Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14
📐 Mathematics Chapter 5

Arithmetic Progressions | अनुक्रम एवं श्रेणी

Complete NCERT Solutions | Exercise 5.1, 5.2, 5.3, 5.4

🔗 Quick Resources

📝

Revision Notes

 ⭐

Important Ques

 🎯

MCQs

 📐

Formulas

 🎥

Video Lectures

 📋

Sample Paper

📖 Exercise 5.1 (Introduction to AP)

Q1(i). Taxi fare: ₹15 for first km, ₹8 for each additional km. Check if it forms an AP. ▼

Step-by-Step Solution:

Step 1: Fare for 1st kilometer = ₹15

Step 2: Fare for 2nd kilometer = ₹15 + ₹8 = ₹23

Step 3: Fare for 3rd kilometer = ₹23 + ₹8 = ₹31

Step 4: Fare for 4th kilometer = ₹31 + ₹8 = ₹39

Step 5: Sequence formed: 15, 23, 31, 39, ...

Step 6: Find differences: 23-15=8, 31-23=8, 39-31=8

Step 7: Since difference is constant (8), it forms an AP.

✅ Yes, it forms an AP with common difference d = 8

Q1(ii). Vacuum pump removes 1/4 of air remaining each time. Check if it forms an AP. ▼

Step-by-Step Solution:

Step 1: Let initial air = V

Step 2: After 1st removal: V - V/4 = 3V/4

Step 3: After 2nd removal: 3V/4 - 1/4(3V/4) = 3V/4 - 3V/16 = 9V/16

Step 4: After 3rd removal: 9V/16 - 1/4(9V/16) = 9V/16 - 9V/64 = 27V/64

Step 5: Sequence: V, 3V/4, 9V/16, 27V/64, ...

Step 6: Check differences: 3V/4 - V = -V/4, 9V/16 - 3V/4 = -3V/16 (not equal)

Step 7: Ratio is constant (3/4), so it's a GP, not AP.

❌ No, it forms a GP, not an AP

Q1(iii). Well digging: ₹150 for first metre, ₹50 extra each subsequent metre. Check if it forms an AP. ▼

Step-by-Step Solution:

Step 1: Cost for 1st metre = ₹150

Step 2: Cost for 2nd metre = ₹150 + ₹50 = ₹200

Step 3: Cost for 3rd metre = ₹200 + ₹50 = ₹250

Step 4: Cost for 4th metre = ₹250 + ₹50 = ₹300

Step 5: Sequence: 150, 200, 250, 300, ...

Step 6: Differences: 200-150=50, 250-200=50, 300-250=50

Step 7: Since difference is constant (50), it forms an AP.

✅ Yes, it forms an AP with common difference d = 50

Q1(iv). Compound interest: ₹10000 at 8% per annum. Check if it forms an AP. ▼

Step-by-Step Solution:

Step 1: Amount after 1 year = 10000(1+8/100) = 10000 × 1.08 = 10800

Step 2: Amount after 2 years = 10800 × 1.08 = 11664

Step 3: Amount after 3 years = 11664 × 1.08 = 12597.12

Step 4: Sequence: 10000, 10800, 11664, 12597, ...

Step 5: Differences: 10800-10000=800, 11664-10800=864, 12597-11664=933

Step 6: Differences are not equal, so it's not an AP.

❌ No, it forms a GP (compound interest), not AP

Q2(i). Write first four terms: a = 10, d = 10 ▼

Step-by-Step Solution:

Step 1: First term a₁ = a = 10

Step 2: Second term a₂ = a + d = 10 + 10 = 20

Step 3: Third term a₃ = a₂ + d = 20 + 10 = 30

Step 4: Fourth term a₄ = a₃ + d = 30 + 10 = 40

✅ First four terms: 10, 20, 30, 40

Q2(ii). Write first four terms: a = -2, d = 0 ▼

Step-by-Step Solution:

Step 1: a₁ = -2

Step 2: a₂ = -2 + 0 = -2

Step 3: a₃ = -2 + 0 = -2

Step 4: a₄ = -2 + 0 = -2

✅ First four terms: -2, -2, -2, -2

Q2(iii). Write first four terms: a = 4, d = -3 ▼

Step-by-Step Solution:

Step 1: a₁ = 4

Step 2: a₂ = 4 + (-3) = 1

Step 3: a₃ = 1 + (-3) = -2

Step 4: a₄ = -2 + (-3) = -5

✅ First four terms: 4, 1, -2, -5

Q2(iv). Write first four terms: a = -1, d = 1/2 ▼

Step-by-Step Solution:

Step 1: a₁ = -1

Step 2: a₂ = -1 + 1/2 = -1/2

Step 3: a₃ = -1/2 + 1/2 = 0

Step 4: a₄ = 0 + 1/2 = 1/2

✅ First four terms: -1, -1/2, 0, 1/2

Q2(v). Write first four terms: a = -1.25, d = -0.25 ▼

Step-by-Step Solution:

Step 1: a₁ = -1.25

Step 2: a₂ = -1.25 + (-0.25) = -1.50

Step 3: a₃ = -1.50 + (-0.25) = -1.75

Step 4: a₄ = -1.75 + (-0.25) = -2.00

✅ First four terms: -1.25, -1.50, -1.75, -2.00

Q3(i). For AP: 3, 1, -1, -3,... find first term and common difference. ▼

Step-by-Step Solution:

Step 1: First term is the first number in the sequence = 3

Step 2: Common difference = second term - first term = 1 - 3 = -2

Step 3: Verify: -1 - 1 = -2, -3 - (-1) = -2 ✓

✅ First term a = 3, Common difference d = -2

Q3(ii). For AP: -5, -1, 3, 7,... find first term and common difference. ▼

Step-by-Step Solution:

Step 1: First term a = -5

Step 2: d = second term - first term = (-1) - (-5) = -1 + 5 = 4

Step 3: Verify: 3 - (-1) = 4, 7 - 3 = 4 ✓

✅ a = -5, d = 4

Q3(iii). For AP: 1/3, 5/3, 9/3, 13/3,... find first term and common difference. ▼

Step-by-Step Solution:

Step 1: First term a = 1/3

Step 2: d = 5/3 - 1/3 = 4/3

Step 3: Verify: 9/3 - 5/3 = 4/3, 13/3 - 9/3 = 4/3 ✓

✅ a = 1/3, d = 4/3

Q3(iv). For AP: 0.6, 1.7, 2.8, 3.9,... find first term and common difference. ▼

Step-by-Step Solution:

Step 1: First term a = 0.6

Step 2: d = 1.7 - 0.6 = 1.1

Step 3: Verify: 2.8 - 1.7 = 1.1, 3.9 - 2.8 = 1.1 ✓

✅ a = 0.6, d = 1.1

Q4(i). Check if 2,4,8,16,... is AP. If yes, find d and three more terms. ▼

Step-by-Step Solution:

Step 1: Find differences: 4-2=2, 8-4=4, 16-8=8

Step 2: Differences are not equal (2 ≠ 4 ≠ 8)

Step 3: Since common difference is not constant, it is not an AP.

