Arithmetic Progressions — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Arithmetic Progressions" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Arithmetic Progressions" — 9 important questions with detailed answers for CBSE bo…
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NCERT Solutions for Class 10 Mathematics — Arithmetic Progressions. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Arithmetic Progressions — All 9 Questions Solved
Q1. Define an arithmetic progression and state the conditions for a sequence to be an AP.
Arithmetic Progression (AP):
An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant.
General form: a, a + d, a + 2d, a + 3d, ...
where a = first term
d = common difference (constant)
Condition for AP:
A sequence with terms t₁, t₂, t₃, ... is an AP if and only if:
tₙ₊₁ - tₙ = d (constant) for all positive integers n
Or equivalently: t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = ... = d
Examples of AP:
1) 2, 5, 8, 11, 14, ... (a = 2, d = 3)
2) 10, 7, 4, 1, -2, ... (a = 10, d = -3)
3) 3, 3, 3, 3, ... (a = 3, d = 0) — constant sequence
Examples NOT in AP:
1) 1, 2, 4, 8, ... (not AP; this is a geometric progression)
2) 1, 4, 9, 16, ... (not AP; these are perfect squares)
Key formulas:
- nth term: tₙ = a + (n - 1)d
- Common difference: d = tₙ - tₙ₋₁
- Sum of n terms: Sₙ = n/2 × (2a + (n - 1)d) or Sₙ = n/2 × (first term + last term)
Q2. Find the 15th term of the AP: 4, 7, 10, 13, ...
Given AP: 4, 7, 10, 13, ...
Step 1: Identify the parameters
First term: a = 4
Common difference: d = 7 - 4 = 3
(Verify: 10 - 7 = 3, 13 - 10 = 3 ✓)
We need to find the 15th term, so n = 15.
Step 2: Use the formula for nth term
tₙ = a + (n - 1)d
t₁₅ = 4 + (15 - 1) × 3
t₁₅ = 4 + 14 × 3
t₁₅ = 4 + 42
t₁₅ = 46
Step 3: Verification
First few terms:
t₁ = 4
t₂ = 7
t₃ = 10
Following the pattern, each term increases by 3.
From t₁ to t₁₅, we add 3 fourteen times: 4 + 14(3) = 46 ✓
Therefore, the 15th term is 46
Q3. The sum of the first 20 terms of an AP is 650 and the common difference is 3. Find the first term.
Given:
Sum of first 20 terms: S₂₀ = 650
Common difference: d = 3
Number of terms: n = 20
Find: First term a = ?
Step 1: Use the sum formula for AP
Sₙ = n/2 × (2a + (n - 1)d)
Substitute the known values:
650 = 20/2 × (2a + (20 - 1) × 3)
650 = 10 × (2a + 19 × 3)
650 = 10 × (2a + 57)
Step 2: Solve for a
650 = 20a + 570
20a = 650 - 570
20a = 80
a = 4
Step 3: Verification
With a = 4 and d = 3:
S₂₀ = 20/2 × (2(4) + (20 - 1) × 3)
= 10 × (8 + 57)
= 10 × 65
= 650 ✓
Therefore, the first term is a = 4
Q4. Is 100 a term of the AP: 5, 8, 11, 14, ...? If yes, find which term it is.
Given AP: 5, 8, 11, 14, ...
Step 1: Identify the parameters
First term: a = 5
Common difference: d = 8 - 5 = 3
Step 2: Assume 100 is the nth term and find n
Using tₙ = a + (n - 1)d:
100 = 5 + (n - 1) × 3
100 = 5 + 3n - 3
100 = 2 + 3n
3n = 98
n = 98/3 = 32.67...
Since n is not a positive integer, 100 is not a term of this AP.
Alternative verification:
For a number to be in an AP with a = 5 and d = 3:
term = 5 + 3k (where k ≥ 0 is an integer)
If 100 is a term: 100 = 5 + 3k
3k = 95
k = 95/3 = 31.67...
Since k is not an integer, 100 is not a term of this AP.
Note: The terms of this AP that are close to 100 are:
t₃₂ = 5 + (32 - 1) × 3 = 5 + 93 = 98
t₃₃ = 5 + (33 - 1) × 3 = 5 + 96 = 101
Therefore, 100 is NOT a term of the given AP
Q5. Find the sum of all three-digit numbers divisible by 7.
We need to find the sum of all three-digit numbers divisible by 7.
Step 1: Find the first three-digit number divisible by 7
Smallest three-digit number: 100
100 ÷ 7 = 14.28...
