What is the formula for the nth term of an Arithmetic Progression?

The nth term formula is aₙ = a + (n − 1)d, where ‘a’ is the first term, ‘d’ is the common difference, and ‘n’ is the term number. This formula helps find any term in the sequence without writing all previous terms.

What is the sum formula for Arithmetic Progression?

The sum of the first n terms of an AP is Sₙ = n/2 [2a + (n − 1)d]. It is used to calculate the total of a given number of terms in an AP. This formula is frequently used in board exams and real-life applications like salary or distance problems.

Why is learning the Arithmetic Progression formula important?rtant?

AP is an important topic in Class 10 Maths and forms the foundation for higher-level topics in algebra and statistics. Understanding it also improves problem-solving skills and logical thinking both essential for board exams and competitive tests.

Is there a way to learn Arithmetic Progression faster and smarter?

Yes! Consistent practice, conceptual understanding, and guided learning make a huge difference. PlanetSpark offers personalised Maths sessions that focus on concept mastery rather than memorisation helping Class 10 students score higher with less stress.

How can PlanetSpark help Class 10 students master AP?

PlanetSpark’s interactive Maths course simplifies AP through visual explanations, live classes, and AI-based practice modules. Students learn concepts like nth term and sum formulas through real-life examples, ensuring stronger conceptual clarity and exam confidence.

Arithmetic Progression Formula Explained For Class 10 Students

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Struggling to remember the arithmetic progression formula before every test? You’re not alone! Many Class 10 students find AP confusing because they try to memorise instead of understanding its logic. The good news? Once the pattern clicks, solving AP questions becomes effortless.

In this blog, we’ll break down what arithmetic progression is, explain each formula step-by-step, share NCERT-based solved examples, and give tips to master AP for board exams. And if Maths still feels overwhelming, PlanetSpark’s interactive Maths Course helps Class 10 learners build clarity and confidence—turning fear of formulas into love for problem-solving!

What is Arithmetic Progression (AP)?

In mathematics, a sequence in which each term after the first is obtained by adding a fixed constant (the common difference) to the previous term is called an arithmetic progression (AP).
For example: 5, 8, 11, 14, 17… here the fixed jump is 3.
The reason the arithmetic progression formula is so useful is that it allows efficient calculation of any term in the sequence and the sum of many terms, without listing them one by one.

Understanding the Arithmetic Progression Formula

When students work through NCERT Chapter 5 (Class 10 Maths), they will find repeated references to the arithmetic progression formula. That’s because both the nth-term formula and the sum formula for arithmetic progression are essential building-blocks. Understanding them deeply means less rote learning, and more confident application in board-style questions.
In this section:

The primary formula for the nth term: $a_n = a + (n - 1)d$ $a$ $n$ $$ $=$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$
The sum formula: $S_n = \frac n2 [2a + (n - 1)d]$ $S$ $n$ $$ $=$ $2$ $n$ $$ $[$ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $]$ or equivalently $S_n = \frac n2 [a + a_n]$ $S$ $n$ $$ $=$ $2$ $n$ $$ $[$ $a$ $+$ $a$ $n$ $$ $]$
These are the heart of solving Chapter 5 questions of AP in Class 10.

Key Terms in Arithmetic Progression Formula

Before applying formulas, clarity on key terms is crucial:

First term (a): the initial number in the AP.
Common difference (d): the fixed amount added (or subtracted) each time to move from one term to the next.
nth term (Tₙ or aₙ): the term at position n, given by the formula.
This understanding prevents confusion when working with different sequences and ensures correct use of the arithmetic progression formula in exercises.