Step 4: This is actually a GP with common ratio 2.

❌ Not an AP (it is a GP)

Q4(ii). Check if 2, 5/2, 3, 7/2,... is AP. If yes, find d and three more terms. ▼

Step-by-Step Solution:

Step 1: Convert to decimals: 2, 2.5, 3, 3.5, ...

Step 2: Differences: 2.5-2=0.5, 3-2.5=0.5, 3.5-3=0.5

Step 3: Common difference d = 0.5 (constant)

Step 4: Next three terms: 3.5+0.5=4, 4+0.5=4.5, 4.5+0.5=5

✅ Yes, AP with d = 0.5, next terms: 4, 4.5, 5

Q4(iii). Check if -1.2, -3.2, -5.2, -7.2,... is AP. If yes, find d and three more terms. ▼

Step-by-Step Solution:

Step 1: Differences: (-3.2) - (-1.2) = -2

Step 2: (-5.2) - (-3.2) = -2, (-7.2) - (-5.2) = -2

Step 3: Common difference d = -2 (constant)

Step 4: Next terms: -7.2-2=-9.2, -9.2-2=-11.2, -11.2-2=-13.2

✅ Yes, AP with d = -2, next terms: -9.2, -11.2, -13.2

Q4(iv). Check if -10, -6, -2, 2,... is AP. If yes, find d and three more terms. ▼

Step-by-Step Solution:

Step 1: Differences: (-6) - (-10) = 4

Step 2: (-2) - (-6) = 4, 2 - (-2) = 4

Step 3: Common difference d = 4 (constant)

Step 4: Next terms: 2+4=6, 6+4=10, 10+4=14

✅ Yes, AP with d = 4, next terms: 6, 10, 14

Q4(v). Check if 3, 3+√2, 3+2√2, 3+3√2,... is AP. If yes, find d and three more terms. ▼

Step-by-Step Solution:

Step 1: Differences: (3+√2) - 3 = √2

Step 2: (3+2√2) - (3+√2) = √2, (3+3√2) - (3+2√2) = √2

Step 3: Common difference d = √2 (constant)

Step 4: Next terms: 3+4√2, 3+5√2, 3+6√2

✅ Yes, AP with d = √2, next terms: 3+4√2, 3+5√2, 3+6√2

Q4(vi). Check if 0.2, 0.22, 0.222, 0.2222,... is AP. ▼

Step-by-Step Solution:

Step 1: Differences: 0.22 - 0.2 = 0.02

Step 2: 0.222 - 0.22 = 0.002, 0.2222 - 0.222 = 0.0002

Step 3: Differences are not equal (0.02 ≠ 0.002 ≠ 0.0002)

Step 4: Since common difference is not constant, it is not an AP.

❌ Not an AP

Q4(vii). Check if 0, -4, -8, -12,... is AP. If yes, find d and three more terms. ▼

Step-by-Step Solution:

Step 1: Differences: (-4) - 0 = -4

Step 2: (-8) - (-4) = -4, (-12) - (-8) = -4

Step 3: Common difference d = -4 (constant)

Step 4: Next terms: -12-4=-16, -16-4=-20, -20-4=-24

✅ Yes, AP with d = -4, next terms: -16, -20, -24

Q4(viii). Check if -1/2, -1/2, -1/2, -1/2,... is AP. If yes, find d and three more terms. ▼

Step-by-Step Solution:

Step 1: Differences: (-1/2) - (-1/2) = 0

Step 2: All differences = 0 (constant)

Step 3: Common difference d = 0

Step 4: Next terms: -1/2, -1/2, -1/2 (all same)

✅ Yes, AP with d = 0, next terms: -1/2, -1/2, -1/2

Q4(ix). Check if 1, 3, 9, 27,... is AP. ▼

Step-by-Step Solution:

Step 1: Differences: 3-1=2, 9-3=6, 27-9=18

Step 2: Differences are not equal (2 ≠ 6 ≠ 18)

Step 3: This is actually a GP with common ratio 3.

❌ Not an AP (it is a GP)

Q4(x). Check if a, 2a, 3a, 4a,... is AP. If yes, find d and three more terms. ▼

Step-by-Step Solution:

Step 1: Differences: 2a - a = a, 3a - 2a = a, 4a - 3a = a

Step 2: Common difference d = a (constant)

Step 3: Next three terms: 5a, 6a, 7a

✅ Yes, AP with d = a, next terms: 5a, 6a, 7a

Q4(xi). Check if a, a², a³, a⁴,... is AP. ▼

Step-by-Step Solution:

Step 1: Differences: a² - a = a(a-1), a³ - a² = a²(a-1)

Step 2: These are equal only if a = 0 or a = 1

Step 3: For general a, differences are not equal.

❌ Not an AP (it is a GP)

Q4(xii). Check if √2, √8, √18, √32,... is AP. If yes, find d and three more terms. ▼

Step-by-Step Solution:

Step 1: Simplify: √8 = 2√2, √18 = 3√2, √32 = 4√2

Step 2: Sequence: √2, 2√2, 3√2, 4√2, ...

Step 3: Differences: 2√2 - √2 = √2, 3√2 - 2√2 = √2, 4√2 - 3√2 = √2

Step 4: Common difference d = √2 (constant)

Step 5: Next terms: 5√2, 6√2, 7√2 = √50, √72, √98

✅ Yes, AP with d = √2, next terms: √50, √72, √98

Q4(xiii). Check if √3, √6, √9, √12,... is AP. ▼

Step-by-Step Solution:

Step 1: Approximate values: √3≈1.732, √6≈2.449, √9=3, √12≈3.464

Step 2: Differences: 2.449-1.732=0.717, 3-2.449=0.551, 3.464-3=0.464

Step 3: Differences are not equal

❌ Not an AP

Q4(xiv). Check if 1², 3², 5², 7²,... is AP. ▼

Step-by-Step Solution:

Step 1: Calculate squares: 1, 9, 25, 49, ...

