Sequences and Series – JEE Main Maths Formula Sheet & Class 11 Notes | AP, GP, HP, AGP, AM-GM-HM Formulas

Definitions — Sequence, Progression, and Series

Why Precise Definitions Underpin All Sequences and Series JEE Main Questions

A sequence in Sequences and Series is a function from the set of natural numbers ℕ to the set of real numbers ℝ (or complex numbers ℂ): each natural number n maps to a term tₙ. The sequence t₁, t₂, t₃, …, tₙ is finite if it has a finite number of terms, infinite otherwise. The general term tₙ (also written Tₙ or aₙ) represents the nth term. If tₙ can be expressed as an explicit function of n, the sequence is well-defined (e.g., tₙ = 2n gives 2, 4, 6, 8, …).

A progression is a sequence in which terms follow a definite pattern: each successive term is obtained by a rule applied to the preceding term. Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP) are the three standard progression types in Sequences and Series JEE Main.

A series in Sequences and Series is formed by adding the terms of a sequence: S = t₁ + t₂ + t₃ + … + tₙ. The partial sum Sₙ = t₁+t₂+…+tₙ is the sum of the first n terms. Crucially: tₙ = Sₙ – Sₙ₋₁ (for n ≥ 2; t₁ = S₁) — this formula converts between a given Sₙ and the individual terms of the sequence. If Sₙ is quadratic in n → the sequence is an AP. If Sₙ is a polynomial of degree k in n, the kth differences of the sequence are constant.

Arithmetic Progression (AP) — All Formulas for JEE Main

Why AP Formulas Are the Foundation of All Sequences and Series JEE Main Questions

An Arithmetic Progression (AP) in Sequences and Series is a sequence in which consecutive terms differ by a constant amount d (common difference): a, a+d, a+2d, a+3d, …. The AP is determined by its first term a and common difference d = aₙ – aₙ₋₁.

Key AP formulas in Sequences and Series (JEE Main):

nth term of AP: aₙ = a + (n–1)d = l (last term)

nth term from end of AP: aᵣ (rth from end) = l – (r–1)d = a + (n–r)d = (n–r+1)th term from beginning

Sum of first n terms of AP: Sₙ = n/2 · [2a + (n–1)d] = n/2 · [a + l] = n/2 · (first + last)

Middle term of AP: If n is odd, middle term = a₍ₙ₊₁₎/₂; if n is even, two middle terms = aₙ/₂ and aₙ/₂₊₁ with average = (a+l)/2.

aₙ from Sₙ: aₙ = Sₙ – Sₙ₋₁ (for n ≥ 2); a₁ = S₁.

Properties of AP in Sequences and Series (JEE Main):

1. If a₁, a₂, …, aₙ are in AP: (a) a₁±k, a₂±k, …, aₙ±k are in AP (adding constant preserves AP). (b) ka₁, ka₂, …, kaₙ are in AP (multiplying by constant preserves AP). (c) a₁/k, a₂/k, …, aₙ/k are in AP.

2. Sum of two APs term-by-term is also an AP. Difference of two APs term-by-term is also an AP.

3. Equidistant terms sum: a₁+aₙ = a₂+aₙ₋₁ = a₃+aₙ₋₂ = … = 2a + (n–1)d. In general: aᵣ₋ₖ + aᵣ₊ₖ = 2aᵣ (symmetric about rth term).

4. If nth term of a sequence is linear in n → sequence is AP.

5. If Sₙ is quadratic in n → sequence is AP.

6. Three numbers a, b, c are in AP iff a+c = 2b (b is the arithmetic mean).

Choosing standard terms in AP (Sequences and Series JEE Main shortcut):

3 numbers in AP: a–d, a, a+d (sum = 3a; product involves a²–d²)

4 numbers in AP: a–3d, a–d, a+d, a+3d (sum = 4a; easier when sum is given)

5 numbers in AP: a–2d, a–d, a, a+d, a+2d (sum = 5a)

This choice simplifies equations when sum and product are both given.

Arithmetic Mean (AM) in Sequences and Series: AM of two numbers a and b = (a+b)/2. AM of n numbers a₁, a₂, …, aₙ = (a₁+a₂+…+aₙ)/n. Insertion of n AMs between a and b: common difference d = (b–a)/(n+1); kth AM = a + k(b–a)/(n+1). Sum of n AMs = n·(a+b)/2. Download the Free PDF for Sequences and Series for all AP formula examples for JEE Main.

