📐 Arithmetic Progressions

NCERT Class 10 Mathematics — Chapter 5 (Exercises 5.1 to 5.4)

📝 Important Notes Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 🧮 Calculators
📍 Quick Jump: Notes Ex 5.1 Ex 5.2 Ex 5.3 Ex 5.4 Calculators ⬆ Top

Important Notes & Formulas

🔑 Definition of AP

A sequence a₁, a₂, a₃, ... is an AP if the difference between consecutive terms is constant.

Condition: a₂ − a₁ = a₃ − a₂ = ... = d

📐 nth Term Formula

The nth term of an AP with first term a and common difference d:

aₙ = a + (n − 1)d

➕ Sum of n Terms

Two equivalent formulas:

Sₙ = n/2 [2a + (n−1)d]

Sₙ = n/2 [a + l] (where l = last term)

🎯 Key Properties

• If terms are in AP, then aₙ − aₘ = (n−m)d

• d = (aₙ − aₘ)/(n − m)

• If Sₙ is given, aₙ = Sₙ − Sₙ₋₁

📊 Common Difference

Always calculate:

d = a₂ − a₁ = a₃ − a₂

If differences are not equal, the sequence is not an AP.

💡 Quick Tips

• For word problems, identify a and d first

• "Sum of first n terms" → use Sₙ formula

• "Which term" → set aₙ = value and solve for n

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Exercise 5.1

1
In which of the following situations does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.

A
Yes, it forms an AP with a = 15, d = 8
B
No, it does not form an AP
C
Yes, with a = 15, d = 15
D
Yes, with a = 8, d = 15

✅ Step-by-Step Solution

Step 1: Fare for 1st km = ₹15
Step 2: Fare for 2nd km = ₹15 + ₹8 = ₹23
Step 3: Fare for 3rd km = ₹23 + ₹8 = ₹31
Step 4: Sequence: 15, 23, 31, ...
Step 5: d = 23 − 15 = 8 and 31 − 23 = 8
Conclusion: Yes, it forms an AP with first term a = 15 and common difference d = 8.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

A
Yes, it forms an AP
B
No, it does not form an AP (it forms a GP)
C
Yes, with decreasing difference
D
Cannot be determined

✅ Step-by-Step Solution

Step 1: Let initial air = V
Step 2: After 1st pump: V − V/4 = 3V/4
Step 3: After 2nd pump: 3V/4 × 3/4 = 9V/16
Step 4: Sequence: V, 3V/4, 9V/16, ...
Step 5: Differences: −V/4, −3V/16 (not equal)
Conclusion: No, it does not form an AP. It forms a Geometric Progression (GP) with common ratio r = 3/4.
(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.

A
Yes, with a = 150, d = 50
B
No, cost increases non-uniformly
C
Yes, with a = 50, d = 150
D
Yes, with a = 150, d = 200

✅ 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: Sequence: 150, 200, 250, ...
Step 5: d = 200 − 150 = 50
Conclusion: Yes, it forms an AP with a = 150, d = 50.
(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.

A
Yes, it forms an AP with d = 800
B
No, compound interest does not form an AP
C
Yes, with constant ratio
D
Only simple interest forms AP

