Sequences & Series

Arithmetic, geometric, harmonic progressions, special series, method of differences, and powerful AM-GM-HM inequalities — a JEE favourite with high weightage.

AP GP HP & AGP Special Series Method of Differences AM-GM-HM

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Arithmetic Progression (AP)

An AP is a sequence where the difference between consecutive terms is constant. It is the most fundamental progression tested in JEE.

◆ nth Term & Sum of AP

Definition — Arithmetic Progression A sequence \(a_1, a_2, a_3, \ldots\) is an AP if \(a_{n+1} - a_n = d\) (constant) for all \(n \geq 1\). The common difference is \(d\).

Formulae

nth term: \(a_n = a + (n-1)d\)
Sum of first \(n\) terms: \(S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)\) where \(l = a_n\) is the last term
Arithmetic Mean: If \(a, b, c\) are in AP, then \(b = \frac{a+c}{2}\)

★ Example

The sum of the first \(n\) terms of an AP is \(S_n = 3n^2 + 5n\). Find the common difference.

\(a_n = S_n - S_{n-1} = 3n^2 + 5n - 3(n-1)^2 - 5(n-1) = 6n + 2\).
So \(d = a_2 - a_1 = 14 - 8 = 6\).
Shortcut: If \(S_n = An^2 + Bn\), then \(d = 2A = 6\).

◆ Inserting Arithmetic Means

Insertion of \(n\) AMs between \(a\) and \(b\) If \(n\) arithmetic means \(A_1, A_2, \ldots, A_n\) are inserted between \(a\) and \(b\), then \(d = \frac{b-a}{n+1}\) and \(A_k = a + k \cdot \frac{b-a}{n+1}\). Also, \(\sum_{k=1}^{n} A_k = \frac{n(a+b)}{2}\).

★ Example

Insert 4 AMs between 3 and 23.

\(d = \frac{23 - 3}{4 + 1} = 4\). The AMs are \(7, 11, 15, 19\).

Geometric Progression (GP)

A GP is a sequence where each term is obtained by multiplying the previous term by a fixed ratio. Infinite GP sums appear frequently in JEE.

◆ nth Term, Finite & Infinite Sum

Definition — Geometric Progression A sequence \(a, ar, ar^2, \ldots\) where \(r \neq 0\) is the common ratio. The nth term is \(a_n = ar^{n-1}\).

Sum Formulae

Finite sum: \(S_n = a\,\frac{1-r^n}{1-r}\) for \(r \neq 1\); \(S_n = na\) for \(r=1\)
Infinite sum (\(|r|<1\)): \(S_\infty = \frac{a}{1-r}\)
Geometric Mean: If \(a, b, c\) are in GP, then \(b^2 = ac\)

★ Example

Find the sum to infinity: \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\)

Here \(a = 1\), \(r = \frac{1}{3}\). Since \(|r| < 1\): \[S_\infty = \frac{1}{1 - 1/3} = \frac{3}{2}\]

◆ Properties & JEE Tricks

Useful Properties

For 3 terms in GP, take them as \(\frac{a}{r}, a, ar\) (product = \(a^3\)).
For 4 terms, take \(\frac{a}{r^3}, \frac{a}{r}, ar, ar^3\) (product = \(a^4\)).
If each term of a GP is raised to power \(k\), the result is again a GP with ratio \(r^k\).
The product of terms equidistant from the beginning and end is constant: \(a_k \cdot a_{n+1-k} = a_1 \cdot a_n\).

★ Example

Three numbers are in GP. Their sum is 21 and their product is 216. Find them.

Let the terms be \(\frac{a}{r}, a, ar\). Product \(= a^3 = 216 \Rightarrow a = 6\).
Sum \(= \frac{6}{r} + 6 + 6r = 21 \Rightarrow \frac{6}{r} + 6r = 15 \Rightarrow 6r^2 - 15r + 6 = 0\).
\(2r^2 - 5r + 2 = 0 \Rightarrow r = 2\) or \(r = \frac{1}{2}\).
The numbers are 3, 6, 12 (or 12, 6, 3).

