Class 10 Mathematics Exercise 5.4 NCERT Solutions Olympiad Board Exam

Chapter 5 — ARITHMETIC PROGRESSIONS

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

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Concept Used

An Arithmetic Progression (AP) has general term:

\[ a_n = a + (n-1)d \]

If we want a term to be negative, we impose condition:

\[ a_n < 0 \]

We then solve the inequality to find the smallest integer value of \( n \)

Solution Roadmap

Identify first term \( a \) and common difference \( d \)
Write general term formula
Apply condition for negativity
Solve inequality step-by-step
Find smallest integer satisfying condition

Step-by-Step Solution

Given AP: 121, 117, 113, ...

\[ a = 121 \] \[ d = 117 - 121 = -4 \]

Let the first negative term be \( a_n \)

\[ a_n = a + (n-1)d \]

Substituting values:

\[ a_n = 121 + (n-1)(-4) \]

Simplify:

\[ a_n = 121 - 4(n-1) \] \[ a_n = 121 - 4n + 4 \] \[ a_n = 125 - 4n \]

For first negative term:

\[ a_n < 0 \] \[ 125 - 4n < 0 \] \[ 125 < 4n \] \[ \frac{125}{4} < n \] \[ 31.25 < n \]

Since \( n \) must be a natural number, the smallest integer satisfying this is:

\[ n = 32 \]

Therefore, the 32nd term is the first negative term.

Verification (Important for Exams)

\[ a_{32} = 125 - 4(32) = 125 - 128 = -3 \]

Since it is negative, the result is correct.

Why This Question is Important

Tests inequality handling in AP — very common in CBSE board exams
Direct application of nth term formula
Frequently appears in competitive exams like NTSE, SSC, Banking
Concept of "first term satisfying condition" is foundational for higher algebra

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Q2 →

Concept Used

nth term of AP:

\[ a_n = a + (n-1)d \]

Sum of n terms:

\[ S_n = \frac{n}{2} \left[ 2a + (n-1)d \right] \]

Form equations using given conditions (sum & product)
Solve system to find \(a\) and \(d\)

Solution Roadmap

Write expressions for \(a_3\) and \(a_7\)
Form equation using sum condition
Form equation using product condition
Substitute and solve for \(d\)
Find \(a\)
Use sum formula to compute \(S_{16}\)

Step-by-Step Solution

Let first term = \(a\), common difference = \(d\)

Third term:

\[ a_3 = a + 2d \]

Seventh term:

\[ a_7 = a + 6d \]

Given sum:

\[ a_3 + a_7 = 6 \] \[ (a + 2d) + (a + 6d) = 6 \] \[ 2a + 8d = 6 \] Divide both sides by 2: \[ a + 4d = 3 \] \[ a = 3 - 4d \tag{1} \]

Given product:

\[ a_3 \cdot a_7 = 8 \] \[ (a + 2d)(a + 6d) = 8 \] Expand carefully: \[ a^2 + 6ad + 2ad + 12d^2 = 8 \] \[ a^2 + 8ad + 12d^2 = 8 \tag{2} \]

Substitute (1) into (2):

\[ (3 - 4d)^2 + 8(3 - 4d)d + 12d^2 = 8 \] Expand step-by-step: \[ (3 - 4d)^2 = 9 - 24d + 16d^2 \] \[ 8(3 - 4d)d = 24d - 32d^2 \] Substitute: \[ 9 - 24d + 16d^2 + 24d - 32d^2 + 12d^2 = 8 \] Combine like terms: \[ -24d + 24d = 0 \] \[ 16d^2 - 32d^2 + 12d^2 = -4d^2 \] So: \[ 9 - 4d^2 = 8 \] \[ 9 - 8 = 4d^2 \] \[ 1 = 4d^2 \] \[ d^2 = \frac{1}{4} \] \[ d = \pm \frac{1}{2} \]

Find \(a\):