✅ Step-by-Step Solution

Step 1: Principal = ₹10000, Rate = 8% p.a.
Step 2: After 1 year: 10000 × 1.08 = 10800
Step 3: After 2 years: 10800 × 1.08 = 11664
Step 4: After 3 years: 11664 × 1.08 = 12597.12
Step 5: Differences: 800, 864, 933.12 (not equal)
Conclusion: No, it does not form an AP. Compound interest grows exponentially, not linearly. Simple interest would form an AP.
2
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
Formula: aₙ = a + (n−1)d
a₁ = 10 + 0 = 10
a₂ = 10 + 10 = 20
a₃ = 10 + 20 = 30
a₄ = 10 + 30 = 40
Answer: 10, 20, 30, 40
(ii) a = −2, d = 0
a₁ = −2 + 0 = −2
a₂ = −2 + 0 = −2
a₃ = −2 + 0 = −2
a₄ = −2 + 0 = −2
Answer: −2, −2, −2, −2 (constant AP)
(iii) a = 4, d = −3
a₁ = 4 + 0 = 4
a₂ = 4 + (−3) = 1
a₃ = 4 + (−6) = −2
a₄ = 4 + (−9) = −5
Answer: 4, 1, −2, −5
(iv) a = −1, d = 1/2
a₁ = −1 + 0 = −1
a₂ = −1 + 1/2 = −1/2
a₃ = −1 + 1 = 0
a₄ = −1 + 3/2 = 1/2
Answer: −1, −1/2, 0, 1/2
(v) a = −1.25, d = −0.25
a₁ = −1.25 + 0 = −1.25
a₂ = −1.25 + (−0.25) = −1.50
a₃ = −1.25 + (−0.50) = −1.75
a₄ = −1.25 + (−0.75) = −2.00
Answer: −1.25, −1.50, −1.75, −2.00
3
For the following APs, write the first term and the common difference:
(i) 3, 1, −1, −3, ...
First term: a = 3
d = 1 − 3 = −2
Verify: −1 − 1 = −2 ✓
(ii) −5, −1, 3, 7, ...
First term: a = −5
d = −1 − (−5) = 4
Verify: 3 − (−1) = 4 ✓
(iii) 1/3, 5/3, 9/3, 13/3, ...
First term: a = 1/3
d = 5/3 − 1/3 = 4/3
Verify: 9/3 − 5/3 = 4/3 ✓
(iv) 0.6, 1.7, 2.8, 3.9, ...
First term: a = 0.6
d = 1.7 − 0.6 = 1.1
Verify: 2.8 − 1.7 = 1.1 ✓
4
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, ...
Check: 4 − 2 = 2, 8 − 4 = 4
Since 2 ≠ 4, Not an AP (It's a GP with r = 2)
(ii) 2, 5/2, 3, 7/2, ...
Check: 5/2 − 2 = 1/2, 3 − 5/2 = 1/2, 7/2 − 3 = 1/2
Yes, AP with d = 1/2
Next terms: 7/2 + 1/2 = 4, 4 + 1/2 = 9/2, 9/2 + 1/2 = 5
(iii) −1.2, −3.2, −5.2, −7.2, ...
Check: −3.2 − (−1.2) = −2, −5.2 − (−3.2) = −2
Yes, AP with d = −2
Next terms: −9.2, −11.2, −13.2
(iv) −10, −6, −2, 2, ...
Check: −6 − (−10) = 4, −2 − (−6) = 4
Yes, AP with d = 4
Next terms: 6, 10, 14
(v) 3, 3+√2, 3+2√2, 3+3√2, ...
Check: (3+√2) − 3 = √2, (3+2√2) − (3+√2) = √2
Yes, AP with d = √2
Next terms: 3+4√2, 3+5√2, 3+6√2
(vi) 0.2, 0.22, 0.222, 0.2222, ...
Check: 0.22 − 0.2 = 0.02, 0.222 − 0.22 = 0.002
Since 0.02 ≠ 0.002, Not an AP
(vii) 0, −4, −8, −12, ...
Check: −4 − 0 = −4, −8 − (−4) = −4
Yes, AP with d = −4
Next terms: −16, −20, −24
(viii) −1/2, −1/2, −1/2, −1/2, ...
Check: −1/2 − (−1/2) = 0
Yes, AP with d = 0 (constant AP)
Next terms: −1/2, −1/2, −1/2
(ix) 1, 3, 9, 27, ...
Check: 3 − 1 = 2, 9 − 3 = 6
Since 2 ≠ 6, Not an AP (It's a GP with r = 3)
(x) a, 2a, 3a, 4a, ...
Check: 2a − a = a, 3a − 2a = a
Yes, AP with d = a
Next terms: 5a, 6a, 7a
(xi) a, a², a³, a⁴, ...
Check: a² − a = a(a−1), a³ − a² = a²(a−1)
These are equal only if a = 0 or a = 1. In general, Not an AP
(xii) √2, √8, √18, √32, ...
Simplify: √2, 2√2, 3√2, 4√2
Check: 2√2 − √2 = √2, 3√2 − 2√2 = √2
Yes, AP with d = √2
Next terms: 5√2, 6√2, 7√2
(xiii) √3, √6, √9, √12, ...
Check: √6 − √3 ≈ 2.45 − 1.73 = 0.72
√9 − √6 = 3 − 2.45 = 0.55
Since 0.72 ≠ 0.55, Not an AP
(xiv) 1², 3², 5², 7², ... i.e., 1, 9, 25, 49, ...
Check: 9 − 1 = 8, 25 − 9 = 16
Since 8 ≠ 16, Not an AP
(xv) 1², 5², 7², 73, ... i.e., 1, 25, 49, 73, ...
Check: 25 − 1 = 24, 49 − 25 = 24, 73 − 49 = 24
Yes, AP with d = 24
Next terms: 97, 121, 145
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Exercise 5.2