HP & Arithmetico-Geometric Progression (AGP)

HP is the reciprocal of AP, while AGP combines AP and GP. The infinite sum of AGP is a classic JEE Advanced problem.

◆ Harmonic Progression

Definition — HP A sequence \(a_1, a_2, \ldots\) is in HP if \(\frac{1}{a_1}, \frac{1}{a_2}, \ldots\) is in AP. There is no direct formula for the sum of an HP.

Harmonic Mean The HM of \(a\) and \(b\) is \(H = \frac{2ab}{a+b}\). If \(n\) HMs are inserted between \(a\) and \(b\), take the reciprocals, insert AMs, then take reciprocals again.

◆ Arithmetico-Geometric Progression

Definition — AGP A series of the form \(\sum_{k=1}^{n}(a + (k-1)d)\,r^{k-1}\) where the AP part is \(a + (k-1)d\) and the GP part is \(r^{k-1}\).

Sum of Infinite AGP (\(|r| < 1\)) \[S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}\] Method: Write \(S\), multiply by \(r\) to get \(rS\), subtract to reduce the AP part to a constant-difference GP.

★ Example

Find \(S = 1 + 3 \cdot \frac{1}{2} + 5 \cdot \frac{1}{4} + 7 \cdot \frac{1}{8} + \cdots\)

Here \(a = 1\), \(d = 2\), \(r = \frac{1}{2}\).
\(S = \frac{1}{1-\frac{1}{2}} + \frac{2 \cdot \frac{1}{2}}{(1-\frac{1}{2})^2} = 2 + \frac{1}{1/4} = 2 + 4 = 6\).
Alternatively (multiply-subtract method):
\(S = 1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \cdots\)
\(\frac{S}{2} = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \cdots\)
\(S - \frac{S}{2} = 1 + 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + \cdots \Rightarrow \frac{S}{2} = 1 + \frac{1/2}{1-1/2} = 1 + 1 = 2 + 1 = 3 \Rightarrow S = 6\).

Special Series & Telescoping

Standard summation formulae and telescoping techniques are essential for JEE. Memorise these results — they save crucial time.

◆ Standard Summation Formulae

Power Sum Formulae

\(\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)
\(\displaystyle\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)
\(\displaystyle\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2\) (equals square of \(\sum k\))

★ Example

Find \(\sum_{k=1}^{20}(2k-1)^2\).

\(\sum_{k=1}^{20}(2k-1)^2 = \sum_{k=1}^{20}(4k^2 - 4k + 1) = 4 \cdot \frac{20 \cdot 21 \cdot 41}{6} - 4 \cdot \frac{20 \cdot 21}{2} + 20\) \(= 4 \cdot 2870 - 4 \cdot 210 + 20 = 11480 - 840 + 20 = \mathbf{10660}\).
Shortcut: Sum of first \(n\) odd squares \(= \frac{n(2n-1)(2n+1)}{3} = \frac{20 \cdot 39 \cdot 41}{3} = \frac{31980}{3} = 10660\).

◆ Telescoping Series

Definition — Telescoping A series where consecutive terms cancel, leaving only the first and last terms. The key technique is partial fraction decomposition.

★ Example

Find \(\displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)}\).

Using partial fractions: \(\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\).
\(S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}\).

Method of Differences

When the nth term of a series is not directly identifiable, use the method of differences to find a pattern in \(a_{n} - a_{n-1}\).

◆ Finding nth Term via Differences

Method If \(S_n = a_1 + a_2 + \cdots + a_n\), compute the first differences \(b_k = a_{k+1} - a_k\). If \(\{b_k\}\) forms an AP or GP, sum it to find \(a_n\), then sum \(a_n\) to get \(S_n\).

★ Example

Find the sum: \(3 + 7 + 13 + 21 + 31 + \cdots\) to \(n\) terms.

