Sequence & Series — Complete JEE Mains 2026 Revision Capsule
A compact but exhaustive capsule covering: AP, GP, HP, mixed series, telescoping, AGP, recurrence relations, inequalities, patterns, tricks, worked examples, and a practice set with answers. Memorize formulas, practice patterns, and use the quick-checks to score reliably.
$a_n = a + (n-1)d$
$S_n = \dfrac{n}{2}[2a+(n-1)d]$
$a_n = ar^{n-1}$
$S_n = a\dfrac{r^n-1}{r-1}$, $S_\infty=\dfrac{a}{1-r}$ ($|r|<1 div="">
$HM \le GM \le AM$
$HM=\dfrac{2ab}{a+b}$
1. Basics & Notation
Sequence: ordered list $\\{a_n\\}$, Series: sum $S_n=\\sum_{k=1}^n a_k$. General term often written $a_n$ or $T_n$. Use index shifts carefully.
2. Arithmetic Progression (AP)
nth term: $a_n = a + (n-1)d$. Sum: $S_n = \\dfrac{n}{2}[2a + (n-1)d] = n\\cdot \\dfrac{a_1 + a_n}{2}$.
Common results & tricks:
- Average of terms in AP = arithmetic mean of first & last term.
- If number of terms is odd, middle term $= $ mean of sequence.
- If $a_m$ and $a_n$ are known: $a_m - a_n = (m-n)d$ → solves for $d$.
AP Examples
Q: 5th term = 22, 15th term = 62. Find $a$ and $d$.
Ans: $10d = 62-22 = 40 \\Rightarrow d=4$. Then $a_5 = a + 4d = 22 \\Rightarrow a = 6$.
3. Geometric Progression (GP)
nth term: $a_n = ar^{n-1}$. Sum: $S_n = a\\dfrac{r^n - 1}{r-1}$ for $r \\ne 1$. Infinite sum: $S_\\infty = \\dfrac{a}{1-r}$ if $|r|<1 p="">
Key tricks:
- If $a_m$ and $a_n$ known: $\\dfrac{a_m}{a_n} = r^{m-n}$ → solve for $r$.
- When $r$ negative → alternate signs; handle absolute values carefully.
- For sums like $\\sum ar^{n-1}$, use formula directly or multiply by $r$ and subtract.
GP Examples
Q: Insert 3 GMs between 4 and 108.
Ans: total 5 terms in GP → $r^{4} = 108/4 = 27 \\Rightarrow r = 27^{1/4} = 3^{3/4}$. Terms: $4, 4r, 4r^2, 4r^3, 108$.
4. Harmonic Progression (HP)
HP: reciprocals form an AP. If $a_n$ in HP, then $1/a_n$ in AP. Useful to convert to AP for computations.
Means: $AM = \\dfrac{a+b}{2}$, $GM = \\sqrt{ab}$, $HM = \\dfrac{2ab}{a+b}$ with $AM\\ge GM\\ge HM$.
5. Mixed & Special Sequences
Quadratic sequences: if second differences constant, $a_n = An^2 + Bn + C$. Use three initial terms to solve for A,B,C.
Power sums:
- $\\sum_{k=1}^n k = \\dfrac{n(n+1)}{2}$
- $\\sum_{k=1}^n k^2 = \\dfrac{n(n+1)(2n+1)}{6}$
- $\\sum_{k=1}^n k^3 = \\left(\\dfrac{n(n+1)}{2}\\right)^2$
Telescoping series: Recognize expressions like $\\frac{1}{n(n+1)} = \\frac{1}{n} - \\frac{1}{n+1}$ to collapse sums quickly.
Telescoping Example
$\\sum_{n=1}^{100} \\dfrac{1}{n(n+1)} = \\sum (\\dfrac{1}{n} - \\dfrac{1}{n+1}) = 1 - \\dfrac{1}{101}$.
6. Arithmetic–Geometric Progression (AGP)
General term: $T_n = (a + (n-1)d) r^{n-1}$. Sum methods often use multiplication by $r$ and subtraction, or treat as sum of two series.
Infinite AGP sum (|r|<1):
$S_\\infty = \\dfrac{a}{1-r} + \\dfrac{dr}{(1-r)^2}$ (useful when series is convertible to this standard form).
AGP Trick (derivative method)
Consider $S(r)=\\sum_{n\\ge1} r^{n-1}$ and differentiate w.r.t. $r$ to generate $nr^{n-1}$ terms; use linear combos to handle $(an+b)r^{n-1}$ forms.
7. Recurrence Relations & Methods
Linear recurrences of order 1: $a_{n+1} = pa_n + q$. Solve by finding fixed point:
Particular solution: $a = \\dfrac{q}{1-p}$ (if $p \\ne 1$). General solution: $a_n = C p^{n-1} + \\dfrac{q}{1-p}$, choose $C$ using initial condition.
Order-2 linear recurrence (e.g., Fibonacci-type) → characteristic equation method.
Recurrence Example
If $a_{n+1}=3a_n+2$ and $a_1=1$, then steady state $A=\\dfrac{2}{1-3}=-1$. Write $b_n = a_n +1$ → $b_{n+1} = 3b_n$, so $b_n = b_1 3^{n-1}$ → find $a_n$.
8. Important JEE Patterns (Complete)
- Term comparison: AP subtract, GP divide. Use to get $d$ or $r$ quickly.
- Ratio of sums: Use $\\dfrac{S_m}{S_n}$ to form equations in $a,d$ (AP) or $r$ (GP).
- Means insertion: Insert $k$ AMs/ GMs between $a,b$. Solve $d$ or $r$ as $d=(b-a)/(k+1)$ or $r=(b/a)^{1/(k+1)}$. HP via reciprocals.