Case 1: \[ d = \frac{1}{2} \] \[ a = 3 - 4\left(\frac{1}{2}\right) = 3 - 2 = 1 \] Case 2: \[ d = -\frac{1}{2} \] \[ a = 3 - 4\left(-\frac{1}{2}\right) = 3 + 2 = 5 \]

Now find \(S_{16}\):

Case 1: \[ S_{16} = \frac{16}{2} [2(1) + 15(\tfrac{1}{2})] \] \[ = 8 \left[ 2 + \frac{15}{2} \right] \] \[ = 8 \cdot \frac{19}{2} = 76 \] Case 2: \[ S_{16} = \frac{16}{2} [2(5) + 15(-\tfrac{1}{2})] \] \[ = 8 \left[ 10 - \frac{15}{2} \right] \] \[ = 8 \cdot \frac{5}{2} = 20 \]

Therefore, possible sums are: \(S_{16} = 76\) or \(S_{16} = 20\)

Exam Importance

Classic CBSE pattern: forming equations from AP terms
Tests algebraic expansion accuracy (common mistake area)
Concept of multiple solutions (very important for boards)
Frequently appears in NTSE, Olympiads, SSC exams

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Concept Used

Uniform decrease forms an Arithmetic Progression (AP)
Total wood required = sum of all rung lengths

\[ S_n = \frac{n}{2}(a + l) \]

Number of terms when spacing is equal:

\[ n = \frac{\text{Total distance}}{\text{gap}} + 1 \]

Solution Roadmap

Convert all units to cm
Find number of rungs
Model lengths as AP
Apply sum formula

Step-by-Step Solution

Distance between rungs = 25 cm
Total distance = \(2.5 \text{ m} = 250 \text{ cm}\)

Step 1: Number of rungs

\[ n = \frac{250}{25} + 1 \] \[ n = 10 + 1 = 11 \]

Step 2: Form AP

First rung (bottom): \(a = 45\) cm
Last rung (top): \(l = 25\) cm
Number of terms: \(n = 11\)

Step 3: Use sum formula

\[ S_n = \frac{n}{2}(a + l) \] \[ S_{11} = \frac{11}{2}(45 + 25) \] \[ = \frac{11}{2} \times 70 \] \[ = 11 \times 35 \] \[ = 385 \]

Total length of wood required = 385 cm

Quick Check

Since values decrease uniformly from 45 to 25 over 11 terms, the average length is:

\[ \frac{45 + 25}{2} = 35 \] \[ \text{Total} = 11 \times 35 = 385 \]

Verified correct.

Why This Question Matters

Real-life modelling using AP (very important CBSE competency)
Tests understanding of term count logic \(+1\)
Common mistake: wrong \(d\) or skipping AP structure
Asked frequently in board case-study questions
Foundation for engineering entrance problems involving sequences

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Concept Used

Sequence 1 to 49 forms an AP:

\[ a = 1,\quad d = 1 \]

Sum of first \(n\) natural numbers:

\[ S_n = \frac{n(n+1)}{2} \]

Balance condition:

\[ \text{Sum before } x = \text{Sum after } x \]

Solution Roadmap

Write sum before \(x\)
Write total sum
Write sum after \(x\)
Form equation
Solve for \(x\)

Step-by-Step Solution

Numbers form AP: \(1, 2, 3, \dots, 49\)

Let required house number be \(x\)

Step 1: Sum before \(x\)

\[ S_{x-1} = \frac{(x-1)x}{2} \]

Step 2: Total sum

\[ S_{49} = \frac{49 \times 50}{2} = 1225 \]

Step 3: Sum after \(x\)

\[ S_{49} - S_x \] where \[ S_x = \frac{x(x+1)}{2} \]