1
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:
(i) a = 7, d = 3, n = 8, find aₙ
Formula: aₙ = a + (n−1)d
Substitute: a₈ = 7 + (8−1)×3 = 7 + 21 = 28
Answer: 28
(ii) a = −18, n = 10, aₙ = 0, find d
Formula: 0 = −18 + (10−1)d
Solve: 9d = 18 → d = 2
Answer: d = 2
(iii) d = −3, n = 18, aₙ = −5, find a
Formula: −5 = a + (18−1)×(−3)
Solve: −5 = a − 51 → a = 46
Answer: a = 46
(iv) a = −18.9, d = 2.5, aₙ = 3.6, find n
Formula: 3.6 = −18.9 + (n−1)×2.5
Solve: 22.5 = (n−1)×2.5 → n−1 = 9 → n = 10
Answer: n = 10
(v) a = 3.5, d = 0, n = 105, find aₙ
Formula: a₁₀₅ = 3.5 + (105−1)×0
Answer: 3.5 (constant AP)
2
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, ... , is

A
97
B
77
C
−77
D
−87

✅ Step-by-Step Solution

Step 1: a = 10, d = 7 − 10 = −3
Step 2: Using aₙ = a + (n−1)d
Step 3: a₃₀ = 10 + (30−1)×(−3)
Step 4: a₃₀ = 10 + 29×(−3) = 10 − 87 = −77
Answer: Option C (−77)
(ii) 11th term of the AP: −3, −1/2, 2, ... , is