Step 2: Differences: 9-1=8, 25-9=16, 49-25=24

Step 3: Differences are not equal (8 ≠ 16 ≠ 24)

❌ Not an AP

Q4(xv). Check if 1², 5², 7², 73,... is AP. ▼

Step-by-Step Solution:

Step 1: Calculate: 1²=1, 5²=25, 7²=49, 73

Step 2: Differences: 25-1=24, 49-25=24, 73-49=24

Step 3: Common difference d = 24 (constant)

Step 4: Next three terms: 73+24=97, 97+24=121, 121+24=145

✅ Yes, AP with d = 24, next terms: 97, 121, 145

📖 Exercise 5.2 (nth Term of AP)

Q1. Fill in the blanks in the following table, given that a is the first term, d the common difference and aₙ the nth term of the AP. ▼

Step-by-Step Solution:

(i) Given: a = 7, d = 3, n = 8, find aₙ

Step 1: Formula for nth term: aₙ = a + (n-1)d

Step 2: Substitute the values: a₈ = 7 + (8-1) × 3

Step 3: Simplify: a₈ = 7 + 7 × 3 = 7 + 21 = 28

✅ a₈ = 28

(ii) Given: a = -18, n = 10, aₙ = 0, find d

Step 1: Formula: aₙ = a + (n-1)d

Step 2: 0 = -18 + (10-1)d

Step 3: 0 = -18 + 9d

Step 4: 9d = 18

Step 5: d = 18/9 = 2

✅ d = 2

(iii) Given: d = -3, n = 18, aₙ = -5, find a

Step 1: Formula: aₙ = a + (n-1)d

Step 2: -5 = a + (18-1)(-3)

Step 3: -5 = a + 17 × (-3)

Step 4: -5 = a - 51

Step 5: a = -5 + 51 = 46

✅ a = 46

(iv) Given: a = -18.9, d = 2.5, aₙ = 3.6, find n

Step 1: Formula: aₙ = a + (n-1)d

Step 2: 3.6 = -18.9 + (n-1) × 2.5

Step 3: 3.6 + 18.9 = (n-1) × 2.5

Step 4: 22.5 = (n-1) × 2.5

Step 5: n-1 = 22.5 / 2.5 = 9

Step 6: n = 9 + 1 = 10

✅ n = 10

(v) Given: a = 3.5, d = 0, n = 105, find aₙ

Step 1: Formula: aₙ = a + (n-1)d

Step 2: a₁₀₅ = 3.5 + (105-1) × 0

Step 3: a₁₀₅ = 3.5 + 104 × 0 = 3.5 + 0 = 3.5

✅ a₁₀₅ = 3.5

Q2(i). 30th term of the AP: 10, 7, 4, ... is ▼

Step-by-Step Solution:

Step 1: Identify first term a = 10

Step 2: Common difference d = 7 - 10 = -3

Step 3: Formula: aₙ = a + (n-1)d

Step 4: a₃₀ = 10 + (30-1) × (-3)

Step 5: a₃₀ = 10 + 29 × (-3) = 10 - 87 = -77

✅ Answer: (C) -77

Q2(ii). 11th term of the AP: -3, -1/2, 2, ... is ▼

Step-by-Step Solution:

Step 1: First term a = -3

Step 2: Common difference d = (-1/2) - (-3) = -0.5 + 3 = 2.5 = 5/2

Step 3: Formula: aₙ = a + (n-1)d

Step 4: a₁₁ = -3 + (11-1) × (5/2)

Step 5: a₁₁ = -3 + 10 × 5/2 = -3 + 25 = 22

✅ Answer: (B) 22

Q3(i). Find the missing term: 2, □, 26 ▼

Step-by-Step Solution:

Step 1: Let the missing term be x

Step 2: Since it's an AP, the common difference is constant

Step 3: d = x - 2 and also d = 26 - x

Step 4: Equate: x - 2 = 26 - x

Step 5: 2x = 28 → x = 14

✅ Missing term = 14

Q3(ii). Find the missing terms: □, 13, □, 3 ▼

Step-by-Step Solution:

Step 1: Let the AP be: a, 13, b, 3 (4 terms)

Step 2: Common difference d = 13 - a and also d = b - 13 and d = 3 - b

Step 3: Since it has 4 terms, 3 = a + 3d

Step 4: Also 13 = a + d

Step 5: Subtract: (a+3d) - (a+d) = 3 - 13 → 2d = -10 → d = -5

Step 6: a = 13 - d = 13 - (-5) = 18

Step 7: b = 13 + d = 13 + (-5) = 8

✅ Missing terms: 18, 8

Q3(iii). Find the missing terms: 5, □, □, □, 9/2 ▼

Step-by-Step Solution:

Step 1: AP has 5 terms: a₁=5, a₅=9/2=4.5

Step 2: Formula: aₙ = a + (n-1)d

Step 3: 4.5 = 5 + (5-1)d = 5 + 4d

Step 4: 4d = 4.5 - 5 = -0.5 → d = -0.125 = -1/8

Step 5: a₂ = 5 + (-1/8) = 39/8 = 4.875

Step 6: a₃ = 39/8 + (-1/8) = 38/8 = 19/4 = 4.75

Step 7: a₄ = 19/4 + (-1/8) = 38/8 - 1/8 = 37/8 = 4.625

✅ Missing terms: 39/8, 19/4, 37/8

Q3(iv). Find the missing terms: -4, □, □, □, □, 6 ▼

Step-by-Step Solution:

Step 1: AP has 6 terms: a₁=-4, a₆=6

Step 2: Formula: aₙ = a + (n-1)d

Step 3: 6 = -4 + (6-1)d = -4 + 5d

Step 4: 5d = 10 → d = 2

Step 5: a₂ = -4 + 2 = -2

Step 6: a₃ = -2 + 2 = 0

Step 7: a₄ = 0 + 2 = 2

Step 8: a₅ = 2 + 2 = 4

✅ Missing terms: -2, 0, 2, 4

Q3(v). Find the missing terms: □, 38, □, □, □, -22 ▼

Step-by-Step Solution:

Step 1: AP has 6 terms: a₂=38, a₆=-22

Step 2: a₂ = a + d = 38 ...(1)

Step 3: a₆ = a + 5d = -22 ...(2)

Step 4: Subtract (2)-(1): 4d = -60 → d = -15

Step 5: From (1): a = 38 - d = 38 - (-15) = 53

Step 6: a₃ = 38 + (-15) = 23

Step 7: a₄ = 23 + (-15) = 8

Step 8: a₅ = 8 + (-15) = -7

✅ Missing terms: 53, 23, 8, -7

Q4. Which term of the AP: 3, 8, 13, 18, ... is 78? ▼

Step-by-Step Solution:

Step 1: First term a = 3

Step 2: Common difference d = 8 - 3 = 5

Step 3: Let the nth term be 78

Step 4: Formula: aₙ = a + (n-1)d

Step 5: 78 = 3 + (n-1) × 5

Step 6: 78 - 3 = (n-1) × 5

Step 7: 75 = 5(n-1)

Step 8: n-1 = 75/5 = 15

Step 9: n = 16

✅ 78 is the 16th term

Q5(i). Find number of terms: 7, 13, 19, ..., 205 ▼

Step-by-Step Solution:

