What is the nth term formula of an Arithmetic Progression?

The nth term a_n of an AP is given by a_n = a + (n − 1) d, where a is the first term and d is the common difference.

How do you find the sum of the first n terms of an AP?

The sum S_n of the first n terms of an AP is S_n = n/2 [2a + (n − 1) d] where a is the first term and d is the common difference.

How can I tell whether a sequence is an AP?

A sequence is an AP if the difference between consecutive terms is constant (i.e., a2−a1 = a3−a2 = ...).

How to find the number of terms when the sum is given?

Use the sum formula S_n = n/2 [2a + (n−1)d], put S_n and solve the resulting quadratic equation for n (take the positive integer root).

What is the sum of the first n odd numbers?

The sum of the first n odd numbers is n².

Mastering Arithmetic Progression in Class 10 – Smart Learning with FUZY MATH ACADEMY

📗Introduction

Mathematics has always been a subject that tests logic, consistency, and practice. Among the various chapters of Class 10 Maths, Arithmetic Progression (AP) holds a special place. It is not only a scoring chapter but also forms the base for higher mathematical studies and real-life applications like calculating savings, installments, or growth patterns.

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At FUZY MATH ACADEMY, we understand that students from Classes 5 to 12 need not just notes but also interactive learning tools, AI-powered LMS (Learning Management System), and 24/7 chatbot support for instant doubt-solving. That’s why our platform ensures that no student feels stuck while learning AP or any other topic.

In this blog, we will dive deep into Arithmetic Progression, understand its concepts, formulas, real-life examples, and most importantly, how FUZY MATH ACADEMY makes mastering AP easy, fun, and stress-free.

✍️What is Arithmetic Progression?

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms remains constant.

For example:
2, 4, 6, 8, 10 … is an AP where the common difference (d) = 2.

Key components of an AP:

First term (a): The first number of the sequence.
Common difference (d): The difference between two consecutive terms.
n-th term (an): The term at position "n".
Sum of n terms (Sn): Total of the first "n" terms.

⌛Important Formulas of Arithmetic Progression

n-th Term Formula

$a_{n} = a + (n - 1) d$

Sum of First n Terms $S_{n} = \frac{n/}{2} [2 a + (n - 1) d]$

Sum of First n Terms using Last Term (l) $S_{n} = \frac{n/}{2} (a + l)$

📌Real-Life Applications of Arithmetic Progression

Banking & Finance: Calculating fixed deposits, EMIs, or installment amounts.
Sports: Analyzing player scores or lap timings.
Business: Profit and loss progressions over months.
Daily Life: Stacking chairs, arranging bricks, or counting steps in staircases.

When students realize that AP is not just a chapter but also a life skill, their curiosity and motivation increase.

📍Why Students Struggle with Arithmetic Progression

Despite being a simple chapter, students often make mistakes because:

They memorize formulas without understanding.
They fail to connect AP with real-life situations.
Lack of consistent practice leads to silly errors in board exams.

📐How FUZY MATH ACADEMY Helps You Master AP

At FUZY MATH ACADEMY, we don’t just teach; we create a learning experience.

✅ AI-Powered LMS: Interactive lessons, step-by-step problem solving, and quizzes.
✅ 24/7 Chatbot Support: Students can clear doubts anytime without waiting for the next class.
✅ Engaging Practice Tests: Real-time scoring and instant feedback.
✅ Expert Faculty: Teachers who simplify AP concepts with tricks and shortcuts.
✅ Affordable Fees: Quality education at half the price of offline coaching.

This combination ensures that even the most complex AP problems feel easy to solve.

📅Tips & Tricks to Score Full Marks in Arithmetic Progression

Understand the Concept – Never just memorize; visualize AP in real life.
Revise Formulas Daily – Write and practice formulas for 10 minutes every day.
Solve NCERT + PYQs – Almost 90% of AP questions in exams are based on NCERT and previous years.
Attempt Mock Tests – At FUZY MATH ACADEMY, our LMS provides timed quizzes to simulate exam pressure.
Focus on Word Problems – Application-based questions are where students lose marks; practice them the most.

📢Sample Problem

👉Q: The 11th term of an AP is 45, and the 16th term is 65. Find the first term and the common difference.

Solution:
Let first term = a, common difference = d.

$a + 10 d = 45 \dots (1) ,$ $a + 15 d = 65 \dots (2)$

Subtracting (1) from (2):

$5 d = 20 \Rightarrow d = 4$

Substitute in (1):

$a + 40 = 45$

Answer: First term (a) = 5, Common difference (d) = 4.

