 10 Mathematics?
Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool is a key concept covered in NCERT Class 10 Mathematics, Chapter 5: Arithmetic Progressions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool step by step?
To solve problems on Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 5: Arithmetic Progressions?
The essential formulas of Chapter 5 (Arithmetic Progressions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool important for the Class 10 board exam?
Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool?
Common mistakes in Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool?
End-of-chapter NCERT exercises for Part 3 — AP Applications & Exercises | Class 10 Maths Ch 5 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.
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