Geometric Progression (GP) — All Formulas for JEE Main

Why GP Formulas in Sequences and Series Are Among the Most Tested in JEE Main

A Geometric Progression (GP) in Sequences and Series is a sequence of non-zero terms in which every term bears a constant ratio r to its preceding term: a, ar, ar², ar³, …. Common ratio r = aₙ/aₙ₋₁.

Key GP formulas in Sequences and Series (JEE Main):

nth term of GP: aₙ = arⁿ⁻¹ = l (last term)

nth term of GP from end: = l/rⁿ⁻¹ = l·(1/r)ⁿ⁻¹. rth term from end = a·r^(n–r) = (n–r+1)th term from beginning.

Sum of first n terms of GP (Sₙ): Sₙ = a(rⁿ–1)/(r–1) = a(1–rⁿ)/(1–r) for r ≠ 1; Sₙ = na for r = 1.

Sum of infinite GP (S∞): S∞ = a/(1–r) for |r| < 1. Diverges if |r| ≥ 1.

Also: S∞ = first term / (1 – common ratio) = a/(1–r) [valid only when |r| < 1].

Properties of GP in Sequences and Series (JEE Main):

1. Multiplying or dividing all terms of a GP by a non-zero constant gives another GP (same ratio r).

2. If aₙ are in GP and all aᵢ > 0, then log aₙ are in AP. Converse also holds.

3. Reciprocals of a GP form a GP with ratio 1/r.

4. Product of two GPs term-by-term is a GP; quotient of two GPs term-by-term is a GP.

5. Equidistant terms in GP: a₁·aₙ = a₂·aₙ₋₁ = a₃·aₙ₋₂ = a² · r^(n–1). In general: aᵣ₋ₖ · aᵣ₊ₖ = aᵣ² (product of equidistant terms = square of middle term).

6. If a₁, a₂, …, a₂ₘ₊₁ are in GP, the product of all terms = aₘ₊₁²ᵐ⁺¹ (middle term raised to total power).

Choosing standard terms in GP (Sequences and Series JEE Main shortcut):

3 numbers in GP: a/r, a, ar (product = a³, easier when product is given)

4 numbers in GP: a/r³, a/r, ar, ar³ (product = a⁴)

5 numbers in GP: a/r², a/r, a, ar, ar² (product = a⁵)

Geometric Mean (GM) in Sequences and Series: GM of two positive numbers a and b = √(ab). GM of n numbers = (a₁a₂…aₙ)^(1/n). Insertion of n GMs between a and b: common ratio R = (b/a)^(1/(n+1)); kth GM = a·(b/a)^(k/(n+1)). Product of n GMs = (ab)^(n/2). Download the Free PDF for Sequences and Series for all GP examples and infinite GP applications for JEE Main.

Harmonic Progression (HP) and Harmonic Mean — Sequences and Series Formulas

Why HP and HM Are the Most Misunderstood Sequences and Series JEE Main Topics

A Harmonic Progression (HP) in Sequences and Series is a sequence whose reciprocals form an Arithmetic Progression. So if a₁, a₂, …, aₙ are in HP, then 1/a₁, 1/a₂, …, 1/aₙ are in AP. There is no direct sum formula for HP — all HP problems in Sequences and Series are solved by converting to the corresponding AP of reciprocals.

nth term of HP (Sequences and Series formula): If 1/a₁ = a (first term of AP) and common difference of AP = d, then 1/aₙ = a + (n–1)d → aₙ = 1/[a + (n–1)d]. Common difference of the reciprocal AP: d = 1/a₂ – 1/a₁ = (a₁–a₂)/(a₁a₂).

Harmonic Mean (HM) in Sequences and Series: HM of two positive numbers a and b = 2ab/(a+b). HM is the reciprocal of the AM of the reciprocals: 1/HM = (1/a + 1/b)/2 → HM = 2ab/(a+b). HM of n numbers a₁, a₂, …, aₙ = n / (1/a₁ + 1/a₂ + … + 1/aₙ).

Insertion of n Harmonic Means between a and b in Sequences and Series: a, H₁, H₂, …, Hₙ, b are in HP, so 1/a, 1/H₁, …, 1/Hₙ, 1/b are in AP. Common difference of this AP: d = (1/b – 1/a)/(n+1) = (a–b)/[ab(n+1)]. kth Harmonic Mean: 1/Hₖ = 1/a + k(1/b – 1/a)/(n+1) = 1/a + k(a–b)/[ab(n+1)·(–1)]... → Hₖ = ab(n+1) / [ka + (n+1–k)b].