💹 Important Arithmetic Progression Questions with Solutions Arithmetic Progression Class 10 – Important Questions with Solutions Arithmetic Progression – 20 Important Questions with Step-by-Step Solutions

Arithmetic Progression – 20 Important Questions (Step-by-Step Solutions)

Q1. Find the 10^th term of the AP: 2, 5, 8, 11, …

Step 1: first term `a = 2`, common difference `d = 5−2 = 3`.

Step 2: formula for n-th term: `a_n = a + (n−1)d`.

Step 3: put `n=10`: `a₁₀ = 2 + (10−1)×3 = 2 + 27 = 29`.

Answer: 29

Q2. Find the sum of first 20 terms of the AP: 7, 10, 13, …

Step 1: `a = 7`, `d = 3`, `n = 20`.

Step 2: sum formula: `S_n = n/2 [2a + (n−1)d]`.

Step 3: `S₂₀ = 20/2 [2×7 + 19×3] = 10 [14 + 57] = 10 × 71 = 710`.

Answer: 710

Q3. Which term of the AP 3, 8, 13, … is 253?

Step 1: `a=3`, `d=5`, suppose `a_n=253`.

Step 2: use `a_n=a+(n−1)d` ⇒ `253=3+(n−1)5`.

Step 3: `253−3=5(n−1) ⇒ 250=5(n−1) ⇒ n−1=50 ⇒ n=51`.

Answer: 51^st term

Q4. Find the 15^th term of the AP: 10, 7, 4, …

Step 1: `a=10`, `d=7−10=−3`.

Step 2: `a₁₅=10+(15−1)(−3)=10+14×(−3)=10−42=−32`.

Answer: −32

Q5. Find the sum of first 30 multiples of 5.

Step 1: sequence is 5, 10, 15, … → `a=5`, `d=5`, `n=30`.

Step 2: `S₃₀=30/2 [2×5 + 29×5] = 15 [10 + 145] = 15×155 = 2325`.

Answer: 2325

Q6. How many terms of the AP 9, 17, 25, … must be taken to get sum 636?

Step 1: `a=9`, `d=8`, let number of terms be `n`.

Step 2: `S_n=n/2 [2a + (n−1)d] = n/2 [18 + 8(n−1)] = n/2 (8n +10)`.

Step 3: set equal to 636: `n/2 (8n+10)=636 ⇒ 8n²+10n−1272=0` (multiply both sides by 2).

Step 4: Solve quadratic `8n²+10n−1272=0`. Discriminant `D=10²−4×8×(−1272)=100+40704=40804`. √D = 202? (check) — better factor: divide equation by 2 → `4n²+5n−636=0`, D = 5² − 4×4×(−636)=25+10176=10201, √D=101.

Step 5: `n = [−5 ± 101]/8` ⇒ positive root `n = (96)/8 = 12`.

Answer: 12 terms

Q7. Find the 25^th term of AP: 7, 13, 19, …

Step 1: `a=7`, `d=6`, `n=25`.

Step 2: `a₂₅=7+(25−1)×6=7+24×6=7+144=151`.

Answer: 151

Q8. If the 7^th term of an AP is 20 and the 13^th term is 50, find the AP (first term and d).

Step 1: use formula `a + 6d = 20` and `a + 12d = 50`.

Step 2: subtract: `(a+12d) − (a+6d) = 50 − 20 ⇒ 6d = 30 ⇒ d = 5`.

Step 3: put `d=5` into `a+6d=20 ⇒ a + 30 = 20 ⇒ a = −10`.

Answer: AP is −10, −5, 0, 5, … (a = −10, d = 5)

Q9. Find the sum of first 15 terms of AP: 5, 8, 11, …

Step 1: `a=5`, `d=3`, `n=15`.

Step 2: `S₁₅=15/2 [2×5 + 14×3] = 7.5 [10 + 42] = 7.5 × 52 = 390`.

Answer: 390

Q10. Which term of AP 21, 18, 15, … is −81?

Step 1: `a = 21`, `d = −3`, want `a_n = −81`.

Step 2: `−81 = 21 + (n−1)(−3)` ⇒ `−81−21 = −3(n−1) ⇒ −102 = −3(n−1)`.

Step 3: `n−1 = 34 ⇒ n = 35`.

Answer: 35^th term

Q11. Find the sum of first 40 natural numbers.

Step 1: natural numbers are AP with `a=1`, `d=1`, `n=40`.

Step 2: `S₄₀ = 40/2 [1 + 40] = 20 × 41 = 820`.