- Second-difference test: Constant second difference → quadratic $\\Rightarrow a_n = An^2 + Bn + C$.
- Recurrence → AGP: Convert first-order linear recurrence to AGP and use sum formulas.
- Telescoping: Recognize pairs that split via partial fractions; collapse sums to first–last terms.
- Symmetric relations: AP: $2b=a+c$; GP: $b^2=ac$; HP: $\\dfrac{2}{b}=\\dfrac{1}{a}+\\dfrac{1}{c}$; use as constraints.
- Difference method: If $T_n = f(n)-f(n-1)$ then $S_n = f(n)-f(0)$ (useful for telescoping sums).
- AM–GM for bounds: Use to find maxima/minima of expressions involving sums & products.
- AGP tricks: Multiply by $r$, subtract to eliminate geometric part, solve for $S_n$. Use derivative-of-geometric-sum for $nr^{n}$ terms.
9. Common JEE Traps & How to Avoid Them
- Infinite GP without checking: Always ensure $|r|<1$ for convergence.
- Sign errors with negative r: Alternate signs flip sums; track parity of n carefully.
- Sequence vs series mix-up: Problem may ask term (sequence) but student sums (series) — read carefully.
- Telescoping splits incorrectly: Partial fraction decomposition must be exact.
- Recurrence base case misuse: Always apply initial condition after general solution.
- Assuming AP/GP where none exist: Verify via two/three terms or differences.
10. Worked Illustrations (Quick Solves)
Illustration 1 — Find n given sum:
Find $n$ if $1+2+\\dots+n = 210$. Use $\\dfrac{n(n+1)}{2} = 210 \\Rightarrow n^2 + n - 420=0$. Solve: $n=20$ (positive root).
Illustration 2 — Sum with parameter:
Solve for $r$ if $S_3 : S_2 = 7:3$ for GP with first term 1. Here $S_3 = 1+r+r^2$, $S_2 = 1+r$. So $(1+r+r^2)/(1+r) = 7/3$. Cross-multiply: $3+3r+3r^2 = 7+7r$ → $3r^2 -4r -4 =0$ → solve quadratic for r.
Illustration 3 — AGP infinite sum:
Sum $\\sum_{n=1}^{\\infty} n r^{n-1}$ for $|r|<1$. Use derivative trick: $\\sum_{n=0}^{\\infty} r^n = 1/(1-r)$. Differentiate both sides w.r.t. $r$ and shift index → $\\sum_{n=1}^{\\infty} nr^{n-1} = 1/(1-r)^2$.
11. Practice Problems (Try First — Answers Hidden)
- Find sum of first 50 natural numbers.
- Find common ratio r if $a_3 = 12$ and $a_6 = 96$ in a GP with positive r.
- Insert 4 AMs between 3 and 23.
- Evaluate $\\sum_{n=1}^{50} \\dfrac{1}{n(n+1)}$.
- Given $a_{n+1} = 2a_n + 3$ and $a_1 = 1$, find $a_n$ formula.
- Find $\\sum_{n=1}^{\\infty} (3 + 2(n-1))\\left(\\dfrac{1}{4}\\right)^{n-1}$.
- For AP with $S_5 = 40$ and $S_{10} = 125$, find first term and common difference.
- Show that $1^3 + 2^3 + \\dots + n^3 = (1 + 2 + \\dots + n)^2$.
- Find values of $k$ such that $x^2 + kx + 4$ has real roots whose sum is 6.
Show Answers & Solutions
- $\\dfrac{50\\cdot51}{2} = 1275$.
- In GP, $a_6/a_3 = r^{3} = 96/12 = 8 \\Rightarrow r = 2$.
- Insert 4 AMs → total terms 6. $d = (23-3)/5 = 4$. Sequence: $3,7,11,15,19,23$.
- Telescopes: $\\sum_{1}^{50} (1/n - 1/(n+1)) = 1 - 1/51 = 50/51$.
- Solve recurrence: steady state $A = -3/(2-1) = -3$. Let $b_n = a_n +3$ → $b_{n+1} = 2 b_n$, $b_1 = 4$. So $b_n = 4\\cdot 2^{n-1}$ → $a_n = 4\\cdot 2^{n-1} - 3$.
- Use AGP infinite formula with $a=3$, $d=2$, $r=1/4$: $S_\\infty=\\dfrac{3}{1-1/4} + \\dfrac{2(1/4)}{(1-1/4)^2} = \\dfrac{3}{3/4} + \\dfrac{1/2}{(9/16)} = 4 + \\dfrac{1/2}{9/16} = 4 + \\dfrac{8}{9} = \\dfrac{44}{9}$.
- Use $S_n$ formula: system $S_5=5/2[2a+4d]=40$ and $S_{10}=10/2[2a+9d]=125$. Solve: From first $2a+4d = 16 \\Rightarrow a+2d=8$. From second $2a+9d=25 \\Rightarrow a+4.5d=12.5$. Subtract: $2.5d=4.5 \\Rightarrow d=1.8$. Then $a=8-2d=8-3.6=4.4$.
- Proof: Use formulas: $\\sum k^3 = (\\dfrac{n(n+1)}{2})^2$ (standard identity; can be proved by induction or using telescoping of binomial expansions).
- If sum of roots = 6, then by Vieta $-k = 6 \\Rightarrow k = -6$ (since $a=1$, sum = $-b/a$). Check discriminant for reality: $D = k^2 -16 = 36 - 16 = 20 >0$ so OK.
12. One-Line Takeaways
- AP: differences constant → linear formulas; GP: ratios constant → multiplicative formulas.
- Telescoping and partial fractions are your fastest tools for many sums.
- Check convergence for infinite sums: $|r|<1$ is necessary for GP.
- Recurrences often reduce to AGP or geometric forms — use steady-state + homogeneous solution.
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