A
28
B
22
C
−38
D
−48½

✅ Step-by-Step Solution

Step 1: a = −3, d = −1/2 − (−3) = 5/2
Step 2: Using aₙ = a + (n−1)d
Step 3: a₁₁ = −3 + (11−1)×(5/2)
Step 4: a₁₁ = −3 + 10×(5/2) = −3 + 25 = 22
Answer: Option B (22)
3
In the following APs, find the missing terms in the boxes:
(i) 2, ⬜, 26
Step 1: Let missing term be x. Then a = 2, a₃ = 26
Step 2: a₃ = a + 2d → 26 = 2 + 2d → d = 12
Step 3: Missing term = 2 + 12 = 14
Answer: 14
(ii) ⬜, 13, ⬜, 3
Step 1: Let terms be a, 13, a₃, 3
Step 2: a₄ = a + 3d = 3 and a₂ = a + d = 13
Step 3: From a + d = 13 and a + 3d = 3: subtract to get 2d = −10 → d = −5
Step 4: a = 13 − (−5) = 18, a₃ = 13 + (−5) = 8
Answer: 18, 8
(iii) 5, ⬜, ⬜, 9½
Step 1: a = 5, a₄ = 19/2
Step 2: 5 + 3d = 19/2 → 3d = 9/2 → d = 3/2
Step 3: a₂ = 5 + 3/2 = 13/2 = 6½, a₃ = 13/2 + 3/2 = 8
Answer: 6½, 8
(iv) −4, ⬜, ⬜, ⬜, ⬜, 6
Step 1: a = −4, a₆ = 6
Step 2: −4 + 5d = 6 → 5d = 10 → d = 2
Step 3: Terms: −4, −2, 0, 2, 4, 6
(v) ⬜, 38, ⬜, ⬜, ⬜, −22
Step 1: a₂ = 38, a₆ = −22
Step 2: a + d = 38 and a + 5d = −22
Step 3: Subtract: 4d = −60 → d = −15
Step 4: a = 38 − (−15) = 53
Answer: 53, 23, 8, −7
4
Which term of the AP: 3, 8, 13, 18, ... is 78?
Step 1: a = 3, d = 5
Step 2: Let aₙ = 78
Step 3: 3 + (n−1)×5 = 78
Step 4: (n−1)×5 = 75 → n−1 = 15 → n = 16
Answer: 16th term
5
Find the number of terms in each of the following APs:
(i) 7, 13, 19, ..., 205
Step 1: a = 7, d = 6, l = 205
Step 2: 205 = 7 + (n−1)×6
Step 3: 198 = (n−1)×6 → n−1 = 33 → n = 34
Answer: 34 terms
(ii) 18, 15½, 13, ..., −47
Step 1: a = 18, d = 15½ − 18 = −5/2, l = −47
Step 2: −47 = 18 + (n−1)×(−5/2)
Step 3: −65 = (n−1)×(−5/2) → n−1 = 26 → n = 27
Answer: 27 terms
6
Check whether −150 is a term of the AP: 11, 8, 5, 2, ...
Step 1: a = 11, d = −3
Step 2: Let aₙ = −150
Step 3: 11 + (n−1)×(−3) = −150
Step 4: (n−1)×(−3) = −161 → n−1 = 161/3 ≈ 53.67
Step 5: n = 54.67 which is not a positive integer
Answer: No, −150 is not a term of this AP.
7
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Step 1: a₁₁ = a + 10d = 38 ... (i)
Step 2: a₁₆ = a + 15d = 73 ... (ii)
Step 3: (ii) − (i): 5d = 35 → d = 7
Step 4: From (i): a + 70 = 38 → a = −32
Step 5: a₃₁ = −32 + 30×7 = −32 + 210 = 178
Answer: 178
8
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Step 1: a₃ = a + 2d = 12 ... (i)
Step 2: a₅₀ = a + 49d = 106 ... (ii)
Step 3: (ii) − (i): 47d = 94 → d = 2
Step 4: From (i): a + 4 = 12 → a = 8
Step 5: a₂₉ = 8 + 28×2 = 8 + 56 = 64
Answer: 64
9
If the 3rd and the 9th terms of an AP are 4 and −8 respectively, which term of this AP is zero?
Step 1: a₃ = a + 2d = 4 ... (i)
Step 2: a₉ = a + 8d = −8 ... (ii)
Step 3: (ii) − (i): 6d = −12 → d = −2
Step 4: From (i): a + 2(−2) = 4 → a = 8
Step 5: Let aₙ = 0: 8 + (n−1)×(−2) = 0
Step 6: (n−1)×(−2) = −8 → n−1 = 4 → n = 5
Answer: 5th term
10