Step 4: Apply condition

\[ S_{x-1} = S_{49} - S_x \] Substitute: \[ \frac{(x-1)x}{2} = 1225 - \frac{x(x+1)}{2} \] Multiply both sides by 2: \[ x(x-1) = 2450 - x(x+1) \] Bring all terms to one side: \[ x(x-1) + x(x+1) = 2450 \] Expand: \[ (x^2 - x) + (x^2 + x) = 2450 \] \[ 2x^2 = 2450 \] \[ x^2 = 1225 \] \[ x = \sqrt{1225} \] \[ x = 35 \]

Therefore, the required house number is \(x = 35\)

Verification Insight

\[ \text{Total sum} = 1225 \] Middle split: \[ \frac{1225 - 35}{2} = 595 \]

Sum before 35 = 595 and sum after 35 = 595 → balanced

Why This Question is Important

Classic symmetry-based AP problem (very high CBSE frequency)
Tests algebra + series understanding together
Shortcut insight: middle balancing concept
Frequently appears in Olympiads and reasoning sections
Builds foundation for partition problems in higher maths

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Q5 →

Concept Used

Each step forms a cuboid
Height increases uniformly → forms AP
Total volume = sum of volumes of all steps

\[ \text{Volume} = \text{Length} \times \text{Breadth} \times \text{Height} \]

Solution Roadmap

Find volume of nth step
Observe AP pattern
Use sum of AP

Step-by-Step Solution

Length = 50 m
Tread (breadth) = \( \frac{1}{2} \) m
Rise (height increment) = \( \frac{1}{4} \) m

Step 1: Volume of nth step

Height of nth step = \( n \times \frac{1}{4} \)

\[ V_n = 50 \times \frac{1}{2} \times \left(n \times \frac{1}{4}\right) \] \[ V_n = 25 \times \frac{n}{4} \] \[ V_n = \frac{25n}{4} \]

Step 2: Form AP

First term: \[ V_1 = \frac{25}{4} \] Second term: \[ V_2 = \frac{50}{4} \] Third term: \[ V_3 = \frac{75}{4} \]

So AP is:

\[ \frac{25}{4}, \frac{50}{4}, \frac{75}{4}, \dots \] \[ a = \frac{25}{4}, \quad d = \frac{25}{4}, \quad n = 15 \]

Step 3: Total volume

\[ S_n = \frac{n}{2} \left[ 2a + (n-1)d \right] \] \[ S_{15} = \frac{15}{2} \left[ 2 \cdot \frac{25}{4} + 14 \cdot \frac{25}{4} \right] \] \[ = \frac{15}{2} \left[ \frac{50}{4} + \frac{350}{4} \right] \] \[ = \frac{15}{2} \cdot \frac{400}{4} \] \[ = \frac{15}{2} \cdot 100 \] \[ = 15 \cdot 50 \] \[ = 750 \]

Total volume of concrete required = \(750 \text{ m}^3\)

Quick Insight

Since heights form AP (1/4, 2/4, ..., 15/4), total volume is proportional to:

\[ 1 + 2 + \cdots + 15 = \frac{15 \cdot 16}{2} = 120 \] \[ \text{Total volume} = \frac{25}{4} \times 120 = 750 \]

Confirms result instantly.

Why This Question is Important

Combines mensuration + AP (very high-weight concept)
Tests modelling skill from geometry to algebra
Common CBSE case-study pattern question
Frequently asked in engineering entrance reasoning
Shortcut via summation insight gives topper advantage

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Chapter Complete!

All 5 solutions for ARITHMETIC PROGRESSIONS covered.

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Chapter 5 — ARITHMETIC PROGRESSIONS

Concept Used

Solution Roadmap

Step-by-Step Solution

Verification (Important for Exams)

Why This Question is Important

Concept Used

Solution Roadmap

Step-by-Step Solution

Exam Importance

Concept Used

Solution Roadmap

Step-by-Step Solution

Quick Check

Why This Question Matters

Concept Used

Solution Roadmap

Step-by-Step Solution

Verification Insight

Why This Question is Important

Concept Used

Solution Roadmap

Step-by-Step Solution

Quick Insight

Why This Question is Important

Chapter Complete!

Arithmetic Progressions

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