Step 1: a = 7, d = 13-7 = 6, last term l = 205

Step 2: Formula: l = a + (n-1)d

Step 3: 205 = 7 + (n-1) × 6

Step 4: 205 - 7 = 6(n-1)

Step 5: 198 = 6(n-1)

Step 6: n-1 = 198/6 = 33

Step 7: n = 34

✅ Number of terms = 34

Q5(ii). Find number of terms: 18, 15½, 13, ..., -47 ▼

Step-by-Step Solution:

Step 1: a = 18, d = 15.5 - 18 = -2.5 = -5/2, l = -47

Step 2: Formula: l = a + (n-1)d

Step 3: -47 = 18 + (n-1) × (-2.5)

Step 4: -47 - 18 = -2.5(n-1)

Step 5: -65 = -2.5(n-1)

Step 6: n-1 = 65/2.5 = 26

Step 7: n = 27

✅ Number of terms = 27

Q6. Check whether -150 is a term of the AP: 11, 8, 5, 2, ... ▼

Step-by-Step Solution:

Step 1: a = 11, d = 8-11 = -3

Step 2: Let the nth term be -150

Step 3: Formula: aₙ = a + (n-1)d

Step 4: -150 = 11 + (n-1) × (-3)

Step 5: -150 - 11 = -3(n-1)

Step 6: -161 = -3(n-1)

Step 7: n-1 = 161/3 = 53.666... (not an integer)

❌ -150 is not a term of this AP

Q7. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73. ▼

Step-by-Step Solution:

Step 1: Let first term = a, common difference = d

Step 2: a₁₁ = a + 10d = 38 ...(1)

Step 3: a₁₆ = a + 15d = 73 ...(2)

Step 4: Subtract (2) - (1): 5d = 35 → d = 7

Step 5: From (1): a + 10(7) = 38 → a + 70 = 38 → a = -32

Step 6: a₃₁ = a + 30d = -32 + 30×7 = -32 + 210 = 178

✅ 31st term = 178

Q8. An AP consists of 50 terms of which 3rd term is 12 and last term is 106. Find the 29th term. ▼

Step-by-Step Solution:

Step 1: a₃ = a + 2d = 12 ...(1)

Step 2: a₅₀ = a + 49d = 106 ...(2)

Step 3: Subtract (2) - (1): 47d = 94 → d = 2

Step 4: From (1): a + 4 = 12 → a = 8

Step 5: a₂₉ = a + 28d = 8 + 28×2 = 8 + 56 = 64

✅ 29th term = 64

Q9. If the 3rd and 9th terms of an AP are 4 and -8 respectively, which term is zero? ▼

Step-by-Step Solution:

Step 1: a₃ = a + 2d = 4 ...(1)

Step 2: a₉ = a + 8d = -8 ...(2)

Step 3: Subtract (2) - (1): 6d = -12 → d = -2

Step 4: From (1): a + 2(-2) = 4 → a - 4 = 4 → a = 8

Step 5: Let nth term be 0: a + (n-1)d = 0

Step 6: 8 + (n-1)(-2) = 0 → 8 - 2n + 2 = 0 → 10 - 2n = 0

Step 7: 2n = 10 → n = 5

✅ 5th term is zero

Q10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference. ▼

Step-by-Step Solution:

Step 1: a₁₇ = a + 16d

Step 2: a₁₀ = a + 9d

Step 3: Given: a₁₇ - a₁₀ = 7

Step 4: (a+16d) - (a+9d) = 7

Step 5: 7d = 7

Step 6: d = 1

✅ Common difference = 1

Q11. Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term? ▼

Step-by-Step Solution:

Step 1: a = 3, d = 15-3 = 12

Step 2: a₅₄ = 3 + (54-1)×12 = 3 + 53×12 = 3 + 636 = 639

Step 3: Let the required term be aₙ = a₅₄ + 132 = 639 + 132 = 771

Step 4: 3 + (n-1)×12 = 771

Step 5: (n-1)×12 = 768

Step 6: n-1 = 768/12 = 64

Step 7: n = 65

✅ 65th term

Q12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms? ▼

Step-by-Step Solution:

Step 1: Let the two APs have first terms A and B, same common difference d

Step 2: 100th term difference = (A+99d) - (B+99d) = A - B = 100

Step 3: 1000th term difference = (A+999d) - (B+999d) = A - B

Step 4: Since A - B = 100, the difference remains 100

✅ Difference = 100

Q13. How many three-digit numbers are divisible by 7? ▼

Step-by-Step Solution:

Step 1: Smallest 3-digit number = 100

Step 2: First number divisible by 7 after 100: 105 (since 7×15=105)

Step 3: Largest 3-digit number = 999

Step 4: Last number divisible by 7 before 999: 994 (since 7×142=994)

Step 5: Sequence: 105, 112, 119, ..., 994

Step 6: a = 105, d = 7, l = 994

Step 7: 994 = 105 + (n-1)×7

Step 8: 889 = (n-1)×7

Step 9: n-1 = 889/7 = 127

Step 10: n = 128

✅ 128 three-digit numbers are divisible by 7

Q14. How many multiples of 4 lie between 10 and 250? ▼

Step-by-Step Solution:

Step 1: First multiple of 4 greater than 10: 12 (4×3=12)

Step 2: Last multiple of 4 less than 250: 248 (4×62=248)

Step 3: Sequence: 12, 16, 20, ..., 248

Step 4: a = 12, d = 4, l = 248

Step 5: 248 = 12 + (n-1)×4

Step 6: 236 = 4(n-1)

Step 7: n-1 = 236/4 = 59

Step 8: n = 60

✅ 60 multiples of 4

Q15. For what value of n, are the nth terms of two APs: 63,65,67,... and 3,10,17,... equal? ▼

Step-by-Step Solution:

Step 1: First AP: a₁ = 63, d₁ = 65-63 = 2

Step 2: Second AP: a₂ = 3, d₂ = 10-3 = 7

Step 3: nth term of first AP = 63 + (n-1)×2 = 2n + 61

Step 4: nth term of second AP = 3 + (n-1)×7 = 7n - 4

Step 5: Set equal: 2n + 61 = 7n - 4

Step 6: 61 + 4 = 7n - 2n

Step 7: 65 = 5n

Step 8: n = 13

✅ n = 13

Q16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. ▼

Step-by-Step Solution:

Step 1: a₃ = a + 2d = 16 ...(1)

Step 2: a₇ - a₅ = 12

Step 3: (a+6d) - (a+4d) = 12

Step 4: 2d = 12 → d = 6

Step 5: From (1): a + 12 = 16 → a = 4

Step 6: AP = 4, 10, 16, 22, 28, ...

✅ AP: 4, 10, 16, 22, 28, ...

