RD Sharma Class 10 Chapter 9 Solutions focuses on this concept in a simple and clear way. Whether you're a student preparing for board exams or someone aiming to strengthen your basics, this chapter helps build a strong foundation.
The NCERT Solutions for Class 10 maths RD Sharma book explains how numbers in a sequence follow a pattern and how you can find missing terms, the sum of terms, and common differences. The RD Sharma Class 10 Chapter 9 Solutions are made to match the latest CBSE Class 10 maths syllabus 2025-26. They help students solve questions step by step using the easiest method.
These RD Sharma Class 10 Solutions 2025-26 are useful for daily homework, exam revision, and better practice. If you're looking for easy and accurate help, the RD Sharma Arithmetic Progressions Class 10 Solutions are perfect for you. You can also get the RD Sharma Class 10 Arithmetic Progressions PDF download to study offline anytime.
RD Sharma Class 10 Chapter 9 – Arithmetic Progressions PDF with Solutions
Arithmetic Progressions is a important topic in Class 10 Maths. The RD Sharma Class 10 Arithmetic Progressions PDF helps students understand this chapter in easy steps. With the help of RD Sharma Class 10 Solutions 2025-26 PDF, you can practice all questions and get better marks in exams.
This RD Sharma Class 10 Arithmetic Progressions PDF download covers solved examples, exercises, and formulas that are very usefull for revision. Many students find it hard to solve AP questions, but with this PDF, the problems becomes simple and fun.
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Important Questions RD Sharma Class 10 Chapter 9 – Arithmetic Progressions with Solutions
Find the first five terms of the sequence with nth term an = 3n + 2.
Solutions: For n = 1 to 5:
n = 1: a₁ = 3 × 1 + 2 = 5
n = 2: a₂ = 8
n = 3: a₃ = 11
n = 4: a₄ = 14
n = 5: a₅ = 17
Given an AP: –5, –1, 3, 7,… Write the first term a and common difference d.
Solutions: a = –5
d = –1 – (–5) = 4
For the AP 1, 4, 7, 10…, find the 10th term.
Solutions: a = 1, d = 3, n = 10
a₁₀ = a + (n–1)d = 1 + 9×3 = 28
In the AP √2, 3√2, 5√2, ..., find the 18th term.
Solutions: a = √2, d = 2√2, n = 18
a₁₈ = √2 + 17×2√2 = 35√2
Find nth term and 20th term of AP: 13, 8, 3, –2, …
Solutions: a = 13, d = 8 – 13 = –5
aₙ = a + (n–1)d
a₂₀ = 13 + 19×(–5) = –82
If the 12th term of an AP is 54 and the 13th term is 58, find the first term and common difference.
Solutions: a₁₂ = a + 11d = 54
a₁₃ = a + 12d = 58
Subtract:
d = 4
Substitute:
a + 44 = 54 ⇒ a = 10
Which term of the sequence 114, 109, 104, … is the first negative term?
Solutions: a = 114, d = –5
aₙ = 114 + (n–1)(–5) < 0
114 – 5(n–1) < 0 ⇒ n–1 > 22.8 ⇒ n > 23.8
First negative term is the 24th term.
Find sum of the first 10 terms of the AP: 50, 46, 42, …
Solutions: a = 50, d = –4, n = 10
S₁₀ = 10/2 [2×50 + 9×(–4)] = 5×64 = 320
Find the sum of the AP: 1, 3, 5, ..., to 12 terms.
Solutions: a = 1, d = 2, n = 12
S₁₂ = 6 × 24 = 144
Find the sum of first n odd natural numbers.
Solutions: 1, 3, 5, ... (AP with a = 1, d = 2)
Sn = n²
If the sum of the first n terms of an AP is Sn = 3n² + 5n, find the nth term.
Solutions: aₙ = Sn – Sn–1
= [3n² + 5n] – [3(n–1)² + 5(n–1)]
= 6n + 2
If x + 1, 3x, 4x + 2 are in AP, find x.
Solutions: 3x – (x + 1) = (4x + 2) – 3x
2x – 1 = x + 2
x = 3
Show that a – b, a, a + b are in AP.
Solutions: Common differences:
a – (a–b) = b; (a + b) – a = b
Both equal ⇒ Proof complete.
Find the three terms in AP whose sum is 21 and the product of the first and third is 6 more than the second term.