Formula for the nth Term of an AP

The standard expression for the nth term of an AP (with first term a and common difference d) is:

a_n = a + (n - 1)\,d

$a$ $n$ $$ $=$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$

Here:

$a_n$ $a$ $n$ $$ denotes the nth term.
$n$ $n$ is the term number in the sequence.
Using this formula allows finding any specific term in the sequence, a vital skill for board exam questions

Formula for the Sum of n Terms ( $S_n$ $S$ $n$ $$ ) in an AP

When the task is to find the sum of the first n terms of an AP, the sum formula for arithmetic progression is used:

S_n = \frac{n}{2} [2a + (n - 1)\,d]

$S$ $n$ $$ $=$ $2$ $n$ $$ $[$ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $]$

Alternatively, if the last term (l) is known:

S_n = \frac{n}{2} (a + l)

$S$ $n$ $$ $=$ $2$ $n$ $$ $($ $a$ $+$ $l$ $)$

Applying these formulas accurately enables efficient solutions of sum-type questions in Class 10 maths and ensures comprehension beyond mere memorisation.

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Derivation: Step-by-Step Explanation of the Arithmetic Progression Formula

This section explains how the two key results of an arithmetic progression (AP) follow logically: the formula for the nth term and the sum formula for arithmetic progression. Clear, algebraic steps make the rules easy to remember and apply in Class 10 problems.

1. Deriving the nth-term formula $a_n = a + (n-1)d$ $a$ $n$ $$ $=$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$

Start with the definition of an AP.
If the first term is $a$ $a$ and the common difference is $d$ $d$ , the sequence of terms is:
$a,\; a+d,\; a+2d,\; a+3d,\; \dots$ $a$ $,$ $a$ $+$ $d$ $,$ $a$ $+$ $2$ $d$ $,$ $a$ $+$ $3$ $d$ $,$ $\dots$
Write the first few terms with their term numbers:
$\begin{aligned} a_1 &= a \\ a_2 &= a + d \\ a_3 &= a + 2d \\ a_4 &= a + 3d \\ \end{aligned}$ $a$ $1$ $$ $a$ $2$ $$ $a$ $3$ $$ $a$ $4$ $$ $$ $=$ $a$ $=$ $a$ $+$ $d$ $=$ $a$ $+$ $2$ $d$ $=$ $a$ $+$ $3$ $d$ $$
Notice the pattern: each time the term index increases by 1, $d$ $d$ is added once more.
Generalise the pattern. For the $n$ $n$ th term, the common difference $d$ $d$ has been added $(n-1)$ $($ $n$ $-$ $1$ $)$ times (because the first term needs zero additions):
$a_n = a + (n - 1)\,d.$ $a$ $n$ $$ $=$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $.$
Interpretation: this formula gives the exact value at position $n$ $n$ when $a$ $a$ and $d$ $d$ are known. It answers the “which value sits at place $n$ $n$ ?” question directly.

Short numeric check (example): if $a=5$ $a$ $=$ $5$ and $d=3$ $d$ $=$ $3$ , then the 10th term is
$a$ $10$ $$ $=$ $5$ $+$ $($ $10$ $-$ $1$ $)$ $\times$ $3$ $=$ $5$ $+$ $9$ $\times$ $3$ $=$ $5$ $+$ $27$ $=$ $32.$

2. Deriving the sum formula $S_n = \dfrac{n}{2}\,[2a + (n-1)d]$ $S$ $n$ $$ $=$ $2$ $n$ $$ $[$ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $]$

Two equivalent derivations appear commonly; both are shown because each helps build intuition.

A — Pairing forwards and backwards (Gauss style)