Q17. Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253 ▼

Step-by-Step Solution:

Step 1: From the last, a = 253, d = -5 (reverse order)

Step 2: 20th term from last = 253 + (20-1)×(-5)

Step 3: = 253 + 19 × (-5) = 253 - 95 = 158

✅ 20th term from last = 158

Q18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms. ▼

Step-by-Step Solution:

Step 1: a₄ + a₈ = (a+3d) + (a+7d) = 2a + 10d = 24 → a + 5d = 12 ...(1)

Step 2: a₆ + a₁₀ = (a+5d) + (a+9d) = 2a + 14d = 44 → a + 7d = 22 ...(2)

Step 3: Subtract (2)-(1): 2d = 10 → d = 5

Step 4: From (1): a + 25 = 12 → a = -13

Step 5: First three terms: a = -13, a+d = -8, a+2d = -3

✅ AP: -13, -8, -3, ...

Q19. Subba Rao started work in 1995 at salary ₹5000 and increment of ₹200 each year. In which year did his income reach ₹7000? ▼

Step-by-Step Solution:

Step 1: a = 5000, d = 200, aₙ = 7000

Step 2: 7000 = 5000 + (n-1)×200

Step 3: 2000 = 200(n-1)

Step 4: n-1 = 10 → n = 11

Step 5: Year = 1995 + 10 = 2005

✅ Year: 2005

Q20. Ramkali saved ₹5 in first week and increased weekly savings by ₹1.75. If in nth week, savings become ₹20.75, find n. ▼

Step-by-Step Solution:

Step 1: a = 5, d = 1.75, aₙ = 20.75

Step 2: 20.75 = 5 + (n-1)×1.75

Step 3: 15.75 = 1.75(n-1)

Step 4: n-1 = 15.75/1.75 = 9

Step 5: n = 10

✅ n = 10 weeks

📖 Exercise 5.3 (Sum of n Terms of AP)

Q1(i). Find the sum of AP: 2, 7, 12, ... to 10 terms. ▼

Step-by-Step Solution:

Step 1: First term a = 2

Step 2: Common difference d = 7 - 2 = 5

Step 3: Number of terms n = 10

Step 4: Formula for sum of n terms: Sₙ = n/2 [2a + (n-1)d]

Step 5: S₁₀ = 10/2 [2×2 + (10-1)×5]

Step 6: S₁₀ = 5 [4 + 9×5] = 5 [4 + 45] = 5 × 49 = 245

✅ Sum = 245

Q1(ii). Find the sum of AP: -37, -33, -29, ... to 12 terms. ▼

Step-by-Step Solution:

Step 1: a = -37

Step 2: d = (-33) - (-37) = -33 + 37 = 4

Step 3: n = 12

Step 4: S₁₂ = 12/2 [2×(-37) + (12-1)×4]

Step 5: S₁₂ = 6 [-74 + 11×4] = 6 [-74 + 44] = 6 × (-30) = -180

✅ Sum = -180

Q1(iii). Find the sum of AP: 0.6, 1.7, 2.8, ... to 100 terms. ▼

Step-by-Step Solution:

Step 1: a = 0.6

Step 2: d = 1.7 - 0.6 = 1.1

Step 3: n = 100

Step 4: S₁₀₀ = 100/2 [2×0.6 + (100-1)×1.1]

Step 5: S₁₀₀ = 50 [1.2 + 99×1.1] = 50 [1.2 + 108.9] = 50 × 110.1 = 5505

✅ Sum = 5505

Q1(iv). Find the sum of AP: 1/15, 1/12, 1/10, ... to 11 terms. ▼

Step-by-Step Solution:

Step 1: a = 1/15

Step 2: d = 1/12 - 1/15 = (5-4)/60 = 1/60

Step 3: n = 11

Step 4: S₁₁ = 11/2 [2×(1/15) + (11-1)×(1/60)]

Step 5: S₁₁ = 11/2 [2/15 + 10/60] = 11/2 [2/15 + 1/6]

Step 6: LCM = 30: 2/15 = 4/30, 1/6 = 5/30, sum = 9/30 = 3/10

Step 7: S₁₁ = 11/2 × 3/10 = 33/20 = 1.65

✅ Sum = 33/20 = 1.65

Q2(i). Find the sum: 7 + 10½ + 14 + ... + 84 ▼

Step-by-Step Solution:

Step 1: a = 7, d = 10.5 - 7 = 3.5, last term l = 84

Step 2: Find n: l = a + (n-1)d

Step 3: 84 = 7 + (n-1)×3.5

Step 4: 77 = 3.5(n-1)

Step 5: n-1 = 77/3.5 = 22 → n = 23

Step 6: Formula: Sₙ = n/2 (a + l)

Step 7: S₂₃ = 23/2 (7 + 84) = 23/2 × 91 = 23 × 45.5 = 1046.5

✅ Sum = 1046.5 or 1046½

Q2(ii). Find the sum: 34 + 32 + 30 + ... + 10 ▼

Step-by-Step Solution:

Step 1: a = 34, d = 32-34 = -2, l = 10

Step 2: Find n: l = a + (n-1)d

Step 3: 10 = 34 + (n-1)×(-2)

Step 4: 10 - 34 = -2(n-1) → -24 = -2(n-1)

Step 5: n-1 = 12 → n = 13

Step 6: S₁₃ = 13/2 (34 + 10) = 13/2 × 44 = 13 × 22 = 286

✅ Sum = 286

Q2(iii). Find the sum: -5 + (-8) + (-11) + ... + (-230) ▼

Step-by-Step Solution:

Step 1: a = -5, d = (-8) - (-5) = -3, l = -230

Step 2: Find n: l = a + (n-1)d

Step 3: -230 = -5 + (n-1)×(-3)

Step 4: -230 + 5 = -3(n-1) → -225 = -3(n-1)

Step 5: n-1 = 75 → n = 76

Step 6: S₇₆ = 76/2 (-5 - 230) = 38 × (-235) = -8930

✅ Sum = -8930

Q3(i). Given a=5, d=3, aₙ=50, find n and Sₙ. ▼

Step-by-Step Solution:

Step 1: aₙ = a + (n-1)d

Step 2: 50 = 5 + (n-1)×3

Step 3: 45 = 3(n-1) → n-1 = 15 → n = 16

Step 4: S₁₆ = n/2 (a + aₙ) = 16/2 (5 + 50) = 8 × 55 = 440

✅ n = 16, Sₙ = 440

Q3(ii). Given a=7, a₁₃=35, find d and S₁₃. ▼

Step-by-Step Solution:

Step 1: a₁₃ = a + 12d

Step 2: 35 = 7 + 12d

Step 3: 28 = 12d → d = 28/12 = 7/3

Step 4: S₁₃ = 13/2 (a + a₁₃) = 13/2 (7 + 35) = 13/2 × 42 = 13 × 21 = 273

✅ d = 7/3, S₁₃ = 273

Q3(iii). Given a₁₂=37, d=3, find a and S₁₂. ▼

Step-by-Step Solution:

