JEE Main Maths Sequence & Series 2027: AP, GP, Infinite Series & AM-GM

A bouncing ball that never quite settles, a chessboard doubling in grains of rice: both are geometric series in disguise. This is a short, formula-driven chapter, and once the AP and GP toolkit is automatic, most JEE Main questions turn into direct substitution.

An infinite geometric series only converges when the common ratio is smaller than one in size.

---

Chapter at a Glance

Priority reflects past-paper trends, not an official NTA weightage.

What the NTA Syllabus Covers

• Arithmetic and geometric progressions
• Insertion of arithmetic and geometric means between two given numbers
• Relation between A.M and G.M

---

Master These Topics

1. Arithmetic Progression

For an AP with first term $a$ and common difference $d$:

$$a_n = a + (n-1)d \qquad S_n = \frac{n}{2}\big(2a + (n-1)d\big) = \frac{n}{2}(a + l)$$

where $l$ is the last term. The $n$th term from the end of a finite AP is $l - (n-1)d$.

Worked example: For the AP $3, 8, 13, \ldots$ with $a = 3$, $d = 5$, the sum of the first 20 terms is $S_{20} = \dfrac{20}{2}\big(6 + 19 \cdot 5\big) = 10 \times 101 = 1010$.

2. Geometric Progression and Infinite Series

$$a_n = ar^{n-1} \qquad S_n = \frac{a(r^n - 1)}{r - 1} \; (r \ne 1) \qquad S_\infty = \frac{a}{1 - r} \; (\lvert r \rvert \lt 1)$$

Worked example: For $a = 8$ and $r = \tfrac{1}{2}$, the series $8 + 4 + 2 + 1 + \cdots$ has $S_\infty = \dfrac{8}{1 - 1/2} = 16$.

Trap: The infinite-sum formula only applies when $\lvert r \rvert \lt 1$. For $r \ge 1$, the series does not converge, and using the formula anyway gives a meaningless answer.

3. Inserting Means Between Two Numbers

Arithmetic means: to insert $n$ AMs between $a$ and $b$, the common difference is $d = \dfrac{b - a}{n + 1}$.

Worked example: Insert 3 AMs between 2 and 18. Here $d = \dfrac{18 - 2}{4} = 4$, so the means are $6, 10, 14$.

Geometric means: to insert $n$ GMs between $a$ and $b$ (both positive), the common ratio is $r = \left(\dfrac{b}{a}\right)^{1/(n+1)}$.

Worked example: Insert 2 GMs between 2 and 16. Here $r = (8)^{1/3} = 2$, so the means are $4, 8$.

4. The AM-GM Relation

For two positive numbers $a$ and $b$, the single AM is $A = \dfrac{a+b}{2}$ and the single GM is $G = \sqrt{ab}$, and

$$A \ge G$$

with equality exactly when $a = b$. A useful fact: $a$ and $b$ are the two roots of $x^2 - 2Ax + G^2 = 0$.

Worked example: For $a = 4$ and $b = 9$, $A = 6.5$ and $G = 6$, confirming $A \ge G$. Solving $x^2 - 13x + 36 = 0$ gives back $x = 4, 9$.

Trap: The AM-GM relation applies to positive real numbers. Applying it carelessly to negative numbers can give a false inequality.

---

Common Traps to Avoid

• Using $S_\infty = a/(1-r)$ when $\lvert r \rvert \ge 1$.
• Confusing $n$ terms inserted with $n + 1$ gaps when finding $d$ or $r$.
• Mixing the nth term formula of an AP with that of a GP.
• Applying the AM-GM relation to non-positive numbers.

60-Second Revision Sheet

• AP: $a_n = a + (n-1)d$, $S_n = \tfrac{n}{2}(2a + (n-1)d)$
• GP: $a_n = ar^{n-1}$, $S_n = \tfrac{a(r^n-1)}{r-1}$, $S_\infty = \tfrac{a}{1-r}$ for $\lvert r \rvert \lt 1$
• Insert $n$ AMs: $d = \tfrac{b-a}{n+1}$; insert $n$ GMs: $r = (b/a)^{1/(n+1)}$
• $A \ge G$ for positive numbers, equality when $a = b$

Your Study Plan

1. Day 1: AP nth term, sum and word problems.
2. Day 2: GP nth term, sum and sum to infinity.
3. Day 3: inserting arithmetic and geometric means.
4. Day 4: AM-GM relation and a timed mixed set.

Practice Sequence & Series Questions Free → (opens in a new tab)

---

Continue Your Maths Journey

• Previous chapter: Binomial Theorem
• Next chapter: Limit, Continuity & Differentiability
• All 14 JEE Main Maths chapters
• Complete JEE Main Syllabus 2027 guide

---

Frequently Asked Questions