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Step 1: a₁₇ − a₁₀ = 7
Step 2: (a + 16d) − (a + 9d) = 7
Step 3: 7d = 7 → d = 1
Answer: d = 1
11
Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?
Step 1: a = 3, d = 12
Step 2: a₅₄ = 3 + 53×12 = 3 + 636 = 639
Step 3: Required term = 639 + 132 = 771
Step 4: Let aₙ = 771: 3 + (n−1)×12 = 771
Step 5: (n−1)×12 = 768 → n−1 = 64 → n = 65
Answer: 65th term
12
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Step 1: Let first terms be a₁ and a₂, common difference = d
Step 2: a₁₀₀(1st) − a₁₀₀(2nd) = (a₁ + 99d) − (a₂ + 99d) = a₁ − a₂ = 100
Step 3: a₁₀₀₀(1st) − a₁₀₀₀(2nd) = (a₁ + 999d) − (a₂ + 999d) = a₁ − a₂ = 100
Answer: 100 (The difference remains constant for any term)
13
How many three-digit numbers are divisible by 7?
Step 1: First 3-digit number divisible by 7 = 105, Last = 994
Step 2: AP: 105, 112, 119, ..., 994 with d = 7
Step 3: 994 = 105 + (n−1)×7
Step 4: 889 = (n−1)×7 → n−1 = 127 → n = 128
Answer: 128
14
How many multiples of 4 lie between 10 and 250?
Step 1: First multiple of 4 after 10 = 12, Last before 250 = 248
Step 2: AP: 12, 16, 20, ..., 248 with d = 4
Step 3: 248 = 12 + (n−1)×4
Step 4: 236 = (n−1)×4 → n−1 = 59 → n = 60
Answer: 60
15
For what value of n, are the nth terms of two APs: 63, 65, 67, ... and 3, 10, 17, ... equal?
Step 1: 1st AP: a = 63, d = 2 → aₙ = 63 + (n−1)×2 = 61 + 2n
Step 2: 2nd AP: a = 3, d = 7 → aₙ = 3 + (n−1)×7 = 7n − 4
Step 3: Set equal: 61 + 2n = 7n − 4
Step 4: 65 = 5n → n = 13
Answer: n = 13
16
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Step 1: a₃ = a + 2d = 16 ... (i)
Step 2: a₇ − a₅ = 12 → (a + 6d) − (a + 4d) = 12 → 2d = 12 → d = 6
Step 3: From (i): a + 12 = 16 → a = 4
Answer: AP: 4, 10, 16, 22, ...
17
Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.
Method: Reverse the AP. Last term becomes first.
Step 1: Reverse AP: 253, 248, ..., 13, 8, 3 with d = −5
Step 2: a₂₀ = 253 + (20−1)×(−5) = 253 − 95 = 158
Answer: 158
18
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 of the AP.
Step 1: a₄ + a₈ = (a+3d) + (a+7d) = 2a + 10d = 24 → a + 5d = 12 ... (i)
Step 2: a₆ + a₁₀ = (a+5d) + (a+9d) = 2a + 14d = 44 → a + 7d = 22 ... (ii)
Step 3: (ii) − (i): 2d = 10 → d = 5
Step 4: From (i): a + 25 = 12 → a = −13
Answer: First three terms: −13, −8, −3
19
Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
Step 1: AP: 5000, 5200, 5400, ... with a = 5000, d = 200
Step 2: Let aₙ = 7000
Step 3: 5000 + (n−1)×200 = 7000
Step 4: (n−1)×200 = 2000 → n−1 = 10 → n = 11
Step 5: Year = 1995 + 10 = 2005
Answer: Year 2005
20
Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.
Step 1: AP: 5, 6.75, 8.5, ... with a = 5, d = 1.75
Step 2: Let aₙ = 20.75
Step 3: 5 + (n−1)×1.75 = 20.75
Step 4: (n−1)×1.75 = 15.75 → n−1 = 9 → n = 10
Answer: n = 10
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Exercise 5.3