Differences: \(4, 6, 8, 10, \ldots\) — this is an AP with first term 4 and \(d = 2\).
For \(n \geq 2\): \(a_n = a_1 + \sum_{k=1}^{n-1}b_k = 3 + \sum_{k=1}^{n-1}(2k+2)\)
\(= 3 + 2 \cdot \frac{(n-1)n}{2} + 2(n-1) = 3 + n^2 - n + 2n - 2 = n^2 + n + 1\).
Check: \(a_1 = 1 + 1 + 1 = 3\) ✓. So:
\(S_n = \sum_{k=1}^{n}(k^2 + k + 1) = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n = \frac{n(n^2 + 3n + 5)}{3}\).

AM-GM-HM Inequality & Applications

The AM-GM-HM inequality chain is one of the most powerful tools in JEE for optimization and bounding problems.

◆ The Inequality Chain

AM ≥ GM ≥ HM For positive reals \(a_1, a_2, \ldots, a_n\): \[\underbrace{\frac{a_1 + a_2 + \cdots + a_n}{n}}_{\text{AM}} \geq \underbrace{(a_1 \cdot a_2 \cdots a_n)^{1/n}}_{\text{GM}} \geq \underbrace{\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}}_{\text{HM}}\] Equality holds iff \(a_1 = a_2 = \cdots = a_n\).

★ Example

If \(x > 0\), find the minimum value of \(x + \frac{9}{x}\).

By AM-GM: \(x + \frac{9}{x} \geq 2\sqrt{x \cdot \frac{9}{x}} = 2 \cdot 3 = 6\).
Equality when \(x = \frac{9}{x} \Rightarrow x = 3\).
Minimum value = 6 at \(x = 3\).

◆ Weighted AM-GM & JEE Applications

Weighted AM-GM For positive reals \(a_i\) with positive weights \(w_i\) where \(\sum w_i = 1\): \[w_1 a_1 + w_2 a_2 + \cdots + w_n a_n \geq a_1^{w_1} a_2^{w_2} \cdots a_n^{w_n}\]

★ Example

Find the minimum value of \(2x + 3y + \frac{6}{xy}\) for \(x, y > 0\).

By AM-GM on three terms: \[2x + 3y + \frac{6}{xy} \geq 3\left(2x \cdot 3y \cdot \frac{6}{xy}\right)^{1/3} = 3(36)^{1/3} = 3 \cdot 6^{2/3}\] Wait — let's be more careful. \(2x \cdot 3y \cdot \frac{6}{xy} = 36\). So AM-GM gives: \[\frac{2x + 3y + \frac{6}{xy}}{3} \geq (36)^{1/3} = 6^{2/3}\] Minimum \(= 3 \cdot 6^{2/3}\) when \(2x = 3y = \frac{6}{xy}\).
From \(2x = 3y\): \(x = \frac{3y}{2}\). From \(3y = \frac{6}{xy} = \frac{6}{\frac{3y}{2} \cdot y} = \frac{4}{y}\): \(3y^2 = 4\), \(y = \frac{2}{\sqrt{3}}\), \(x = \frac{\sqrt{3}}{1} = \sqrt{3}\).
Minimum \(= 3 \sqrt[3]{36} = 3 \cdot 6^{2/3}\).

★ Key Takeaways

If \(S_n = An^2 + Bn\), then \(d = 2A\) and \(a_1 = A + B\) — instant AP identification.
For 3 unknowns in GP, always assume \(\frac{a}{r}, a, ar\) to use the product constraint directly.
Infinite AGP sum: \(S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}\). Derive it via the multiply-subtract method.
Telescoping is the go-to technique for sums like \(\sum\frac{1}{k(k+1)\cdots(k+m)}\).
AM-GM with equality condition gives both the extremum and the point where it is attained — essential for JEE optimization.

📝 Practice Problems

Problem 1

The 4th and 10th terms of a GP are \(\frac{1}{3}\) and \(\frac{243}{1}\) respectively. Find the 2nd term.