Step 1: a₁₂ = a + 11d

Step 2: 37 = a + 11×3 = a + 33

Step 3: a = 37 - 33 = 4

Step 4: S₁₂ = 12/2 (a + a₁₂) = 6 (4 + 37) = 6 × 41 = 246

✅ a = 4, S₁₂ = 246

Q3(iv). Given a₃=15, S₁₀=125, find d and a₁₀. ▼

Step-by-Step Solution:

Step 1: a₃ = a + 2d = 15 ...(1)

Step 2: S₁₀ = 10/2 [2a + 9d] = 5(2a + 9d) = 125

Step 3: 2a + 9d = 25 ...(2)

Step 4: Multiply (1) by 2: 2a + 4d = 30

Step 5: Subtract (2) - (2a+4d): (2a+9d) - (2a+4d) = 25 - 30 → 5d = -5 → d = -1

Step 6: From (1): a + 2(-1) = 15 → a - 2 = 15 → a = 17

Step 7: a₁₀ = a + 9d = 17 + 9(-1) = 17 - 9 = 8

✅ d = -1, a₁₀ = 8

Q3(v). Given d=5, S₉=75, find a and a₉. ▼

Step-by-Step Solution:

Step 1: S₉ = 9/2 [2a + 8×5] = 9/2 (2a + 40) = 75

Step 2: 9/2 × 2(a + 20) = 9(a + 20) = 75

Step 3: a + 20 = 75/9 = 25/3

Step 4: a = 25/3 - 20 = 25/3 - 60/3 = -35/3

Step 5: a₉ = a + 8d = -35/3 + 40 = -35/3 + 120/3 = 85/3

✅ a = -35/3, a₉ = 85/3

Q3(vi). Given a=2, d=8, Sₙ=90, find n and aₙ. ▼

Step-by-Step Solution:

Step 1: Sₙ = n/2 [2×2 + (n-1)×8] = n/2 [4 + 8n - 8] = n/2 (8n - 4) = 90

Step 2: n(8n - 4) = 180

Step 3: 8n² - 4n - 180 = 0

Step 4: Divide by 4: 2n² - n - 45 = 0

Step 5: 2n² - 10n + 9n - 45 = 0 → 2n(n-5) + 9(n-5) = 0 → (n-5)(2n+9)=0

Step 6: n = 5 or n = -9/2 (reject)

Step 7: a₅ = 2 + 4×8 = 2 + 32 = 34

✅ n = 5, aₙ = 34

Q3(vii). Given a=8, aₙ=62, Sₙ=210, find n and d. ▼

Step-by-Step Solution:

Step 1: Sₙ = n/2 (a + aₙ) = n/2 (8 + 62) = n/2 × 70 = 35n = 210

Step 2: n = 210/35 = 6

Step 3: aₙ = a + (n-1)d → 62 = 8 + 5d → 54 = 5d → d = 10.8

✅ n = 6, d = 54/5 = 10.8

Q3(viii). Given aₙ=4, d=2, Sₙ=-14, find n and a. ▼

Step-by-Step Solution:

Step 1: aₙ = a + (n-1)×2 = a + 2n - 2 = 4 → a = 6 - 2n ...(1)

Step 2: Sₙ = n/2 (a + aₙ) = n/2 (a + 4) = -14

Step 3: Substitute a: n/2 (6 - 2n + 4) = n/2 (10 - 2n) = -14

Step 4: n(5 - n) = -14 → 5n - n² = -14 → n² - 5n - 14 = 0

Step 5: (n - 7)(n + 2) = 0 → n = 7 (n positive)

Step 6: a = 6 - 2×7 = 6 - 14 = -8

✅ n = 7, a = -8

Q3(ix). Given a=3, n=8, S=192, find d. ▼

Step-by-Step Solution:

Step 1: S₈ = 8/2 [2×3 + 7d] = 4 [6 + 7d] = 192

Step 2: 6 + 7d = 192/4 = 48

Step 3: 7d = 42 → d = 6

✅ d = 6

Q3(x). Given l=28, S=144, and total 9 terms, find a. ▼

Step-by-Step Solution:

Step 1: n = 9, l = 28, S₉ = 144

Step 2: S₉ = 9/2 (a + 28) = 144

Step 3: 9(a + 28) = 288 → a + 28 = 32 → a = 4

✅ a = 4

Q4. How many terms of AP: 9, 17, 25, ... must be taken to give a sum of 636? ▼

Step-by-Step Solution:

Step 1: a = 9, d = 17-9 = 8, Sₙ = 636

Step 2: Sₙ = n/2 [2×9 + (n-1)×8] = n/2 [18 + 8n - 8] = n/2 (8n + 10) = 636

Step 3: n(4n + 5) = 636 → 4n² + 5n - 636 = 0

Step 4: 4n² + 53n - 48n - 636 = 0 → n(4n+53) - 12(4n+53) = 0

Step 5: (4n+53)(n-12) = 0 → n = 12 (n positive)

✅ n = 12 terms

Q5. First term of AP is 5, last term is 45 and sum is 400. Find number of terms and common difference. ▼

Step-by-Step Solution:

Step 1: a = 5, l = 45, Sₙ = 400

Step 2: Sₙ = n/2 (5 + 45) = n/2 × 50 = 25n = 400 → n = 16

Step 3: l = a + (n-1)d → 45 = 5 + 15d → 40 = 15d → d = 40/15 = 8/3

✅ n = 16, d = 8/3

Q6. First and last terms of AP are 17 and 350 respectively. If common difference is 9, how many terms are there and what is their sum? ▼

Step-by-Step Solution:

Step 1: a = 17, l = 350, d = 9

Step 2: l = a + (n-1)d → 350 = 17 + (n-1)×9

Step 3: 333 = 9(n-1) → n-1 = 37 → n = 38

Step 4: S₃₈ = 38/2 (17 + 350) = 19 × 367 = 6973

✅ n = 38, Sum = 6973

Q7. Find the sum of first 22 terms of an AP in which d=7 and 22nd term is 149. ▼

Step-by-Step Solution:

Step 1: d = 7, a₂₂ = 149, n = 22

Step 2: a₂₂ = a + 21d = a + 147 = 149 → a = 2

Step 3: S₂₂ = 22/2 (2 + 149) = 11 × 151 = 1661

✅ Sum = 1661

Q8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. ▼

Step-by-Step Solution:

Step 1: a₂ = 14, a₃ = 18

Step 2: d = a₃ - a₂ = 18 - 14 = 4

Step 3: a = a₂ - d = 14 - 4 = 10

Step 4: S₅₁ = 51/2 [2×10 + 50×4] = 51/2 [20 + 200] = 51/2 × 220 = 51 × 110 = 5610