Answer: 820

Q12. Find the sum of first n terms of AP: 2, 7, 12, … (express S_n in terms of n).

Step 1: `a=2`, `d=5`.

Step 2: `S_n = n/2 [2a + (n−1)d] = n/2 [4 + 5(n−1)] = n/2 [5n −1]`.

Answer: `S_n = (n/2)(5n − 1)`

Q13. The sum of first 14 terms of an AP is 1050, and common difference d = 10. Find the first term.

Step 1: use `S_n = n/2 [2a + (n−1)d]` with `S₁₄=1050`, `n=14`, `d=10`.

Step 2: `1050 = 14/2 [2a + 13×10] ⇒ 1050 = 7 [2a + 130]`.

Step 3: divide by 7: `2a + 130 = 150 ⇒ 2a = 20 ⇒ a = 10`.

Answer: first term `a = 10`

Q14. If the 3^rd term of an AP is 5 and the 8^th term is 20, find the AP.

Step 1: `a + 2d = 5` and `a + 7d = 20`.

Step 2: subtract: `5d = 15 ⇒ d = 3`.

Step 3: substitute back: `a + 2×3 = 5 ⇒ a = −1`.

Answer: AP is −1, 2, 5, 8, … (a = −1, d = 3)

Q15. Find the 11^th term of AP: −7, −4, −1, …

Step 1: `a = −7`, `d = 3`, `n = 11`.

Step 2: `a₁₁ = −7 + (11−1)×3 = −7 + 30 = 23`.

Answer: 23

Q16. Find the sum of first 25 odd numbers.

Step 1: odd numbers form AP 1, 3, 5, … with n terms; known formula: sum of first n odd numbers = n².

Step 2: for `n = 25`, sum = `25² = 625`.

Answer: 625

Q17. If the sum of first 10 terms of an AP is 210 and its first term is 5, find the common difference.

Step 1: `S₁₀ = 210`, `a = 5`, `n = 10`.

Step 2: `S₁₀ = 10/2 [2a + 9d] = 5[10 + 9d]`.

Step 3: set equal: `5(10 + 9d) = 210 ⇒ 10 + 9d = 42 ⇒ 9d = 32 ⇒ d = 32/9`.

Answer: `d = 32/9`

Q18. Find n if the n^th term of AP 7, 13, 19, … is 205.

Step 1: `a = 7`, `d = 6`, `a_n = 205`.

Step 2: `205 = 7 + (n−1)6 ⇒ 198 = 6(n−1) ⇒ n−1 = 33 ⇒ n = 34`.

Answer: 34

Q19. If sum of first n terms of an AP is `S_n = 2n² + 3n`, find its 10^th term.

Step 1: nth term `a_n = S_n − S_n−1.`

Step 2: compute `S_n−1 = 2(n−1)² + 3(n−1) = 2(n²−2n+1) + 3n − 3 = 2n² −4n +2 +3n −3 = 2n² − n −1`.

Step 3: `a_n = (2n² + 3n) − (2n² − n −1) = 4n + 1`.

Step 4: for `n = 10`, `a₁₀ = 4×10 + 1 = 41`.

Answer: 41

Q20. Find the sum of first 50 even numbers.

Step 1: even numbers are AP 2,4,6,… with `a = 2`, `d = 2`, `n = 50`.

Step 2: last even number = 2n = 100, but use sum formula: `S₅₀ = 50/2 [2 + 100] = 25 × 102 = 2550`.

Answer: 2550

FREQUENTLY ASKED QUESTIONS

👉Q1. Find the 20th term of the AP: 3, 7, 11, 15 …

Solution:
Here, $a = 3$ $d = 4$ .

$a_{n} = a + (n - 1) d$

$= 3 + (20 - 1) \times 4 = 3 + 76 = 79$

Answer: 79

👉Q2. Which term of the AP: 5, 11, 17, 23 … will be 95?

Solution:

$a = 5, d = 6$
Let the nth term = 95

$a_{n} = a + (n - 1) d$

$95 = 5 + (n - 1) \times 6$

$95 - 5 = 6 (n - 1)$

$90 = 6 n - 6$

$6 n = 96 \Rightarrow n = 16$ Answer:16th term

👉Q3. The sum of first 25 terms of the AP: 7, 13, 19, … is?

Solution:
$a = 7, d = 6, n = 25$
$S_n = n/2[2a + (n-1)d]$

$= 2 5 [2 (7) + 24 (6)] = 25/2 [14 + 144] = 2 5/2 \times 158 = 1975$
Answer: 1975

👉Q4. The 4th term of an AP is 8 and the 9th term is 23. Find the AP.

