NCERT Solutions for Class 10 Mathematics Chapter 5 Arithmetic Progressions

This worksheet focuses on Class 10 Mathematics Chapter Arithmetic Progressions and helps students understand patterns in numbers where the difference between consecutive terms remains constant. Arithmetic progressions are an important part of algebra and are widely used in solving mathematical sequences, finding unknown terms, and calculating sums of number patterns. This worksheet provides clear and structured practice problems based on NCERT standards. This worksheet includes complete and accurate NCERT Solutions that help students verify their answers and understand the correct steps involved in solving progression problems.

Chapter summary: 

This worksheet is based on mathematical concepts. The focus of the chapter is on understanding number sequences that follow a constant difference pattern. Students learn how to identify arithmetic progressions, determine the first term and common difference, find the nth term of a sequence, and calculate the sum of terms in a progression. The worksheet also includes problem-solving questions where arithmetic progressions are applied to real-life situations such as distances, prizes, and numerical patterns. The chapter is concept-based and problem-solving oriented rather than narrative or activity-based.

What this NCERT chapter covers?

• Understanding arithmetic progressions and identifying sequences with constant differences 
• Finding the first term and common difference of an arithmetic progression 
• Calculating the nth term of an arithmetic progression 
• Determining missing terms within a sequence 
• Calculating the sum of terms in an arithmetic progression 
• Applying arithmetic progression concepts to practical mathematical problems 

How to use these NCERT solutions?

Students should first attempt all the questions in this worksheet independently before referring to the solutions. After completing the questions, they can use the provided answers to check their work and identify any mistakes. Parents and teachers can guide students in reviewing each solution step carefully so that the method of solving becomes clear. The solutions follow the same order as the worksheet and are aligned with NCERT structure, making them useful for revision and concept reinforcement.

Student tips & learning tricks

• Always check whether the difference between consecutive terms is constant before identifying an arithmetic progression 
• Carefully substitute values when applying formulas for nth term or sum of terms 
• Pay attention to negative numbers and fractions while performing calculations 
• Write each step clearly while solving problems to avoid mistakes 
• Practice identifying sequences and calculating terms regularly to improve speed and accuracy 

Why NCERT solutions are important?

NCERT Solutions help students understand the correct methods for solving mathematical problems as expected in school examinations. They follow the official NCERT approach and provide accurate answers aligned with the curriculum. Using these solutions allows students to strengthen their conceptual understanding, improve problem-solving confidence, and prepare effectively for tests and assessments.

Complete answer key – NCERT solutions

Exercise No. 5.1

1. 
(i) Taxi fare: 
Fare = 15, 23, 31, 39, ... 
Difference = 8 (constant) 
→ Yes, it is an AP 

(ii) Air removal: 
Each time multiplied by 3/4 
→ Not constant difference 
→ Not an AP 

(iii) Cost of digging: 
150, 200, 250, ... 
Difference = 50 
→ Yes, AP 

(iv) Compound interest: 
Amount increases multiplicatively 
→ Not constant difference 
→ Not an AP 

2. 
(i) a = 10, d = 10 
First four terms: 
10 
10 + 10 = 20 
20 + 10 = 30 
30 + 10 = 40 
Answer: 10, 20, 30, 40 

(ii) a = –2, d = 0 
First four terms: 
–2 
–2 + 0 = –2 
–2 + 0 = –2 
–2 + 0 = –2 
Answer: –2, –2, –2, –2 

(iii) a = 4, d = –3 
First four terms: 
4 
4 – 3 = 1 
1 – 3 = –2 
–2 – 3 = –5 
Answer: 4, 1, –2, –5 

(iv) a = –1, d = 1/2 
First four terms: 
–1 
–1 + 1/2 = –1/2 
–1/2 + 1/2 = 0 
0 + 1/2 = 1/2 
Answer: –1, –1/2, 0, 1/2 

(v) a = –1.25, d = –0.25 
First four terms: 
–1.25 
–1.25 – 0.25 = –1.5 
–1.5 – 0.25 = –1.75 
–1.75 – 0.25 = –2.0 
Answer: –1.25, –1.5, –1.75, –2.0 

3. 
(i) 3, 1, –1, –3 
a = first term = 3 
d = 1 – 3 = –2 
Answer: a = 3, d = –2 

(ii) –5, –1, 3, 7 
a = –5 
d = –1 – (–5) = 4 
Answer: a = –5, d = 4 

(iii) 1/3, 5/3, 9/3, 13/3 
a = 1/3 
d = 5/3 – 1/3 = 4/3 
Answer: a = 1/3, d = 4/3 

(iv) 0.6, 1.7, 2.8, 3.9 
a = 0.6 
d = 1.7 – 0.6 = 1.1 
Answer: a = 0.6, d = 1.1 

4. 
(i) 2, 4, 8, 16 
Differences: 
4 – 2 = 2 
8 – 4 = 4 
16 – 8 = 8 
Differences not same → Not an AP 

(ii) Given: 5/2, 3, 7/2, 4 
a₂ – a₁ = 3 – 5/2 = 1/2 
a₃ – a₂ = 7/2 – 3 = 1/2 
a₄ – a₃ = 4 – 7/2 = 1/2 
Since the difference is same, it is an AP with d = 1/2 
Next three terms: 9/2, 5, 11/2 

(iii) –1.2, –3.2, –5.2, –7.2 
Differences: 
–3.2 – (–1.2) = –2 
–5.2 – (–3.2) = –2 
–7.2 – (–5.2) = –2 
Same difference → AP, d = –2 
Next terms: –9.2, –11.2, –13.2 

(iv) –10, –6, –2, 2 
Differences: 
–6 – (–10) = 4 
–2 – (–6) = 4 
2 – (–2) = 4 
Same difference → AP, d = 4 
Next terms: 6, 10, 14 

(v) 3, 3+√2, 3+2√2, 3+3√2 
Differences: 
(3+√2) – 3 = √2 
(3+2√2) – (3+√2) = √2 
Same → AP 
d = √2 
Next terms: 3+4√2, 3+5√2, 3+6√2 

Exercise No. 5.2

1. 
(i) a = 7, d = 3, n = 8 
aₙ = a + (n – 1)d 
= 7 + (8 – 1)×3 = 7 + 21 = 28 

(ii) a = –18, n = 10, aₙ = 0 
0 = –18 + (10 – 1)d 
0 = –18 + 9d → 9d = 18 → d = 2 

(iii) d = –3, n = 18, aₙ = –5 
–5 = a + (18 – 1)(–3) 
–5 = a – 51 → a = 46 

(iv) a = –18.9, d = 2.5, aₙ = 3.6 
3.6 = –18.9 + (n – 1)×2.5 
3.6 + 18.9 = (n – 1)×2.5 
22.5 = (n – 1)×2.5 → n – 1 = 9 → n = 10 

(v) a = 3.5, d = 0, n = 105 
aₙ = a + (n – 1)d = 3.5 + 0 = 3.5 

2. 
(i) a = 10, d = –3 
aₙ = 10 + (30 – 1)(–3) = 10 – 87 = –77 → (C) 

(ii) a = –3, d = 5/2 
a₁₁ = –3 + (11 – 1)(5/2) 
= –3 + 10×5/2 = –3 + 25 = 22 → (B) 

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