✅ Sum = 5610

Q9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. ▼

Step-by-Step Solution:

Step 1: S₇ = 7/2 [2a + 6d] = 7(a + 3d) = 49 → a + 3d = 7 ...(1)

Step 2: S₁₇ = 17/2 [2a + 16d] = 17(a + 8d) = 289 → a + 8d = 17 ...(2)

Step 3: Subtract (2)-(1): 5d = 10 → d = 2

Step 4: a + 6 = 7 → a = 1

Step 5: Sₙ = n/2 [2×1 + (n-1)×2] = n/2 [2 + 2n - 2] = n/2 × 2n = n²

✅ Sₙ = n²

Q10(i). Show that aₙ = 3 + 4n forms an AP. Find sum of first 15 terms. ▼

Step-by-Step Solution:

Step 1: a₁ = 3 + 4×1 = 7

Step 2: a₂ = 3 + 8 = 11, a₃ = 3 + 12 = 15

Step 3: d = 11 - 7 = 4 (constant)

Step 4: S₁₅ = 15/2 [2×7 + 14×4] = 15/2 [14 + 56] = 15/2 × 70 = 15 × 35 = 525

✅ AP with d=4, Sum = 525

Q10(ii). Show that aₙ = 9 - 5n forms an AP. Find sum of first 15 terms. ▼

Step-by-Step Solution:

Step 1: a₁ = 9 - 5 = 4

Step 2: a₂ = 9 - 10 = -1, a₃ = 9 - 15 = -6

Step 3: d = -1 - 4 = -5 (constant)

Step 4: S₁₅ = 15/2 [2×4 + 14×(-5)] = 15/2 [8 - 70] = 15/2 × (-62) = 15 × (-31) = -465

✅ AP with d=-5, Sum = -465

Q11. If sum of first n terms of an AP is 4n - n², find first term, sum of first two terms, second term, 3rd, 10th and nth terms. ▼

Step-by-Step Solution:

Step 1: S₁ = 4×1 - 1² = 4 - 1 = 3 → First term a = 3

Step 2: S₂ = 4×2 - 4 = 8 - 4 = 4 → Sum of first two terms = 4

Step 3: Second term = S₂ - S₁ = 4 - 3 = 1

Step 4: Third term = S₃ - S₂ = (4×3 - 9) - 4 = (12-9) - 4 = 3 - 4 = -1

Step 5: Tenth term = S₁₀ - S₉ = (40-100) - (36-81) = (-60) - (-45) = -15

Step 6: nth term = Sₙ - Sₙ₋₁ = (4n - n²) - [4(n-1) - (n-1)²]

Step 7: = 4n - n² - [4n - 4 - (n² - 2n + 1)]

Step 8: = 4n - n² - [4n - 4 - n² + 2n - 1] = 4n - n² - [6n - n² - 5] = 4n - n² - 6n + n² + 5 = 5 - 2n

✅ a=3, S₂=4, 2nd=1, 3rd=-1, 10th=-15, nth=5-2n

Q12. Find the sum of the first 40 positive integers divisible by 6. ▼

Step-by-Step Solution:

Step 1: First positive integer divisible by 6 = 6

Step 2: Sequence: 6, 12, 18, ... (AP with a=6, d=6, n=40)

Step 3: S₄₀ = 40/2 [2×6 + 39×6] = 20 [12 + 234] = 20 × 246 = 4920

✅ Sum = 4920

Q13. Find the sum of the first 15 multiples of 8. ▼

Step-by-Step Solution:

Step 1: First multiple = 8, a=8, d=8, n=15

Step 2: S₁₅ = 15/2 [2×8 + 14×8] = 15/2 [16 + 112] = 15/2 × 128 = 15 × 64 = 960

✅ Sum = 960

Q14. Find the sum of the odd numbers between 0 and 50. ▼

Step-by-Step Solution:

Step 1: Odd numbers between 0 and 50: 1, 3, 5, ..., 49

Step 2: a=1, d=2, l=49

Step 3: Find n: 49 = 1 + (n-1)×2 → 48 = 2(n-1) → n-1=24 → n=25

Step 4: S₂₅ = 25/2 (1 + 49) = 25/2 × 50 = 25 × 25 = 625

✅ Sum = 625

Q15. Penalty for delay: ₹200 for first day, ₹250 for second, ₹300 for third, etc. Find penalty for 30 days. ▼

Step-by-Step Solution:

Step 1: Penalties: 200, 250, 300, ... (AP with a=200, d=50, n=30)

Step 2: S₃₀ = 30/2 [2×200 + 29×50] = 15 [400 + 1450] = 15 × 1850 = 27750

✅ Total penalty = ₹27750

Q16. ₹700 to be used for 7 cash prizes, each prize ₹20 less than preceding prize. Find value of each prize. ▼

Step-by-Step Solution:

Step 1: Let first prize = a, then AP: a, a-20, a-40, ... (d = -20, n=7)

Step 2: S₇ = 7/2 [2a + 6×(-20)] = 7/2 (2a - 120) = 7(a - 60) = 700

Step 3: a - 60 = 100 → a = 160

Step 4: Prizes: 160, 140, 120, 100, 80, 60, 40

✅ Prizes: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40

Q17. Students planting trees: Each section of each class plants same number as class. Three sections each. Classes I to XII. Find total trees. ▼

Step-by-Step Solution:

Step 1: Class I: 3 sections × 1 tree = 3 trees

Step 2: Class II: 3 × 2 = 6 trees, ... Class XII: 3 × 12 = 36 trees

Step 3: Sequence: 3, 6, 9, ..., 36 (AP with a=3, d=3, n=12)

Step 4: S₁₂ = 12/2 (3 + 36) = 6 × 39 = 234

✅ Total trees = 234

Q18. Spiral made of successive semicircles: radii 0.5, 1.0, 1.5, 2.0, ... (13 semicircles). Find total length. (π = 22/7) ▼

Step-by-Step Solution:

Step 1: Length of semicircle = πr

Step 2: Lengths: π×0.5, π×1.0, π×1.5, ..., π×6.5 (13 terms, r from 0.5 to 6.5 step 0.5)

Step 3: Sum = π(0.5 + 1.0 + 1.5 + ... + 6.5)

Step 4: 0.5 + 1.0 + ... + 6.5 = (0.5+6.5)×13/2 = 7 × 13/2 = 91/2 = 45.5

Step 5: Total length = 22/7 × 45.5 = 22 × 6.5 = 143 cm

✅ Total length = 143 cm

Q19. 200 logs stacked: 20 logs in bottom row, 19 in next, 18 in next, etc. Find number of rows and logs in top row. ▼

Step-by-Step Solution:

Step 1: Sequence: 20, 19, 18, ..., (AP with a=20, d=-1)