Show Solution ▼

\(ar^3 = \frac{1}{3}\) and \(ar^9 = 243\). Dividing: \(r^6 = 729 = 3^6 \Rightarrow r = 3\). Then \(a \cdot 27 = \frac{1}{3} \Rightarrow a = \frac{1}{81}\). Second term \(= ar = \frac{1}{81} \cdot 3 = \frac{1}{27}\).

Problem 2

Find the sum \(\displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}\).

Show Solution ▼

Using partial fractions: \(\frac{1}{k(k+1)(k+2)} = \frac{1}{2}\left[\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}\right]\). This telescopes to \(\frac{1}{2}\left[\frac{1}{1 \cdot 2} - \frac{1}{(n+1)(n+2)}\right] = \frac{1}{4} - \frac{1}{2(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}\).

Problem 3

If \(a, b, c\) are in AP and \(a^2, b^2, c^2\) are in GP, find the common ratio of the GP given \(a < b < c\) and \(a + b + c = \frac{3}{2}\).

Show Solution ▼

Since \(a, b, c\) are in AP: \(b = \frac{1}{2}\), \(a = \frac{1}{2} - d\), \(c = \frac{1}{2} + d\). Since \(a^2, b^2, c^2\) are in GP: \(b^4 = a^2 c^2\), i.e., \(\frac{1}{16} = (\frac{1}{4} - d^2)^2\). So \(\frac{1}{4} - d^2 = \pm\frac{1}{4}\). Taking \(d^2 = \frac{1}{2}\) (since \(a < b < c\), \(d > 0\)): \(d = \frac{1}{\sqrt{2}}\). Common ratio \(= \frac{c^2}{b^2} = \frac{(\frac{1}{2}+\frac{1}{\sqrt{2}})^2}{1/4}\). After simplification, the common ratio is \((1+\sqrt{2})^2 = 3 + 2\sqrt{2}\).

Problem 4

Find the sum: \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1)\).

Show Solution ▼

\(\sum_{k=1}^{n}k(k+1) = \sum k^2 + \sum k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6}(2n+1+3) = \frac{n(n+1)(n+2)}{3}\).

Problem 5

For positive reals \(a, b, c\) with \(abc = 8\), find the minimum value of \((1+a)(1+b)(1+c)\).

Show Solution ▼

By AM-GM: \(1 + a \geq 2\sqrt{a}\), \(1 + b \geq 2\sqrt{b}\), \(1 + c \geq 2\sqrt{c}\). Multiplying: \((1+a)(1+b)(1+c) \geq 8\sqrt{abc} = 8\sqrt{8} = 16\sqrt{2}\). Equality when \(a = b = c = 2\). Minimum = \(27\). Wait — let's recalculate with \(a = b = c = 2\): \((3)(3)(3) = 27\). Using AM-GM more carefully on each factor: \(1 + a = 1 + a \geq 2\sqrt{a}\) gives \(3 \geq 2\sqrt{2}\), which is not tight. So minimum by substitution is indeed \(\mathbf{27}\) at \(a=b=c=2\).

🎯 Interactive Quiz

1. The sum of an infinite GP is 4 and the sum of the cubes of its terms is 192. The common ratio is:

A \(\frac{1}{2}\)

B \(-\frac{1}{2}\)

C \(\frac{1}{4}\)

D \(\frac{2}{3}\)

2. If the AM and GM of two positive numbers are 10 and 8 respectively, then the HM is:

A 6

B 6.4

C 8

D 9

3. The value of \(\displaystyle\sum_{k=1}^{10} k \cdot 2^k\) is:

A 10230

B 2046

C 18434

D 20480

4. If the pth, qth, rth terms of an AP are \(a, b, c\) respectively, then \(a(q-r) + b(r-p) + c(p-q)\) equals:

A 0

B 1

C \(abc\)

D \(a + b + c\)

5. The minimum value of \(\frac{(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)}{abc}\) where \(a, b, c > 0\) is:

A 8

B 9

C 27

D 64