Let $S_n$ $S$ $n$ $$ denote the sum of the first $n$ $n$ terms:
$S_n = a_1 + a_2 + a_3 + \dots + a_n.$ $S$ $n$ $$ $=$ $a$ $1$ $$ $+$ $a$ $2$ $$ $+$ $a$ $3$ $$ $+$ $\dots$ $+$ $a$ $n$ $$ $.$
Using the notation of the AP:
$S_n = a + (a+d) + (a+2d) + \dots + \bigl(a+(n-1)d\bigr).$ $S$ $n$ $$ $=$ $a$ $+$ $($ $a$ $+$ $d$ $)$ $+$ $($ $a$ $+$ $2$ $d$ $)$ $+$ $\dots$ $+$ $($ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $)$ $.$
Write the sum again in reverse order:
$S_n = \bigl(a+(n-1)d\bigr) + \bigl(a+(n-2)d\bigr) + \dots + a.$ $S$ $n$ $$ $=$ $($ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $)$ $+$ $($ $a$ $+$ $($ $n$ $-$ $2$ $)$ $d$ $)$ $+$ $\dots$ $+$ $a$ $.$
Add these two expressions term-by-term (first with first, second with second, etc.). Each pair sums to the same value:
$\begin{aligned} 2S_n &= \bigl[a + \bigl(a+(n-1)d\bigr)\bigr] + \bigl[(a+d) + \bigl(a+(n-2)d\bigr)\bigr] + \dots \\ &= \bigl(2a + (n-1)d\bigr) + \bigl(2a + (n-1)d\bigr) + \dots \quad\text{(there are $n$ such pairs)} \\ &= n\bigl(2a + (n-1)d\bigr). \end{aligned}$ $2$ $S$ $n$ $$ $$ $=$ $[$ $a$ $+$ $($ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $)]$ $+$ $[$ $($ $a$ $+$ $d$ $)$ $+$ $($ $a$ $+$ $($ $n$ $-$ $2$ $)$ $d$ $)]$ $+$ $\dots$ $=$ $($ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $)$ $+$ $($ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $)$ $+$ $\dots$ $(there are$ $n$ $such pairs)$ $=$ $n$ $($ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $)$ $.$ $$
Divide both sides by 2:

$S$ $n$ $$ $=$ $2$ $n$ $$ $[$ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $]$ $.$

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Why this works (intuition): pairing the first and last, second and second-last, etc., produces $n$ $n$ identical pairs. Each pair equals the sum of the first and last term, i.e. $a + l$ $a$ $+$ $l$ ; algebraic simplification gives the expression above.

B — Using the average of first and last terms

If the last (nth) term is $l$ $l$ , then the sum of $n$ $n$ terms equals the average of the first and last terms, times the number of terms:
$S_n = n \times \text{(average of first and last)} = n\times\frac{a + l}{2} = \frac{n}{2}(a + l).$ $S$ $n$ $$ $=$ $n$ $\times$ $(average of first and last)$ $=$ $n$ $\times$ $2$ $a$ $+$ $l$ $$ $=$ $2$ $n$ $$ $($ $a$ $+$ $l$ $)$ $.$
Replace $l$ $l$ by its expression from the nth-term formula: $l = a_n = a + (n-1)d$ $l$ $=$ $a$ $n$ $$ $=$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ . Substituting gives:
$S_n = \frac{n}{2}\Bigl(a + \bigl(a + (n-1)d\bigr)\Bigr) = \frac{n}{2}\bigl[2a + (n-1)d\bigr],$ $S$ $n$ $$ $=$ $2$ $n$ $$ $($ $a$ $+$ $($ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $)$ $)$ $=$ $2$ $n$ $$ $[$ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $]$ $,$
which matches the previous result.

This viewpoint emphasises that the sum equals the number of terms times the average term; for arithmetic progressions the average of the first and last equals the average of all terms.

3. Worked Example — full steps

Given $a=8$ $a$ $=$ $8$ , $d=4$ $d$ $=$ $4$ , $n=12$ $n$ $=$ $12$ . Find $a_{12}$ $a$ $12$ $$ and $S_{12}$ $S$ $12$ $$ .

Step 1: nth term

a_{12} = a + (n-1)d = 8 + (12-1)\times 4 = 8 + 11\times 4 = 8 + 44 = 52.

$a$ $12$ $$ $=$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $=$ $8$ $+$ $($ $12$ $-$ $1$ $)$ $\times$ $4$ $=$ $8$ $+$ $11$ $\times$ $4$ $=$ $8$ $+$ $44$ $=$ $52.$

Step 2: Sum of first 12 terms (use the sum formula)

\begin{aligned} S_{12} &= \frac{n}{2}\,[2a + (n-1)d] \\ &= \frac{12}{2}\,[2\times 8 + 11\times 4] \\ &= 6\,[16 + 44] = 6\times 60 = 360. \end{aligned}

$S$ $12$ $$ $$ $=$ $2$ $n$ $$ $[$ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $]$ $=$ $212$ $$ $[$ $2$ $\times$ $8$ $+$ $11$ $\times$ $4$ $]$ $=$ $6$ $[$ $16$ $+$ $44$ $]$ $=$ $6$ $\times$ $60$ $=$ $360.$ $$

So the 12th term is 52 and the sum of the first 12 terms is 360.

