Arithmetic Progressions

What is an arithmetic progression?

An arithmetic progression (AP) is a list of numbers in which each term differs from the one before it by the same amount, the common difference $d$:

It only counts as an AP if every consecutive pair has that same difference, not just the first pair. $d$ can be negative (24, 21, 18, …) or zero.

Worked example: does a situation form an AP?

Which of these produce an arithmetic progression?

• Taxi fare after each km, at ₹20 for the first km and ₹8 for each further km: ₹20, ₹28, ₹36, …. A constant difference of 8, so it is an AP.
• Air left in a cylinder when a pump removes $\tfrac{1}{4}$ of what remains each time. Each step keeps $\tfrac{3}{4}$ of the previous amount, a constant ratio, so the gaps shrink. It is not an AP (it's a geometric progression).
• Cost of digging a well, at ₹150 for the first metre and ₹50 more for each further metre: ₹150, ₹200, ₹250, …. It is an AP with $d = 50$.
• ₹10,000 at 8% compound interest. Each year's amount is 1.08 times the last, a constant ratio again. It is not an AP.

The nth term

Starting from $a$ and adding $d$ each time, the $n$th term has had $d$ added $n - 1$ times:

Worked example. In 7, 11, 15, …, $a = 7$ and $d = 4$. The 20th term is $7 + 19 \times 4 = 83$.

Which term is 135? Solve $7 + (n-1)4 = 135$: $n - 1 = 32$, so 135 is the 33rd term.

Is 300 a term? $7 + (n-1)4 = 300$ gives $n - 1 = 73.25$. That isn't a whole number, so 300 is not in the list.

The sum of the first n terms

Here $l$ is the last term. The second form is Gauss's trick: pair the first and last terms, the second and second-last, and so on. Every pair has the same sum.

Worked example. The first 20 terms of 7, 11, 15, … end at 83, so $S_{20} = \dfrac{20}{2}(7 + 83) = 900$. The first 100 natural numbers add to $\dfrac{100}{2}(1 + 100) = 5050$.

Worked example: how many terms give a sum?

How many terms of 24, 21, 18, … must be added to get 78?

Solution: With $a = 24$ and $d = -3$:

So $n = 4$ or $n = 13$. Both are right: the first 4 terms give $24 + 21 + 18 + 15 = 78$, and terms 5 to 13 (12 down to −12) add to zero, so the first 13 terms also total 78.

Worked example: three numbers in AP, given sum and product

Find three numbers in AP whose sum is 15 and product is 80.

Solution: Centre them on the middle term: $a - d$, $a$, $a + d$. The sum is $3a = 15$, so $a = 5$. The product is $a(a^2 - d^2) = 5(25 - d^2) = 80$, so $d^2 = 9$ and $d = \pm 3$. Either sign gives the same numbers: 2, 5 and 8.

Common mistakes

• Using $a + nd$ for the nth term. The first term has had no $d$ added, so it's $a + (n-1)d$.
• Checking only the first difference. 2, 4, 8 starts with a difference of 2 but isn't an AP.
• Rejecting a valid answer. When solving for $n$ from a sum, both roots can be valid, as with 4 and 13 above. Reject a root only if it isn't a positive whole number.
• Mixing up $S_n$ and $a_n$. $a_n$ is one term; $S_n$ is the total of the first $n$.

Practice questions

1. Find the 25th term of 3, 8, 13, ….
2. Find the sum of the first 15 terms of 2, 5, 8, ….
3. How many three-digit numbers are multiples of 3, and what is their sum?
4. Which is the first term of −5, −1, 3, … that is greater than 50?

Answers

1. $3 + 24 \times 5 = 123$.
2. $\dfrac{15}{2}\left[4 + 14 \times 3\right] = \dfrac{15}{2} \times 46 = 345$.
3. They run 102, 105, …, 999, which is $\dfrac{999 - 102}{3} + 1 = 300$ numbers. Their sum is $\dfrac{300}{2}(102 + 999) = 165{,}150$.
4. $-5 + (n-1)4 > 50$ needs $n - 1 > 13.75$, so $n = 15$: the 15th term, 51.

Where this comes up in exams

The 2025-26 CBSE Class X syllabus has a chapter on arithmetic progressions, covering the nth term and the sum of the first n terms, and JEE Main 2026 lists sequences and series, arithmetic progressions included.

Checked against CBSE Mathematics curriculum, Classes IX-X (2025-26) and JEE (Main) 2026 syllabus (NTA). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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