1
Find the sum of the following APs:
(i) 2, 7, 12, ..., to 10 terms.
Step 1: a = 2, d = 5, n = 10
Step 2: S₁₀ = 10/2 [2×2 + (10−1)×5] = 5[4 + 45] = 5×49 = 245
Answer: 245
(ii) −37, −33, −29, ..., to 12 terms.
Step 1: a = −37, d = 4, n = 12
Step 2: S₁₂ = 12/2 [2×(−37) + 11×4] = 6[−74 + 44] = 6×(−30) = −180
Answer: −180
(iii) 0.6, 1.7, 2.8, ..., to 100 terms.
Step 1: a = 0.6, d = 1.1, n = 100
Step 2: S₁₀₀ = 100/2 [2×0.6 + 99×1.1] = 50[1.2 + 108.9] = 50×110.1 = 5505
Answer: 5505
(iv) 1/15, 1/12, 1/10, ..., to 11 terms.
Step 1: a = 1/15, d = 1/12 − 1/15 = 1/60, n = 11
Step 2: S₁₁ = 11/2 [2/15 + 10/60] = 11/2 [2/15 + 1/6] = 11/2 × 9/30 = 33/20
Answer: 33/20 = 1.65
2
Find the sums given below:
(i) 7 + 10½ + 14 + ... + 84
Step 1: a = 7, d = 3.5, l = 84
Step 2: Find n: 84 = 7 + (n−1)×3.5 → 77 = (n−1)×3.5 → n = 23
Step 3: S₂₃ = 23/2 [7 + 84] = 23/2 × 91 = 1046.5
Answer: 1046.5 or 2093/2
(ii) 34 + 32 + 30 + ... + 10
Step 1: a = 34, d = −2, l = 10
Step 2: Find n: 10 = 34 + (n−1)×(−2) → n = 13
Step 3: S₁₃ = 13/2 [34 + 10] = 13/2 × 44 = 286
Answer: 286
(iii) −5 + (−8) + (−11) + ... + (−230)
Step 1: a = −5, d = −3, l = −230
Step 2: Find n: −230 = −5 + (n−1)×(−3) → n = 76
Step 3: S₇₆ = 76/2 [(−5) + (−230)] = 38×(−235) = −8930
Answer: −8930
3
In an AP:
(i) Given a = 5, d = 3, aₙ = 50, find n and Sₙ.
Step 1: 50 = 5 + (n−1)×3 → 45 = 3(n−1) → n = 16
Step 2: S₁₆ = 16/2 [5 + 50] = 8×55 = 440
Answer: n = 16, Sₙ = 440
(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
Step 1: 35 = 7 + 12d → d = 28/12 = 7/3
Step 2: S₁₃ = 13/2 [7 + 35] = 13/2 × 42 = 273
Answer: d = 7/3, S₁₃ = 273
(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.
Step 1: 37 = a + 11×3 → a = 37 − 33 = 4
Step 2: S₁₂ = 12/2 [2×4 + 11×3] = 6[8 + 33] = 246
Answer: a = 4, S₁₂ = 246
(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
Step 1: a + 2d = 15 ... (i)
Step 2: S₁₀ = 10/2 [2a + 9d] = 125 → 2a + 9d = 25 ... (ii)
Step 3: From (i): a = 15 − 2d. Substitute: 2(15−2d) + 9d = 25 → d = −1
Step 4: a = 17, a₁₀ = 17 + 9×(−1) = 8
Answer: d = −1, a₁₀ = 8
(v) Given d = 5, S₉ = 75, find a and a₉.
Step 1: S₉ = 9/2 [2a + 8×5] = 75 → 9(a + 20) = 75 → a = 25/3 − 20 = −35/3
Step 2: a₉ = −35/3 + 8×5 = 85/3
Answer: a = −35/3, a₉ = 85/3
(vi) Given a = 2, d = 8, Sₙ = 90, find n and aₙ.
Step 1: 90 = n/2 [4 + (n−1)×8] = n(4n − 2) = 4n² − 2n
Step 2: 4n² − 2n − 90 = 0 → 2n² − n − 45 = 0 → (2n+9)(n−5) = 0 → n = 5
Step 3: a₅ = 2 + 4×8 = 34
Answer: n = 5, aₙ = 34
(vii) Given a = 8, aₙ = 62, Sₙ = 210, find n and d.
Step 1: Sₙ = n/2 [8 + 62] = 210 → 35n = 210 → n = 6
Step 2: 62 = 8 + 5d → d = 54/5 = 10.8
Answer: n = 6, d = 54/5 = 10.8
(viii) Given aₙ = 4, d = 2, Sₙ = −14, find n and a.
Step 1: a + (n−1)×2 = 4 → a = 6 − 2n
Step 2: Sₙ = n/2 [a + 4] = −14 → n(10 − 2n) = −28 → n² − 5n − 14 = 0 → n = 7
Step 3: a = 6 − 14 = −8
Answer: n = 7, a = −8
(ix) Given a = 3, n = 8, S = 192, find d.
Step 1: 192 = 8/2 [6 + 7d] = 4(6 + 7d)
Step 2: 48 = 6 + 7d → d = 6
Answer: d = 6
(x) Given l = 28, S = 144, and there are total 9 terms. Find a.
Step 1: S₉ = 9/2 [a + 28] = 144
Step 2: 9(a + 28) = 288 → a = 4
Answer: a = 4
4
How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
Step 1: a = 9, d = 8
Step 2: 636 = n/2 [18 + (n−1)×8] = n(4n + 5)
Step 3: 4n² + 5n − 636 = 0 → n = 12
Answer: n = 12
5
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Step 1: Sₙ = n/2 [5 + 45] = 400 → 25n = 400 → n = 16
Step 2: 45 = 5 + 15d → d = 40/15 = 8/3
Answer: n = 16, d = 8/3
6
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Step 1: 350 = 17 + (n−1)×9 → 333 = 9(n−1) → n = 38
Step 2: S₃₈ = 38/2 [17 + 350] = 19×367 = 6973
Answer: n = 38, S = 6973
7
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Step 1: a₂₂ = a + 21×7 = 149 → a = 2
Step 2: S₂₂ = 22/2 [2 + 149] = 11×151 = 1661
Answer: 1661
8
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Step 1: d = 18 − 14 = 4
Step 2: a₂ = a + d = 14 → a = 10