Solutions: Terms: a–d, a, a+d
Sum: 3a = 21 ⇒ a = 7
(a–d)(a+d) = a2 – d2 = a + 6 ⇒ 49 – d² = 13 ⇒ d² = 36 ⇒ d = 6 or –6
Required AP: 1, 7, 13 or 13, 7, 1
The sum of three numbers in AP is 12, sum of their cubes is 288. Find the numbers.
Solutions: Let: a–d, a, a+d
3a = 12 ⇒ a = 4
Sum of cubes: (a–d)³ + a³ + (a+d)³ = 288
(a–d)³ + (a+d)³ = 2(a³ + 3ad²) ⇒ 2(64 + 12d²) + 64 = 288
192 + 24d² = 288 ⇒ d² = 4 ⇒ d = 2 or –2
Numbers: 2, 4, 6
Find four numbers in AP whose sum is 50 and the greatest is four times the least.
Solutions: Terms: a, a + d, a + 2d, a + 3d
Sum: 4a + 6d = 50
a + 3d = 4a ⇒ 3d = 3a ⇒ d = a
4a + 6a = 10a = 50 ⇒ a = 5, d = 5
Required numbers: 5, 10, 15, 20
Which term of the AP 4, 9, 14, ... is 254?
Solutions: a = 4, d = 5, aₙ = 254
aₙ = a + (n–1)d
254 = 4 + (n–1)×5 ⇒ n = 51
Find the sum: (x–y)², (x² + y²), (x + y)² to n terms
Solutions: Common difference:
(x² + y²) – (x–y)² = 2xy
Sₙ = n/2 [2a + (n–1)d], where a = (x–y)², d = 2xy
The angles of a quadrilateral are in AP with common difference 10°. Find all angles.
Solutions: Angles: a–15, a–5, a+5, a+15
Sum = 360°
4a = 360 ⇒ a = 90
Angles: 75°, 85°, 95°, 105°
If the 10th term of an AP is 52 and 15th term is 77, find the first term.
Solutions: a₁₀ = a + 9d = 52
a₁₅ = a + 14d = 77
5d = 25 ⇒ d = 5
a + 45 = 52 ⇒ a = 7
RD Sharma Class 10 Solutions Chapter 9 – Arithmetic Progressions PDF Overview
The RD Sharma Class 10 Solutions Chapter 9 PDF helps students learn Arithmetic Progressions (AP) in easy and simple ways. This chapter is a key part of the CBSE Class 10 Maths syllabus and is very important for board exams. It teach how to find patterns in numbers and solve problems using the AP formulas. The RD Sharma Class 10 Arithmetic Progressions PDF is perfect for students who find this topic hard.
Main topics in Chapter 9:
- Understanding the meaning of Arithmetic Progressions
- Formula for nth term and sum of n terms
- How to solve problems based on APs in real life
- Word problems with AP sequences
- Using APs in geometry and daily life questions
The RD Sharma Class 10 Arithmetic Progressions PDF download gives step-by-step solutions to all exercises. Students often gets confused in finding terms or sum, but this PDF makes it very easy to understand. It is helpful for self-study and quick revision before exam.
Why use RD Sharma Class 10 Solutions 2025-26 PDF for Chapter 9?
- Each solution is explained very clearly and easy to understand
- Good for solving homework and practicing for tests
- Formulas and tricks are explained with solved examples
- Can be downloaded to study anytime, even with no internet
- Mistakes can happen, but this PDF help you learn and correct them
Benefits of Solving RD Sharma Class 10 Chapter 9 – Arithmetic Progressions:
Frequently Asked Questions
The RD Sharma Class 10 Chapter 9 – Arithmetic Progressions PDF includes all important concepts related to Arithmetic Progressions (AP). It explains the formula for the nth term, sum of n terms, and helps solve real-life word problems using AP. Each question from the textbook is solved step-by-step, making it easy for students to understand. Whether it's finding common difference or solving exam-type questions, the RD Sharma Class 10 Solutions 2025-26 PDF is very useful for complete practice and revision.
You can easily download the RD Sharma Class 10 Arithmetic Progressions PDF from this article by scrolling to the download section. The file includes full solutions to Chapter 9 questions, explained in a simple way. This RD Sharma Class 10 Arithmetic Progressions PDF download is perfect for offline study and revision before exams. Just click the link, save it to your device, and study without internet anytime.
Chapter 9 is very important because it builds strong mathematical thinking. Arithmetic Progressions are used in higher-level topics in Class 11, 12, and competitive exams like JEE and NTSE. Practicing with the RD Sharma Class 10 Solutions Chapter 9 PDF helps students solve problems faster and with more confidence. It improves accuracy, helps in time management during exams, and strengthens your base for future studies.