This page covers the arithmetic progressions class 10 formulas from NCERT Chapter 5 — how to recognise an AP, find its common difference, locate any term, and add up terms. Every formula below is taken from the official NCERT Class 10 Mathematics textbook.
The formulas are grouped by topic, with a symbol-meaning table, a when-to-use guide, three worked examples using fresh numbers, and chapter-specific mistakes to avoid. For the derivations this sheet does not repeat, see the Arithmetic Progressions Class 10 Notes; other chapters sit in the Class 10 Maths formulas index.
Formulas at a Glance
The table below indexes every formula on this page; the details follow in the grouped list.
| Purpose (what you are finding) | Formula |
|---|---|
| Common difference of an AP | \( d = a_{k+1} – a_k \) |
| nth term (general term) | \( a_n = a + (n – 1)d \) |
| Number of terms (from the nth-term formula), when \( d \neq 0 \) | \( n = \frac{a_n – a}{d} + 1 \) |
| Sum of the first \( n \) terms from \( a \) and \( d \) | \( S_n = \frac{n}{2}[2a + (n – 1)d] \) |
| Sum using the first and last terms | \( S_n = \frac{n}{2}(a + l) \) |
| nth term from two consecutive sums, \( n \geq 2 \) | \( a_n = S_n – S_{n-1} \) |
| Sum of the first \( n \) positive integers | \( S_n = \frac{n(n + 1)}{2} \) |
| Arithmetic mean of \( a \) and \( c \), when \( a, b, c \) are in AP | \( b = \frac{a + c}{2} \) |
All Formulas, Grouped by Topic
This is the chapter’s formula inventory, grouped under the textbook’s own sub-topics. A one-line “why” sits under each box; the full derivations live on the notes page linked above.
Definition of an Arithmetic Progression
A list \( a_1, a_2, a_3, \dots \) is an arithmetic progression (AP) only when every consecutive difference is the same (NCERT, p. 3):
\[ a_2 – a_1 = a_3 – a_2 = a_4 – a_3 = \dots = d \]
Equivalently, for any position \( k \), the common difference is \[ d = a_{k+1} – a_k \]
The general form of an AP with first term \( a \) and common difference \( d \) is \[ a,\ a + d,\ a + 2d,\ a + 3d,\ \dots \]
Checking two or three consecutive differences may reveal a pattern, but the list is an AP only when every available consecutive difference is the same. The fixed number \( d \) may be positive, negative or zero (NCERT, p. 3).
nth Term of an AP
The nth term, also called the general term, of the AP with first term \( a \) and common difference \( d \), for \( n \geq 1 \) (NCERT, p. 8):
\[ a_n = a + (n – 1)d \]
Why: the first term carries no \( d \), the second carries one \( d \), and the \( n \)th carries \( n – 1 \) copies of \( d \).
If the finite AP has \( m \) terms, its last term is often written as \( l \):
\[ l = a_m = a + (m – 1)d \]
Number of terms when the last term and \( d \) are known — rearranged from the nth-term formula, valid when \( d \neq 0 \):
\[ n = \frac{a_n – a}{d} + 1 \]
Sum of First n Terms of an AP
Sum of the first \( n \) terms, from the first term and the common difference (NCERT, p. 15):
\[ S_n = \frac{n}{2}[2a + (n – 1)d] \]
The same sum using the last term \( l \) — this form is useful when the first and last terms of the sum are given and the common difference is not needed (NCERT, p. 15):
\[ S_n = \frac{n}{2}(a + l) \]
Why the formula works: write the sum forwards and backwards, add the two rows, and the \( n \) columns each add to \( 2a + (n – 1)d \); halving the doubled sum gives the formula. The column trick works for odd and even \( n \) alike, so there is no separate case.