4. Common pitfalls to avoid

Confusing a term request with a sum request: the nth-term formula $a + (n-1)d$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ finds one term; the sum formula for arithmetic progression finds total of multiple terms.
Forgetting that the common difference $d$ $d$ can be negative for decreasing sequences. The same formulas hold algebraically.
Arithmetic errors when multiplying by $n$ $n$ or dividing by 2 always simplify step-by-step and check intermediate results.

5. Final note (exam strategy)

Memorise the two core forms and understand their derivations. Knowing why $a_n = a + (n-1)d$ $a$ $n$ $$ $=$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ and why $S_n = \dfrac{n}{2}\,[2a+(n-1)d]$ $S$ $n$ $$ $=$ $2$ $n$ $$ $[$ $2$ $a$ $+$ $($ $n$ $-$ $1$ $)$ $d$ $]$ hold makes it straightforward to adapt the formulas to NCERT Chapter-style problems and to spot which formula applies in each question.

NCERT Exercises Based on Arithmetic Progression (with Hints)

Here’s a detailed look at the exercises along with examples and hints that simplify learning:

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Exercise 5.1 — Identify if a Sequence is an AP

Concept:
A sequence is an Arithmetic Progression if the difference between consecutive terms is constant.

Formula Check:
If a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = d,
then it’s an AP with common difference (d).

Example 1:
Find whether 5, 11, 17, 23, 29 is an AP.
Solution:
d = 11 − 5 = 6, 17 − 11 = 6, 23 − 17 = 6 → Constant difference.
Hence, the sequence is an AP with d = 6.

Example 2:
Check whether 2, 4, 8, 16, 32 is an AP.
Solution:
4 − 2 = 2, 8 − 4 = 4, 16 − 8 = 8 (not constant)
Not an AP.

Hint:
When differences change, it becomes a geometric progression (GP), not an AP.

Exercise 5.2 — Find the nth Term of an AP

Formula:
The nth term formula for arithmetic progression is
aₙ = a + (n − 1)d

Example 1:
Find the 15th term of the AP: 3, 8, 13, 18, …
Solution:
Here, a = 3, d = 5, n = 15
a₁₅ = 3 + (15 − 1) × 5 = 3 + 70 = 73

Example 2:
Which term of the AP 7, 10, 13, 16, … is 67?
Solution:
aₙ = 67, a = 7, d = 3
67 = 7 + (n − 1) × 3
67 − 7 = 3(n − 1)
60 = 3n − 3
n = 21
The 21st term is 67.

Hint:
Always identify a, d, and n carefully before substitution.

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Exercise 5.3 — Find the Sum of n Terms (Sₙ)

Formula:
Sum formula for arithmetic progression:
Sₙ = n/2 [2a + (n − 1)d]

Example 1:
Find the sum of the first 10 terms of 2, 5, 8, 11, 14, …
Solution:
a = 2, d = 3, n = 10
S₁₀ = 10/2 [2(2) + (10 − 1) × 3]
= 5 [4 + 27]
= 5 × 31 = 155

Example 2:
Find how many terms of the AP 9, 17, 25, 33, … must be taken so that the sum equals 636.
Solution:
a = 9, d = 8, Sₙ = 636
Using Sₙ = n/2 [2a + (n − 1)d]
636 = n/2 [18 + (n − 1) × 8]
1272 = n [18 + 8n − 8]
1272 = n (8n + 10)
8n² + 10n − 1272 = 0
n = 12 (positive root)
Hence, 12 terms must be taken.