Step 2: Sₙ = n/2 [40 + (n-1)(-1)] = n/2 (41 - n) = 200

Step 3: n(41 - n) = 400 → 41n - n² = 400 → n² - 41n + 400 = 0

Step 4: n² - 16n - 25n + 400 = 0 → n(n-16) - 25(n-16)=0 → (n-16)(n-25)=0

Step 5: n = 16 or 25. For n=25, last term = a+24d = 20-24 = -4 (invalid)

Step 6: So n = 16, top row = 20 + 15×(-1) = 5 logs

✅ 16 rows, top row = 5 logs

Q20. Potato race: bucket at starting point, first potato 5m away, each next potato 3m apart. Ten potatoes. Find total distance. ▼

Step-by-Step Solution:

Step 1: First potato: distance = 5m to potato + 5m back = 10m

Step 2: Second potato: 5 + 3 = 8m one way, round trip = 16m

Step 3: Third potato: 11m one way, round trip = 22m

Step 4: Sequence: 10, 16, 22, ... (AP with a=10, d=6, n=10)

Step 5: S₁₀ = 10/2 [2×10 + 9×6] = 5 [20 + 54] = 5 × 74 = 370

✅ Total distance = 370 m

📖 Exercise 5.4 (Optional - Advanced Problems)

Q1. Which term of the AP: 121, 117, 113, ... is its first negative term? ▼

Step-by-Step Solution:

Step 1: Identify first term and common difference

a = 121

d = 117 - 121 = -4

Step 2: Let the nth term be the first negative term

We need aₙ < 0

Step 3: Formula for nth term: aₙ = a + (n-1)d

aₙ = 121 + (n-1)(-4)

aₙ = 121 - 4n + 4

aₙ = 125 - 4n

Step 4: Set aₙ < 0

125 - 4n < 0

125 < 4n

n > 125/4

n > 31.25

Step 5: Since n is an integer, the smallest n greater than 31.25 is 32

Step 6: Verify: a₃₂ = 125 - 4×32 = 125 - 128 = -3

✅ First negative term is the 32nd term

Q2. 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. ▼

Step-by-Step Solution:

Step 1: Let first term = a, common difference = d

Third term: a₃ = a + 2d

Seventh term: a₇ = a + 6d

Step 2: Given: a₃ + a₇ = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

Divide by 2: a + 4d = 3 ...(1)

Step 3: Given: a₃ × a₇ = 8

(a + 2d)(a + 6d) = 8 ...(2)

Step 4: From (1), a = 3 - 4d

Substitute in (2): (3 - 4d + 2d)(3 - 4d + 6d) = 8

(3 - 2d)(3 + 2d) = 8

Step 5: Using identity (a-b)(a+b) = a² - b²

9 - 4d² = 8

4d² = 1

d² = 1/4

d = ±1/2

Step 6: Case 1 - d = 1/2

a = 3 - 4(1/2) = 3 - 2 = 1

S₁₆ = 16/2 [2×1 + (16-1)×1/2] = 8 [2 + 15/2] = 8 [4/2 + 15/2] = 8 × 19/2 = 76

Step 7: Case 2 - d = -1/2

a = 3 - 4(-1/2) = 3 + 2 = 5

S₁₆ = 16/2 [2×5 + 15×(-1/2)] = 8 [10 - 15/2] = 8 [20/2 - 15/2] = 8 × 5/2 = 20

✅ Sum of first 16 terms = 76 or 20

Q3. 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 the bottom rungs are 2½ m apart, what is the length of the wood required for the rungs? ▼

Step-by-Step Solution:

Step 1: Given distance between top and bottom rungs = 2½ m = 250 cm

Distance between consecutive rungs = 25 cm

Step 2: Number of gaps between rungs = 250 ÷ 25 = 10

Number of rungs = 10 + 1 = 11 rungs

Step 3: Length of rungs decrease uniformly from 45 cm to 25 cm

So we have an AP with first term a = 45 cm, last term l = 25 cm, n = 11

Step 4: Find common difference d

l = a + (n-1)d

25 = 45 + (11-1)d

25 = 45 + 10d

10d = 25 - 45 = -20

d = -2

Step 5: The lengths are: 45, 43, 41, 39, 37, 35, 33, 31, 29, 27, 25

Step 6: Total length of wood = Sum of all rungs

S₁₁ = n/2 (a + l) = 11/2 (45 + 25) = 11/2 × 70 = 11 × 35 = 385 cm

✅ Total length of wood required = 385 cm

Q4. 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. ▼

Step-by-Step Solution:

Step 1: Houses are numbered 1, 2, 3, ..., 49

We need to find x such that:

Sum of houses before x = Sum of houses after x

1 + 2 + ... + (x-1) = (x+1) + (x+2) + ... + 49

Step 2: Let Sₙ = n(n+1)/2 be the sum of first n natural numbers

Left side: Sₓ₋₁ = (x-1)x/2

Right side: S₄₉ - Sₓ = (49×50/2) - [x(x+1)/2] = 1225 - x(x+1)/2

Step 3: Equate both sides:

(x-1)x/2 = 1225 - x(x+1)/2

Multiply both sides by 2:

x(x-1) = 2450 - x(x+1)

x² - x = 2450 - x² - x

Step 4: Simplify:

x² - x + x² + x = 2450

2x² = 2450

x² = 1225

x = ±35

Step 5: Since x is a house number between 1 and 49, x = 35

Step 6: Verify: Sum before 35 = 1 to 34 = 34×35/2 = 595

Sum after 35 = 36 to 49 = 1225 - (35×36/2) = 1225 - 630 = 595

✅ Required house number x = 35

Q5. 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 ¼ m and a tread of ½ m. Calculate the total volume of concrete required to build the terrace. ▼

Step-by-Step Solution:

Step 1: Each step is 50 m long

Rise of each step = 1/4 m

Tread of each step = 1/2 m

Volume of a step = Length × Rise × Tread

Step 2: Volume of 1st step = 50 × 1/4 × 1/2 = 50 × 1/8 = 50/8 = 6.25 m³

Volume of 2nd step = 50 × 2/4 × 1/2 = 50 × 2/8 = 100/8 = 12.5 m³

Volume of 3rd step = 50 × 3/4 × 1/2 = 50 × 3/8 = 150/8 = 18.75 m³

And so on...

Step 3: Volumes form an AP: 50/8, 100/8, 150/8, ..., (15×50)/8

a = 50/8, d = 50/8, n = 15

Last term l = 15×50/8 = 750/8 = 93.75 m³

Step 4: Total volume = Sum of all 15 steps

S₁₅ = n/2 (a + l) = 15/2 (50/8 + 750/8) = 15/2 × 800/8 = 15/2 × 100 = 15 × 50 = 750 m³

✅ Total volume of concrete = 750 m³

📖 📥 📤