Since \( a_n = a + (n – 1)d \), the sum can also be written \( S_n = \frac{n}{2}(a + a_n) \). The \( n \)th term from two consecutive sums, valid for \( n \geq 2 \):
\[ a_n = S_n – S_{n-1} \]
Sum of the first \( n \) positive integers — the AP \( 1, 2, 3, \dots, n \) has \( a = 1 \) and \( l = n \) (NCERT, p. 15):
\[ S_n = \frac{n(n + 1)}{2} \]
Arithmetic Mean
If \( a, b, c \) are in AP, then \( b \) is the arithmetic mean of \( a \) and \( c \) (NCERT, p. 24):
\[ b = \frac{a + c}{2} \]
Why: equal gaps in the list mean \( b – a = c – b \), which rearranges to \( 2b = a + c \).
What Each Symbol Means
This chapter has no physical units, so every symbol carries the nature of the numbers in the list (a count, a length, an amount).
| Symbol | What it means | Nature / unit |
|---|---|---|
| \( a \) (or \( a_1 \)) | first term of the AP | same quantity as the terms — a length (cm), an amount (₹), a count |
| \( d \) | common difference — the fixed number added to each term to get the next | same quantity as the terms; may be positive, negative or zero |
| \( n \) | position of a term, or the number of terms being added | positive integer (a count) |
| \( k \) | any position used in the difference test | positive integer (a count) |
| \( a_n \) | nth term (general term) | same quantity as the terms |
| \( a_{k+1} – a_k \) | difference of two consecutive terms | same quantity as the terms |
| \( l \) | last term of a finite AP; \( l = a_m \) | same quantity as the terms |
| \( S_n \) | sum of the first \( n \) terms | same quantity/unit as the terms, added over \( n \) of them |
When to Use Each Formula
| Formula | Reach for it when … |
|---|---|
| \( d = a_{k+1} – a_k \) | you have a list of numbers and want to test whether it is an AP or read off its common difference — the difference must be the same for every pair of consecutive terms. |
| \( a_n = a + (n – 1)d \) | the first term and common difference are known and you want the term at position \( n \). |
| \( n = \frac{a_n – a}{d} + 1 \) | you need the number of terms from the first term, common difference and last value — for example, counting the two-digit numbers divisible by 3. Requires \( d \neq 0 \). |
| \( S_n = \frac{n}{2}[2a + (n – 1)d] \) | you must add the first \( n \) terms and both \( a \) and \( d \) are known. |
| \( S_n = \frac{n}{2}(a + l) \) | the first and last terms of the portion to be added are given and \( d \) is not needed — for example, adding 1 to 100. |
| \( a_n = S_n – S_{n-1} \) | you are given a formula for the sum and want a single term; \( n \geq 2 \). |
| \( S_n = \frac{n(n + 1)}{2} \) | you are adding the first \( n \) positive integers — the special case \( a = 1, d = 1 \). |
| \( b = \frac{a + c}{2} \) | the three numbers \( a, b, c \) are in AP and you need the middle one. |
Worked Examples
Three typical lookups: a direct term, a backwards position, and a sum via the first-and-last-term form. To practise the same formulas on the textbook’s own questions, use the NCERT Solutions for Class 10 Maths Chapter 5 (Arithmetic Progressions).
Example 1: Find a term directly
Find the 15th term of the AP: 4, 9, 14, …
- Step 1: Read the list: \( a = 4 \), \( d = 9 – 4 = 5 \), and \( n = 15 \).
- Step 2: Use the nth-term formula \( a_n = a + (n – 1)d \), since a specific term is wanted.
\[ a_{15} = 4 + (15 – 1) \times 5 = 4 + 70 = 74 \]
Final answer: the 15th term is 74.
Example 2: Work backwards to a position
Which term of the AP: 7, 13, 19, … is 151?
Step 1: Here \( a = 7 \), \( d = 13 – 7 = 6 \), and the term value is given as \( a_n = 151 \).
The unknown is \( n \).