First three-digit number divisible by 7: 7 × 15 = 105
Step 2: Find the last three-digit number divisible by 7
Largest three-digit number: 999
999 ÷ 7 = 142.71...
Last three-digit number divisible by 7: 7 × 142 = 994
Step 3: Identify the AP
The AP is: 105, 112, 119, ..., 994
First term: a = 105
Common difference: d = 7
Last term: l = 994
Step 4: Find the number of terms
Using tₙ = a + (n - 1)d:
994 = 105 + (n - 1) × 7
994 = 105 + 7n - 7
994 = 98 + 7n
7n = 896
n = 128
Step 5: Find the sum
Using Sₙ = n/2 × (first term + last term):
S₁₂₈ = 128/2 × (105 + 994)
S₁₂₈ = 64 × 1099
S₁₂₈ = 70,336
Therefore, the sum of all three-digit numbers divisible by 7 is 70,336
Q6. In an AP, the 7th term is 35 and the 15th term is 91. Find the first term and the common difference.
Given:
t₇ = 35
t₁₅ = 91
Find: a and d
Step 1: Set up equations using tₙ = a + (n - 1)d
For the 7th term: t₇ = a + 6d = 35 ... (1)
For the 15th term: t₁₅ = a + 14d = 91 ... (2)
Step 2: Subtract equation (1) from equation (2)
(a + 14d) - (a + 6d) = 91 - 35
8d = 56
d = 7
Step 3: Find a using equation (1)
a + 6(7) = 35
a + 42 = 35
a = -7
Step 4: Verification
For t₇: t₇ = -7 + (7 - 1) × 7 = -7 + 42 = 35 ✓
For t₁₅: t₁₅ = -7 + (15 - 1) × 7 = -7 + 98 = 91 ✓
Therefore, the first term is a = -7 and the common difference is d = 7
Q7. If the sum of the first n terms of an AP is Sₙ = 3n² + 2n, find the AP.
Given: Sₙ = 3n² + 2n
Find: The AP
Step 1: Find the first term (n = 1)
S₁ = a₁ = 3(1)² + 2(1) = 3 + 2 = 5
First term: a = 5
Step 2: Find the second term
S₂ = a₁ + a₂ = 3(2)² + 2(2) = 12 + 4 = 16
a₁ + a₂ = 16
5 + a₂ = 16
a₂ = 11
Step 3: Find the common difference
d = a₂ - a₁ = 11 - 5 = 6
Step 4: Find the general term
tₙ = a + (n - 1)d = 5 + (n - 1) × 6 = 5 + 6n - 6 = 6n - 1
Alternative method for n ≥ 2:
tₙ = Sₙ - Sₙ₋₁
= (3n² + 2n) - (3(n-1)² + 2(n-1))
= 3n² + 2n - (3(n² - 2n + 1) + 2n - 2)
= 3n² + 2n - 3n² + 6n - 3 - 2n + 2
= 6n - 1
For n = 1: t₁ = 6(1) - 1 = 5 ✓
Step 5: Verify with S₂
S₂ = t₁ + t₂ = 5 + 11 = 16 ✓ (matches 3(2)² + 2(2) = 16)
Therefore, the AP is: 5, 11, 17, 23, ... with general term tₙ = 6n - 1
Q8. The sum of three numbers in AP is 24 and their product is 480. Find the three numbers.
Let the three numbers in AP be (a - d), a, and (a + d).
Step 1: Use the sum condition
(a - d) + a + (a + d) = 24
3a = 24
a = 8
So the numbers are (8 - d), 8, and (8 + d).
Step 2: Use the product condition
(8 - d) × 8 × (8 + d) = 480
8(8 - d)(8 + d) = 480
8(64 - d²) = 480
64 - d² = 60
d² = 4
d = ±2
Step 3: Find the three numbers
If d = 2: numbers are (8 - 2), 8, (8 + 2) = 6, 8, 10
If d = -2: numbers are (8 + 2), 8, (8 - 2) = 10, 8, 6 (same set)
Step 4: Verification
Sum: 6 + 8 + 10 = 24 ✓
Product: 6 × 8 × 10 = 480 ✓
AP check: 8 - 6 = 2 and 10 - 8 = 2 ✓
Therefore, the three numbers are 6, 8, and 10
Q9. How many two-digit numbers are divisible by 7?
We need to find all two-digit numbers divisible by 7.
Step 1: Find the first two-digit number divisible by 7
Smallest two-digit number: 10
10 ÷ 7 = 1.43...
First two-digit number divisible by 7: 7 × 2 = 14
Step 2: Find the last two-digit number divisible by 7
Largest two-digit number: 99
99 ÷ 7 = 14.14...