Step 3: S₅₁ = 51/2 [20 + 50×4] = 51×110 = 5610
Answer: 5610
9
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 1: S₇ = 7/2 [2a + 6d] = 49 → a + 3d = 7 ... (i)
Step 2: S₁₇ = 17/2 [2a + 16d] = 289 → a + 8d = 17 ... (ii)
Step 3: (ii) − (i): 5d = 10 → d = 2
Step 4: From (i): a = 1
Step 5: Sₙ = n/2 [2 + (n−1)×2] = n²
Answer: Sₙ = n²
10
Show that a₁, a₂, ..., aₙ, ... form an AP where aₙ is defined as below. Also find the sum of the first 15 terms in each case.
(i) aₙ = 3 + 4n
Step 1: a₁ = 7, a₂ = 11, a₃ = 15
Step 2: d = 11 − 7 = 4 (constant)
Step 3: S₁₅ = 15/2 [14 + 14×4] = 15/2 × 70 = 525
Answer: AP with d = 4, S₁₅ = 525
(ii) aₙ = 9 − 5n
Step 1: a₁ = 4, a₂ = −1, a₃ = −6
Step 2: d = −1 − 4 = −5 (constant)
Step 3: S₁₅ = 15/2 [8 + 14×(−5)] = 15/2 × (−62) = −465
Answer: AP with d = −5, S₁₅ = −465
11
If the sum of the first n terms of an AP is 4n − n², what is the first term? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Step 1: S₁ = 4(1) − 1 = 3 → First term = 3
Step 2: S₂ = 8 − 4 = 4 → Sum of first two terms = 4
Step 3: a₂ = S₂ − S₁ = 4 − 3 = 1
Step 4: S₃ = 12 − 9 = 3 → a₃ = 3 − 4 = −1
Step 5: d = 1 − 3 = −2
Step 6: a₁₀ = 3 + 9×(−2) = −15
Step 7: aₙ = 3 + (n−1)×(−2) = 5 − 2n
Answer: First term = 3, S₂ = 4, a₂ = 1, a₃ = −1, a₁₀ = −15, aₙ = 5 − 2n
12
Find the sum of the first 40 positive integers divisible by 6.
Step 1: AP: 6, 12, 18, ... with a = 6, d = 6, n = 40
Step 2: S₄₀ = 40/2 [12 + 39×6] = 20 × 246 = 4920
Answer: 4920
13
Find the sum of the first 15 multiples of 8.
Step 1: AP: 8, 16, 24, ... with a = 8, d = 8, n = 15
Step 2: S₁₅ = 15/2 [16 + 14×8] = 15/2 × 128 = 960
Answer: 960
14
Find the sum of the odd numbers between 0 and 50.
Step 1: AP: 1, 3, 5, ..., 49 with a = 1, d = 2, l = 49
Step 2: 49 = 1 + (n−1)×2 → n = 25
Step 3: S₂₅ = 25/2 [1 + 49] = 25/2 × 50 = 625
Answer: 625
15
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Step 1: AP: 200, 250, 300, ... with a = 200, d = 50, n = 30
Step 2: S₃₀ = 30/2 [400 + 29×50] = 15 × 1850 = 27750
Answer: ₹27,750
16
A sum of ₹700 is to be used to give seven cash prizes to students. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
Step 1: n = 7, d = −20, S₇ = 700
Step 2: 700 = 7/2 [2a + 6×(−20)] = 7(a − 60)
Step 3: 100 = a − 60 → a = 160
Answer: Prizes: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40
17
In a school, students thought of planting trees. It was decided that the number of trees that each section of each class will plant will be the same as the class in which they are studying (e.g., Class I plants 1 tree, Class II plants 2 trees, etc. till Class XII). There are three sections of each class. How many trees will be planted?
Step 1: Trees per class: 1, 2, 3, ..., 12 (3 sections each)
Step 2: Sum for one section: S₁₂ = 12/2 [1 + 12] = 78
Step 3: Total trees = 78 × 3 = 234
Answer: 234 trees
18
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
Step 1: Length of semicircle = πr
Step 2: Lengths: 0.5π, 1.0π, 1.5π, ... (13 terms)
Step 3: a = 0.5π, d = 0.5π, n = 13
Step 4: S₁₃ = 13/2 [π + 12×0.5π] = 13/2 × 7π = 91π/2 = 91/2 × 22/7 = 143
Answer: 143 cm
19
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
Step 1: AP: 20, 19, 18, ... with a = 20, d = −1, Sₙ = 200
Step 2: 200 = n/2 [40 + (n−1)×(−1)] = n/2 [41 − n]
Step 3: 400 = 41n − n² → n² − 41n + 400 = 0 → (n−16)(n−25) = 0
Step 4: For n = 25: a₂₅ = 20 + 24×(−1) = −4 (invalid)
Step 5: For n = 16: a₁₆ = 20 + 15×(−1) = 5
Answer: 16 rows, 5 logs in top row
20
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, and so on. What is the total distance the competitor has to run?
Step 1: Distance to 1st potato and back = 2×5 = 10 m
Step 2: Distance to 2nd potato and back = 2×(5+3) = 16 m
Step 3: Distances form AP: 10, 16, 22, ... (10 terms) with a = 10, d = 6
Step 4: S₁₀ = 10/2 [20 + 9×6] = 5 × 74 = 370
Answer: 370 metres
📍 Quick Jump: Notes Ex 5.1 Ex 5.2 Ex 5.3 Ex 5.4 Calculators ⬆ Top