Step 2: Substitute into \( a_n = a + (n – 1)d \) and solve for \( n \).
\[ 151 = 7 + (n – 1) \times 6 \]
\[ 151 – 7 = 6(n – 1) \quad \Rightarrow \quad 144 = 6(n – 1) \quad \Rightarrow \quad n – 1 = 24 \]
\[ n = 25 \]
Final answer: 151 is the 25th term. Check: \( 7 + 24 \times 6 = 151 \), which matches the given value.
Example 3: Sum using the first and last terms
Find the sum: 4 + 8 + 12 + … + 100.
Step 1: This is an AP with \( a = 4 \), \( d = 8 – 4 = 4 \), and the last term \( l = 100 \).
Find \( n \) first with the nth-term formula.
\[ 100 = 4 + (n – 1) \times 4 \quad \Rightarrow \quad 96 = 4(n – 1) \quad \Rightarrow \quad n – 1 = 24 \quad \Rightarrow \quad n = 25 \]
Step 2: Use the first-and-last-term sum form \( S_n = \frac{n}{2}(a + l) \), which suits a sum written between its first and last terms.
\[ S_{25} = \frac{25}{2}(4 + 100) = \frac{25}{2} \times 104 = 25 \times 52 = 1300 \]
Final answer: the sum is 1300.
Common Mistakes to Avoid
Each row pairs the error with the correct rule and a quick check you can run on your own answer.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Finding \( d \) with the subtraction reversed — for 6, 3, 0 writing \( d = 6 – 3 = 3 \). | Always subtract the earlier term from the later term: \( d = a_{k+1} – a_k \), so for the list above \( d = 3 – 6 = -3 \). | Rebuild the list from your \( d \): start at the first term and add \( d \) repeatedly; the signs of the terms must match the given list. |
| Dropping a negative common difference inside the sum formula — for \( a = 24, d = -3 \) writing \( [2a + (n – 1) \times 3] \). | Keep \( d \) with its sign: \( 2a + (n – 1)(-3) \). | Test with a small \( n \): add the first two or three terms by hand and compare with the formula’s output. |
| Accepting a fractional \( n \) as a valid term position. | When \( a_n = a + (n – 1)d \) is solved for \( n \), the answer must be a positive integer; a fraction means the value is not a term of this AP. | Put your \( n \) back into \( a + (n – 1)d \) — it must reproduce the given value exactly. |
| Using \( S_n = \frac{n}{2}(a + l) \) with an \( l \) that is not the last of the \( n \) terms being added. | \( l \) must be the \( n \)th term of exactly the portion being summed, i.e. \( l = a + (n – 1)d \) for the same \( n \). | Confirm \( l = a + (n – 1)d \) with the \( n \) you used before writing the answer. |
Frequently Asked Questions
Does the sum formula work for an odd number of terms?
Yes. The derivation writes the sum forwards and backwards, adds the two rows, and collects \( n \) equal columns, each \( 2a + (n – 1)d \), before halving. That works for odd and even \( n \) alike, so \( S_n = \frac{n}{2}[2a + (n – 1)d] \) holds for every \( n \geq 1 \).
When do I use \( S_n = \frac{n}{2}(a + l) \) instead of \( S_n = \frac{n}{2}[2a + (n – 1)d] \)?
Use the \( a + l \) form when the first and last terms of the sum are given and the common difference is not needed. It is the same formula with \( l = a + (n – 1)d \) substituted in — both forms give the same number.
How do I check whether a number is a term of an AP?
Set \( a_n \) equal to that number and solve \( a + (n – 1)d = \text{number} \) for \( n \). If \( n \) is a positive integer, the number is a term; if \( n \) is a fraction, it is not.
Is 3, 3, 3, 3 an AP?
Yes. Each term is obtained by adding 0 to the preceding term, so the list is an AP with first term 3 and common difference \( d = 0 \).
For other chapters, browse the Maths formulas library.
Reference: NCERT Class 10 Mathematics textbook, chapter Arithmetic Progressions.