Three numbers in HP (Sequences and Series): a, b, c are in HP iff 1/a, 1/b, 1/c are in AP iff 1/a + 1/c = 2/b iff b = 2ac/(a+c). For three numbers to be both in AP and HP simultaneously → only if all three are equal. Download the Free PDF for Sequences and Series for HP and HM worked examples for JEE Main.

AM–GM–HM Inequality — The Most Applied Sequences and Series Result

Why AM ≥ GM ≥ HM Is the Most-Used Inequality in JEE Main Sequences and Series

For any two positive real numbers a and b, the three means satisfy: AM ≥ GM ≥ HM, with equality if and only if a = b.

AM-GM inequality for two numbers (Sequences and Series — JEE Main proof):

AM = (a+b)/2; GM = √(ab); HM = 2ab/(a+b).

AM ≥ GM: (a+b)/2 ≥ √(ab) ← from (√a – √b)² ≥ 0 → a + b – 2√(ab) ≥ 0 → (a+b)/2 ≥ √(ab). Equality iff a = b.

GM ≥ HM: √(ab) ≥ 2ab/(a+b) ← multiply both sides by (a+b)/ab: (a+b)/√(ab) ≥ 2 ← same as AM≥GM for 1/a and 1/b. Equality iff a = b.

Important relation: AM × HM = GM² for two positive numbers. This is a direct formula: (a+b)/2 × 2ab/(a+b) = ab = (√ab)² = GM².

AM-GM inequality for n numbers (Sequences and Series — JEE Main formula):

(a₁+a₂+…+aₙ)/n ≥ (a₁a₂…aₙ)^(1/n) for all positive aᵢ, equality iff a₁ = a₂ = … = aₙ.

Equivalently: AM ≥ GM → (a₁+a₂+…+aₙ) ≥ n·(a₁a₂…aₙ)^(1/n).

Applications of AM-GM inequality in Sequences and Series JEE Main:

Minimum of (x + 1/x) for x > 0: by AM-GM, x + 1/x ≥ 2√(x·1/x) = 2. Minimum = 2 when x = 1.

Minimum of (x² + 1/x²): x² + 1/x² = (x + 1/x)² – 2 ≥ 4 – 2 = 2. Minimum = 2 when x = 1.

Minimum of a·x + b/x (a,b,x > 0): by AM-GM, a·x + b/x ≥ 2√(ab). Minimum = 2√(ab) when ax = b/x → x = √(b/a).

Minimum/maximum of a product with constraint sum: if a+b = constant = s, then ab ≤ (s/2)² → maximum ab = s²/4 when a=b=s/2.

If sum of n positive numbers is fixed → product is maximum when all are equal (AM-GM equality condition).

If product of n positive numbers is fixed → sum is minimum when all are equal.

Weighted AM-GM in Sequences and Series: For positive a₁, a₂, …, aₙ with positive weights w₁, w₂, …, wₙ (w₁+w₂+…+wₙ = 1): w₁a₁ + w₂a₂ + … + wₙaₙ ≥ a₁^w₁ · a₂^w₂ · … · aₙ^wₙ. Download the Free PDF for Sequences and Series for AM-GM applications and optimisation examples for JEE Main.

Arithmetic-Geometric Progression (AGP) — Sequences and Series Formula

Why AGP Sum Formula Is a Unique High-Value JEE Main Sequences and Series Result

An Arithmetic-Geometric Progression (AGP) in Sequences and Series is a sequence where each term is the product of corresponding terms of an AP and a GP. If a, (a+d), (a+2d), … is an AP and b, br, br², … is a GP, then the corresponding AGP is: ab, (a+d)br, (a+2d)br², ….

nth term of AGP (Sequences and Series formula): Tₙ = [a + (n–1)d] · brⁿ⁻¹

Sum of n terms of AGP (Sequences and Series — JEE Main derivation by multiply-and-subtract):

Let Sₙ = ab + (a+d)br + (a+2d)br² + … + [a+(n–1)d]brⁿ⁻¹

Multiply by r: r·Sₙ = abr + (a+d)br² + … + [a+(n–2)d]brⁿ⁻¹ + [a+(n–1)d]brⁿ

Subtract: (1–r)Sₙ = ab + dbr + dbr² + … + dbrⁿ⁻¹ – [a+(n–1)d]brⁿ

= ab + dbr(1–rⁿ⁻¹)/(1–r) – [a+(n–1)d]brⁿ

Sₙ = ab/(1–r) + dbr(1–rⁿ⁻¹)/[(1–r)²] – [a+(n–1)d]brⁿ/(1–r) for r ≠ 1

Sum of infinite AGP (Sequences and Series — JEE Main formula, valid for |r| < 1):

As n → ∞, rⁿ → 0 and rⁿ⁻¹ → 0 (since |r| < 1):

S∞ = ab/(1–r) + dbr/(1–r)² = a/(1–r) · b + dr/(1–r)² · b

For the standard AGP starting with a=1, b=1, d=1: S∞ = 1/(1–r) + r/(1–r)² (if |r|<1)

Classic AGP series examples in Sequences and Series (JEE Main):

S = 1 + 2r + 3r² + 4r³ + … (|r|<1): This is AGP with a=1, d=1, b=1. S∞ = 1/(1–r)² [derivative of 1/(1–r) with respect to r gives 1/(1–r)², or use the infinite AGP formula directly].