Hint:
In sum questions, simplify step-by-step; avoid skipping algebraic manipulation.

Exercise 5.4 — Application-Based Questions

Concept:
Real-life scenarios where AP applies daily wages, distances, salaries, or patterns in growth.

Example 1:
The salary of a person is ₹15,000 per month and increases by ₹1,000 every year. Find the salary after 10 years and the total salary earned in 10 years.
Solution:
a = 15000, d = 1000, n = 10
Salary in 10th year = a₁₀ = a + (n − 1)d = 15000 + 9 × 1000 = ₹24,000
Total salary in 10 years = S₁₀ = 10/2 [2a + (n − 1)d]
= 5 [30000 + 9000] = ₹1,95,000

Example 2:
A ball is dropped and rebounds to 4/5 of the height each time. Find the total distance it travels after 5 rebounds.
(Here, students can discuss the difference between AP and GP sequences.)

Hint:
Identify what changes constantly, if the change is by a fixed difference, it’s AP; if by a fixed ratio, it’s GP.

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Quick Tips to Master Arithmetic Progression in Class 10 Maths

Mastering the arithmetic progression formula isn’t about memorising; it’s about understanding patterns and relationships. A few smart strategies can make AP questions much easier:

Understand the concept of ‘d’ (common difference) — The common difference defines how terms in an AP grow or shrink. Practice identifying it quickly in different types of series.
Visualise the sequence — Write down the first few terms to notice how the series progresses. This helps prevent mistakes in sign and order.
Practise formula variations — Be fluent with both the nth term formula (a + (n–1)d) and the sum formula for arithmetic progression (Sn = n/2 [2a + (n–1)d]).
Solve step-by-step — Always write what is given, what is required, and which formula is used. This helps gain marks in board exams.
Mix word problems and numerical problems — AP questions often appear in real-world contexts such as savings, distances, and salaries.
Check answers logically — If the nth term seems too large or small, recheck your difference ‘d’. Logical checking builds mathematical sense.

PlanetSpark’s Maths Course helps learners master these patterns through interactive exercises, concept-based videos, and real-life problem simulations, ensuring every Class 10 student builds speed and accuracy in AP.

Master Arithmetic Progression and More with PlanetSpark’s Maths Course

Understanding the arithmetic progression formula isn’t just about scoring marks — it’s about building mathematical confidence that lasts a lifetime. That’s exactly where PlanetSpark’s Maths Course makes a difference.

Why PlanetSpark Maths Stands Out?

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Every topic, from what is arithmetic progression to sum formula for arithmetic progression, is explained through real-life examples, visuals, and interactive tools helping learners grasp “why” before “how”.
Expert-Led Classes with 1:1 Attention
Experienced educators from top institutions guide students through live, interactive sessions offering instant feedback, doubt resolution, and step-by-step guidance on tricky AP problems.
Gamified Learning Environment
Maths becomes a fun challenge rather than a chore. PlanetSpark’s gamified platform awards badges, points, and achievements for every solved question — motivating learners to practise more.
Curriculum Aligned with NCERT & CBSE
All concepts, including arithmetic progression formula derivation, nth term, and sum of n terms, strictly follow the Class 10 NCERT pattern, ensuring full exam alignment.
Instant Progress Reports & Analytics
Parents and students can track growth through weekly reports, measuring speed, accuracy, and improvement areas.
Board Exam-Focused Modules
PlanetSpark integrates board-specific mock tests, timed quizzes, and previous-year paper discussions, ensuring complete readiness before exams.

Crack the Code of Progressions with Confidence!

Arithmetic Progression isn’t just a formula it’s the rhythm of numbers that builds logical thinking and precision. Mastering concepts like the nth term and sum formula for arithmetic progression helps develop a strong mathematical base for board exams and beyond. Consistent practice and conceptual clarity are the keys to scoring high in Class 10 Maths.

To make learning fun, structured, and exam-ready, PlanetSpark’s Maths Course offers interactive lessons, expert mentors, and AI-based practice tools designed especially for Class 10 students.