Solution:
$a + 3 d = 8 \dots (1)$
$a+8d = 23 …(2)$
Subtract: $5 d = 15 \Rightarrow d = 3$
From (1): $a + 9 = 8 \Rightarrow a = - 1$
So AP = -1, 2, 5, 8, 11 ….....,AP=a,a+d,a+2d,.......
Answer: AP is $-1, 2, 5, 8, …$

👉Q5. If the sum of the first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Solution:
$S_{7} = 7/2 [2 a + 6 d]$

$\Rightarrow 2 a + 6 d = 14 \dots (1)$ $2a+16d=34 …(2)$

Subtract (2) – (1): $10 d = 20 \Rightarrow d = 2$
Put in (1): $2a+12=14$ ⇒a=1
So,
$S_{n} = n [2 (1) + (n - 1) 2] = n^{2}$
Answer: $S_{n} = n^{2}$

👉Q6. The first term of an AP is 5, the last term is 45, and the sum is 400. Find the number of terms.

Solution:
$n=16,$

$400 = n/2 (5 + 45)$

$400 = n/2 (50)$

$400 = 25 n$

$\Rightarrow n = 16$ Answer: 16 terms

👉Q7. If the sum of the first 14 terms of an AP is 1050 and its 14th term is 97, find the AP.

Solution:
$S_{14} = 14 [2 a + 13 d]$ =1050

⇒2a+13d=150…(1)

$a+13d=97 …(2)$
Multiply (2) by 2:
$2a+26d=194$
Subtract (1):
From (2):
$a + 13 (44/13) = 97$
$\Rightarrow a = 53$
Answer: First term = 53, common difference = $44/13$

👉Q8. The sum of first 16 even numbers is?

Solution:
Even numbers: 2, 4, 6, … form AP with $a=2, d=2$
$S$

$= 8 [4 + 30] = 8 \times 34 = 272$
Answer: 272

👉Q9. Which term of the AP: 3, 8, 13, … will be 253?

Solution:
$a = 3, d = 5$

$a_{n} = 253$

$253= 3 + (n - 1) 5$

$253 - 3 = 5 (n - 1)$

$250 = 5 n - 5$

$255 = 5 n \Rightarrow n = 51$ Answer:51st term

$👉 Q10. The first and last terms of an AP are 17 and 350 respectively. If the common difference is 9, find the number of terms and the sum.$

Solution:
$a=17, l=350, d=9$
$l=a+(n-1)d$

$350=17+(n-1)9$

333=(n−1)9

$n - 1 = 37 \Rightarrow n = 38$

$38/2 (17 + 350) = 19 \times 367 = 6973$

👉Why Choose FUZY MATH ACADEMY for Learning Arithmetic Progression?

Personalized Dashboard – Students track their progress in AP.
Gamified Learning – Badges, rewards, and ranks to make maths fun.
Live + Recorded Classes – Flexibility to study anytime.
One-to-One Doubt Clearing – Human + AI support ensures clarity.

This is why thousands of students prefer FUZY MATH ACADEMY over offline coaching.

👓Comparison Table – Offline Coaching vs FUZY MATH ACADEMY

Feature Offline Coaching FUZY MATH ACADEMY

Fees ₹4000 – ₹6000/month ₹1500 – ₹2500/month

Doubt Solving Only in class hours 24/7 AI Chatbot + Faculty

Study Material Printed notes LMS with video, quizzes, analytics

Flexibility Fixed timings Anytime, Anywhere

Progress Tracking Manual Automated reports on LMS

💥Final Words

Feature	Offline Coaching	FUZY MATH ACADEMY
Fees	₹4000 – ₹6000/month	₹1500 – ₹2500/month
Doubt Solving	Only in class hours	24/7 AI Chatbot + Faculty
Study Material	Printed notes	LMS with video, quizzes, analytics
Flexibility	Fixed timings	Anytime, Anywhere
Progress Tracking	Manual	Automated reports on LMS

Arithmetic Progression is one of the most scoring chapters in Class 10 Maths. With clarity of concept, strong practice, and the right mentorship, students can easily score full marks in this chapter.

At FUZY MATH ACADEMY, we empower students with technology-driven learning and round-the-clock support. Whether you are in Class 5 building basics or in Class 12 preparing for competitive exams, our platform ensures that no doubt goes unanswered, and no student feels left behind.

📞 Contact: 6264302661
🌐 Visit: www.fuzymathacademy.com

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