S = 1 + 3x + 5x² + 7x³ + … (|x|<1): AGP with AP = 1,3,5,7,… (a=1, d=2) and GP = 1,x,x²,… (b=1, r=x). S∞ = 1/(1–x) + 2x/(1–x)² = (1+x)/(1–x)².

S = 2 + 5r + 8r² + 11r³ + …: AP = 2,5,8,11,… (a=2, d=3); GP = 1,r,r²,… S∞ = 2/(1–r) + 3r/(1–r)².

Download the Free PDF for Sequences and Series for all AGP sum derivations and JEE Main examples.

Special Series — Σn, Σn², Σn³ and Method of Differences

Why Special Series Formulas Are the Most Calculation-Intensive JEE Main Sequences and Series Section

Standard special series formulas in Sequences and Series (JEE Main — all must be memorised):

Sum of first n natural numbers: Σn = 1+2+3+…+n = n(n+1)/2

Sum of squares of first n natural numbers: Σn² = 1²+2²+3²+…+n² = n(n+1)(2n+1)/6

Sum of cubes of first n natural numbers: Σn³ = 1³+2³+3³+…+n³ = [n(n+1)/2]² = (Σn)²

Sum of first n odd numbers: 1+3+5+…+(2n–1) = n²

Sum of first n even numbers: 2+4+6+…+2n = n(n+1)

Sum of squares of first n odd numbers: 1²+3²+5²+…+(2n–1)² = n(2n–1)(2n+1)/3

Sum of general term Tₙ = aⁿ³+bⁿ²+cn+d (Sequences and Series):

Sₙ = aΣn³ + bΣn² + cΣn + dn = a[n(n+1)/2]² + b·n(n+1)(2n+1)/6 + c·n(n+1)/2 + dn

Method of Differences in Sequences and Series (JEE Main technique):

Used when Tₙ cannot be directly summed but the differences of consecutive terms form a recognisable pattern (AP, GP, etc.). Step 1: Write out T₁, T₂, T₃, … and compute differences: ΔTₙ = Tₙ₊₁ – Tₙ. Step 2: If 1st differences form an AP → Tₙ is quadratic (Tₙ = an²+bn+c). If 1st differences form a GP → Tₙ = arⁿ⁻¹+b. If 2nd differences form an AP → Tₙ is cubic.

Important patterns in Sequences and Series Method of Differences:

1st difference is AP → Tₙ = an²+bn+c (find a,b,c using initial terms)

1st difference is GP → Tₙ = Arⁿ⁻¹ + B (find A,B,r using initial terms)

2nd difference is AP → Tₙ = an³+bn²+cn+d

2nd difference is GP → Tₙ = Arⁿ⁻¹+Bn+C

Telescoping sums in Sequences and Series (JEE Main technique): Express Tₙ = F(n+1) – F(n). Then Sₙ = T₁+T₂+…+Tₙ = F(2)–F(1) + F(3)–F(2) + … + F(n+1)–F(n) = F(n+1) – F(1). Most useful when Tₙ involves products: n(n+1), n(n+1)(n+2), etc. — use partial fractions. Example: Σ 1/[n(n+1)] = Σ(1/n – 1/(n+1)) = 1 – 1/(n+1) = n/(n+1). Download the Free PDF for Sequences and Series for special series and method of differences examples for JEE Main.

Exponential Series — eˣ and All Standard Deductions for JEE Main

Why the Exponential Series Is the Connection Between Sequences and Series and Calculus in JEE Main

The exponential series in Sequences and Series is one of the most important infinite series in all of mathematics:

eˣ = 1 + x/1! + x²/2! + x³/3! + … = Σₙ₌₀^∞ xⁿ/n! (valid for all x ∈ ℝ)

Definition of e (Sequences and Series):

e = lim_{n→∞} (1 + 1/n)ⁿ = 1 + 1/1! + 1/2! + 1/3! + … = Σ 1/n! ≈ 2.71828…

2 < e < 3. e is irrational and transcendental. e is the base of natural logarithm (ln).