Last two-digit number divisible by 7: 7 × 14 = 98
Step 3: Identify the AP
The two-digit numbers divisible by 7 form an AP: 14, 21, 28, ..., 98
First term: a = 14
Common difference: d = 7
Last term: l = 98
Step 4: Find the number of terms
Using tₙ = a + (n - 1)d:
98 = 14 + (n - 1) × 7
98 = 14 + 7n - 7
98 = 7 + 7n
7n = 91
n = 13
Alternative method:
The two-digit multiples of 7 are: 7 × 2, 7 × 3, 7 × 4, ..., 7 × 14
Count: 14 - 2 + 1 = 13
Therefore, there are 13 two-digit numbers divisible by 7
Arithmetic Progressions — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the common difference of the AP: 2, -2, -6, -10, ...?
- A. 4
- B. -4
- C. -2
- D. 2
Answer: B. -4 — The common difference (d) is found by subtracting any term from its succeeding term. Here, -2 - 2 = -4, and -6 - (-2) = -4.
MCQ 2. The first term of an AP is 5 and the common difference is 3. What is its 7th term?
- A. 20
- B. 23
- C. 26
- D. 29
Answer: B. 23 — The nth term of an AP is given by a_n = a + (n-1)d. Substituting a=5, d=3, n=7, we get a_7 = 5 + (7-1)3 = 5 + 6*3 = 5 + 18 = 23.
MCQ 3. Which of the following sequences is an Arithmetic Progression?
- A. 1, 3, 6, 10, ...
- B. 2, 4, 8, 16, ...
- C. -5, -3, -1, 1, ...
- D. 1/2, 1/3, 1/4, 1/5, ...
Answer: C. -5, -3, -1, 1, ... — An Arithmetic Progression has a constant common difference between consecutive terms. In option C, the common difference is -3 - (-5) = 2, -1 - (-3) = 2, and 1 - (-1) = 2.
MCQ 4. For the AP: 3, 8, 13, 18, ..., which term is 78?
- A. 15th
- B. 16th
- C. 17th
- D. 18th
Answer: B. 16th — Here, a=3, d=8-3=5. We need to find n such that a_n = 78. Using a_n = a + (n-1)d, we have 78 = 3 + (n-1)5, which gives n = 16.
MCQ 5. What is the sum of the first 10 terms of the AP: 1, 3, 5, 7, ...?
- A. 90
- B. 100
- C. 110
- D. 120
Answer: B. 100 — Here, a=1, d=2, n=10. Using S_n = n/2 * [2a + (n-1)d], we get S_10 = 10/2 * [2*1 + (10-1)2] = 5 * [2 + 18] = 5 * 20 = 100.
MCQ 6. If x, 13, y, 23 are in AP, what are the values of x and y?
- A. x=8, y=18
- B. x=3, y=18
- C. x=8, y=28
- D. x=3, y=28
Answer: A. x=8, y=18 — Since the terms are in AP, the common difference is constant. From y - 13 = 23 - y, we get 2y = 36, so y = 18. Then, 13 - x = y - 13 = 18 - 13 = 5, which means x = 8.
MCQ 7. A worker is offered an initial salary of Rs 5000 with an annual increment of Rs 500. What will be his salary in the 5th year?
- A. Rs 6500
- B. Rs 7000
- C. Rs 7500
- D. Rs 8000
Answer: B. Rs 7000 — This forms an AP with a = 5000 and d = 500. The salary in the 5th year is the 5th term a_5 = a + (5-1)d = 5000 + 4*500 = 7000.
MCQ 8. If the first term of an AP is 7 and the 13th term is 31, find the common difference.
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — We have a = 7 and a_13 = 31. Using a_n = a + (n-1)d, we get 31 = 7 + (13-1)d, which simplifies to 24 = 12d, so d = 2.
MCQ 9. The first term of an AP is 6 and the common difference is 4. What is the sum of its first 8 terms?
- A. 152
- B. 160
- C. 176
- D. 184
Answer: B. 160 — Given a=6, d=4, n=8. Using S_n = n/2 * [2a + (n-1)d], S_8 = 8/2 * [2*6 + (8-1)4] = 4 * [12 + 28] = 4 * 40 = 160.
MCQ 10. If k+1, 2k+1, 3k+1 are in AP, what is the common difference?
- A. k
- B. 1
- C. k+1
- D. 2k
Answer: A. k — For terms to be in AP, the common difference must be constant. d = (2k+1) - (k+1) = k. Also, d = (3k+1) - (2k+1) = k.
MCQ 11. The 4th term of an AP is 10 and the 10th term is 28. Find the first term and common difference.