Exercise 5.4 Optional

1
Which term of the AP: 121, 117, 113, ... is its first negative term?
Step 1: a = 121, d = −4
Step 2: Let aₙ < 0: 121 + (n−1)×(−4) < 0
Step 3: 121 − 4n + 4 < 0 → 125 < 4n → n > 31.25
Step 4: Smallest integer n = 32
Answer: 32nd term is the first negative term
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.
Step 1: a₃ + a₇ = (a+2d) + (a+6d) = 2a + 8d = 6 → a + 4d = 3
Step 2: a₃ × a₇ = (a+2d)(a+6d) = 8
Step 3: From (i): a = 3 − 4d. Substitute: (3−2d)(3+2d) = 8 → 9 − 4d² = 8 → d² = 1/4 → d = ±1/2
Case 1: d = 1/2 → a = 1 → S₁₆ = 16/2 [2 + 15×1/2] = 8 × 9.5 = 76
Case 2: d = −1/2 → a = 5 → S₁₆ = 16/2 [10 + 15×(−1/2)] = 8 × 2.5 = 20
Answer: S₁₆ = 76 or 20
3
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 1: Distance = 250 cm, gap = 25 cm → Number of rungs = 250/25 + 1 = 11
Step 2: AP: 45, ..., 25 with n = 11
Step 3: S₁₁ = 11/2 [45 + 25] = 11/2 × 70 = 385
Answer: 385 cm of wood required
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.
Step 1: Sum of houses before x: Sₓ₋₁ = (x−1)x/2
Step 2: Sum of houses after x: S₄₉ − Sₓ = 49×50/2 − x(x+1)/2 = 1225 − x(x+1)/2
Step 3: Set equal: (x−1)x/2 = 1225 − x(x+1)/2
Step 4: x² − x = 2450 − x² − x → 2x² = 2450 → x² = 1225 → x = 35
Answer: x = 35
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 ¼ m and a tread of ½ m. Calculate the total volume of concrete required to build the terrace.
Step 1: Volume of 1st step = 1/4 × 1/2 × 50 = 25/4 m³
Step 2: Volume of 2nd step = 2/4 × 1/2 × 50 = 50/4 m³
Step 3: Volumes form AP: 25/4, 50/4, 75/4, ... (15 terms) with a = 25/4, d = 25/4
Step 4: S₁₅ = 15/2 [25/4 + 15×25/4] = 15/2 × 400/4 = 15/2 × 100 = 750
Answer: 750 m³
📍 Quick Jump: Notes Ex 5.1 Ex 5.2 Ex 5.3 Ex 5.4 Calculators ⬆ Top

🧮 AP Calculators

📐 Find nth Term

➕ Sum of n Terms

📊 Find Common Difference (d)

🔢 Find Number of Terms

🏁 Find First Term (a)

➕ Sum (Given First & Last Term)