Exponential function aˣ (Sequences and Series formula):

aˣ = e^(x·ln a) = 1 + x(ln a)/1! + x²(ln a)²/2! + … for any a > 0.

Key deductions from exponential series in Sequences and Series (JEE Main direct formulas):

(i) e⁻ˣ = 1 – x/1! + x²/2! – x³/3! + … = Σ (–1)ⁿ xⁿ/n!

(ii) (eˣ + e⁻ˣ)/2 = 1 + x²/2! + x⁴/4! + x⁶/6! + … = Σ x^(2n)/(2n)! [cosh x — even terms of eˣ]

(iii) (eˣ – e⁻ˣ)/2 = x/1! + x³/3! + x⁵/5! + … = Σ x^(2n+1)/(2n+1)! [sinh x — odd terms]

(iv) (e + e⁻¹)/2 = 1 + 1/2! + 1/4! + 1/6! + … [put x=1 in (ii)]

(v) (e – e⁻¹)/2 = 1/1! + 1/3! + 1/5! + … [put x=1 in (iii)]

Important e-series summation results in Sequences and Series (JEE Main):

Σ 1/n! = e (from x=1 in eˣ)

Σ (–1)ⁿ/n! = e⁻¹ = 1/e (from x=–1 in eˣ)

Σ n/n! = Σ 1/(n–1)! = e (shift index: n/n! = 1/(n–1)!)

Σ n²/n! = Σ(n–1)/(n–1)! + Σ 1/(n–1)! = e + e = 2e [use n² = n(n–1) + n, then n²/n! = 1/(n–2)! + 1/(n–1)!]

Σ n³/n! = 5e (use n³ = n(n–1)(n–2) + 3n(n–1) + n)

More generally: Σ nᵏ/n! = Bell number Bₖ · e, or compute by factoring out using the n! identity method.

From PDF: Key results — Σ 1/n! = e; Σ n·1/n! = e; Σ n²·1/n! = 2e; Σ n³·1/n! = 5e; Σ n⁴·1/n! = 15e. These Sequences and Series exponential results appear in JEE Main as "evaluate the series" direct questions. Download the Free PDF for all exponential series deductions and JEE Main sum examples.

Logarithmic Series — log(1+x), log(1–x), and Key Results

Why the Logarithmic Series Is a Direct JEE Main Sequences and Series Formula Question

The logarithmic series in Sequences and Series provides infinite power series expansions for logarithmic functions, valid for |x| ≤ 1 (except x = –1 for log(1+x)).

Logarithmic series formulas in Sequences and Series (JEE Main — all must be memorised):

(i) log_e(1+x) = x – x²/2 + x³/3 – x⁴/4 + … = Σ (–1)ⁿ⁺¹ xⁿ/n (valid for –1 < x ≤ 1)

(ii) log_e(1–x) = –x – x²/2 – x³/3 – x⁴/4 – … = –Σ xⁿ/n (valid for –1 ≤ x < 1)

(iii) log_e(1+x) + log_e(1–x) = log_e(1–x²) = –2(x²/2 + x⁴/4 + x⁶/6 + …) = –2Σ x^(2n)/(2n)

(iv) log_e(1+x) – log_e(1–x) = log_e((1+x)/(1–x)) = 2(x + x³/3 + x⁵/5 + …) = 2Σ x^(2n–1)/(2n–1) (valid for |x| < 1)

Important logarithmic series values in Sequences and Series:

Put x = 1 in (i): log_e 2 = 1 – 1/2 + 1/3 – 1/4 + … = Σ (–1)ⁿ⁺¹/n

Put x = 1/2 in (iv): log_e 3 = 2(1/2 + 1/(3·8) + 1/(5·32) + …)

Note: log_e(1+x) – log_e(1–x) = 2[x + x³/3 + x⁵/5 + …] → put x = 1/2: log_e 3 = 2[1/2 + 1/24 + 1/160 + …]

For x = –1 in (i): log_e 0 is undefined (–1 not in domain). The series diverges at x = –1.

Relationship between logarithmic and exponential series in Sequences and Series: These are inverse functions: if y = eˣ then x = log_e y. The logarithmic series is obtained by integrating the geometric series 1/(1+x) = 1–x+x²–x³+… term by term: ∫1/(1+x)dx = log_e(1+x) = x – x²/2 + x³/3 – …. Similarly, –∫1/(1–x)dx = log_e(1–x) = –x–x²/2–x³/3–…. Download the Free PDF for Sequences and Series for all logarithmic series applications for JEE Main.