- A. a=1, d=3
- B. a=2, d=3
- C. a=1, d=2
- D. a=2, d=2
Answer: A. a=1, d=3 — We have a_4 = a + 3d = 10 and a_10 = a + 9d = 28. Subtracting the first from the second gives 6d = 18, so d = 3. Substituting d=3 into a + 3d = 10 gives a + 9 = 10, so a = 1.
MCQ 12. Find the sum of all odd numbers between 0 and 50.
- A. 625
- B. 650
- C. 675
- D. 700
Answer: A. 625 — The odd numbers are 1, 3, 5, ..., 49. This is an AP with a=1, d=2. There are 25 terms (n=25). Using S_n = n/2 * (a + a_n), S_25 = 25/2 * (1 + 49) = 25/2 * 50 = 625.
MCQ 13. How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
- A. 10
- B. 12
- C. 14
- D. 16
Answer: B. 12 — Here, a=9, d=8, S_n = 636. Using S_n = n/2 * [2a + (n-1)d], we get 636 = n/2 * [18 + (n-1)8]. This simplifies to 4n^2 + 5n - 636 = 0. Solving for n gives n=12.
MCQ 14. If the sum of the first n terms of an AP is S_n = 3n^2 + 5n, then what is its nth term a_n?
- A. 6n+2
- B. 6n+8
- C. 3n+2
- D. 3n+8
Answer: A. 6n+2 — We know that a_n = S_n - S_{n-1}. Calculating S_{n-1} = 3(n-1)^2 + 5(n-1) = 3n^2 - n - 2. So, a_n = (3n^2 + 5n) - (3n^2 - n - 2) = 6n + 2.
MCQ 15. If three numbers a-d, a, a+d are in AP, and their sum is 27 and product is 720, find the numbers.
- A. 7, 9, 11
- B. 6, 9, 12
- C. 8, 9, 10
- D. 5, 9, 13
Answer: C. 8, 9, 10 — The sum is (a-d) + a + (a+d) = 3a. Given sum = 27, so 3a = 27, which means a = 9. The product is (9-d)*9*(9+d) = 720. This simplifies to 9(81 - d^2) = 720, so 81 - d^2 = 80, which gives d^2 = 1, so d = ±1. The numbers are 8, 9, 10 (or 10, 9, 8).
MCQ 16. Find the sum of the AP: 10, 15, 20, ..., 100.
- A. 1045
- B. 1075
- C. 1100
- D. 1125
Answer: A. 1045 — Here, a=10, d=5, a_n=100. First find n: 100 = 10 + (n-1)5, which gives n=19. Now use S_n = n/2 * (a + a_n): S_19 = 19/2 * (10 + 100) = 19/2 * 110 = 19 * 55 = 1045.
MCQ 17. The sum of the first n natural numbers is 300. Find n.
- A. 20
- B. 24
- C. 25
- D. 28
Answer: B. 24 — The sum of the first n natural numbers is S_n = n(n+1)/2. Given S_n = 300, so n(n+1)/2 = 300, which means n(n+1) = 600. Factoring gives (n+25)(n-24) = 0. Since n must be positive, n = 24.
MCQ 18. If the nth term of an AP is given by a_n = 5n - 3, find its common difference.
- A. 2
- B. 3
- C. 5
- D. -3
Answer: C. 5 — The common difference d can be found by a_2 - a_1. a_1 = 5(1) - 3 = 2. a_2 = 5(2) - 3 = 7. So, d = 7 - 2 = 5. Alternatively, for a_n = An + B, the common difference is A.
MCQ 19. In an AP, S_5 = 100 and S_6 = 120. Find the 6th term.
- A. 10
- B. 15
- C. 20
- D. 25
Answer: C. 20 — The nth term a_n can be found using the sum of terms: a_n = S_n - S_{n-1}. So, a_6 = S_6 - S_5 = 120 - 100 = 20.
MCQ 20. A sum of Rs 1000 is invested at 8% simple interest per year. Calculate the total interest at the end of each year for 10 years. Do these interests form an AP? If so, find the sum of interests for 10 years.
- A. No, it's not an AP
- B. Yes, sum = Rs 4000
- C. Yes, sum = Rs 4400
- D. Yes, sum = Rs 4800
Answer: C. Yes, sum = Rs 4400 — Simple Interest per year is (1000 * 8 * 1) / 100 = Rs 80. The total interest accumulated at the end of each year forms an AP: 80, 160, 240, ... with a=80, d=80. The sum of these 10 terms is S_10 = 10/2 * [2*80 + (10-1)80] = 5 * [160 + 720] = 5 * 880 = Rs 4400.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.