Sequences and Series — AP, GP, AGP & Special Sums for Class 11

What Are Sequences and Series?

A sequence is an ordered list of numbers following a pattern. A series is what you get when you add up the terms of that sequence. The difference matters — students often mix these up in board exams and lose easy marks.

When we write $1, 4, 9, 16, 25, \ldots$ — that’s a sequence. When we write $1 + 4 + 9 + 16 + 25 + \ldots$ — that’s a series.

Chapter 9 of NCERT Class 11 is one of the highest-weightage chapters for JEE Main. Every year, at least 2–3 questions appear directly from AP, GP, or special sums. The concepts here also feed into chapters on limits, binomial theorem, and coordinate geometry — so getting this right early pays dividends.

Key Terms and Definitions

Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. This constant is called the common difference $d$ .

a, \; a+d, \; a+2d, \; a+3d, \ldots

Geometric Progression (GP): A sequence where the ratio between consecutive terms is constant. This constant is the common ratio $r$ .

a, \; ar, \; ar^2, \; ar^3, \ldots

Arithmetico-Geometric Progression (AGP): Each term is a product of a term from an AP and the corresponding term from a GP. These appear frequently in JEE Main.

a \cdot b, \; (a+d) \cdot br, \; (a+2d) \cdot br^2, \ldots

Partial Sum ( $S_n$ ): The sum of the first $n$ terms of a series.

Infinite Series: When we sum infinitely many terms. Only meaningful (finite) when the series converges — for GP, this requires $|r| < 1$ .

Methods and Core Concepts

Arithmetic Progression (AP)

$n$ -th term: $a_n = a + (n-1)d$

Sum of $n$ terms: $S_n = \dfrac{n}{2}[2a + (n-1)d] = \dfrac{n}{2}[a + l]$

where $l$ is the last term.

The second form $S_n = \dfrac{n}{2}(a + l)$ is faster when you know both ends — use it in MCQs to save time.

Why does this formula work? Pair the first and last terms: $a + l$ . Pair the second and second-last: $(a+d) + (l-d) = a + l$ . Every pair gives the same sum. With $n$ terms, we get $\frac{n}{2}$ such pairs.

Finding $d$ and checking AP: If three numbers $a$ , $b$ , $c$ are in AP, then $2b = a + c$ . This is the condition you use in almost every AP-based JEE question.

Geometric Progression (GP)

$n$ -th term: $a_n = ar^{n-1}$

Sum of $n$ terms ( $r \neq 1$ ): $S_n = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{a(1 - r^n)}{1 - r}$

Sum of $n$ terms ( $r = 1$ ): $S_n = na$

Infinite GP ( $|r| < 1$ ): $S_\infty = \dfrac{a}{1 - r}$

Use $\frac{a(1-r^n)}{1-r}$ when $|r| < 1$ and $\frac{a(r^n - 1)}{r-1}$ when $r > 1$ — both are equivalent but one avoids negative numerator, which is cleaner in calculation.

Three terms in GP trick: Assume them as $\frac{a}{r}, a, ar$ . Their product is $a^3$ , so $a = \sqrt[3]{\text{product}}$ directly. This saves enormous algebra in JEE problems.

Similarly, assume four terms in GP as $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ (not $a, ar, ar^2, ar^3$ ). This keeps the product clean.

Arithmetico-Geometric Progression (AGP)

The general AGP looks like:

S = a \cdot b + (a+d) \cdot br + (a+2d) \cdot br^2 + \ldots

Method — Multiply and Subtract:

Write $S$ , multiply both sides by $r$ , subtract. The resulting series is a GP (the AP parts telescope into a simpler structure).

This technique — called the vS trick — is used in JEE problems every single year.

Special Sums

\sum_{k=1}^{n} k = \frac{n(n+1)}{2}

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

\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2

The third formula is beautiful: sum of cubes equals the square of sum of natural numbers. This appears in JEE Advanced more than you’d expect.

Solved Examples

Example 1 — Easy (CBSE Level)

The sum of first 20 terms of an AP is 400 and the 10th term is 40. Find the first term and common difference.

We know $S_{20} = 400$ , so:

\frac{20}{2}[2a + 19d] = 400 \implies 2a + 19d = 40 \quad \ldots(1)

And $a_{10} = a + 9d = 40 \quad \ldots(2)$

From (2): $a = 40 - 9d$ . Substituting into (1):

2(40 - 9d) + 19d = 40 \implies 80 - 18d + 19d = 40 \implies d = -40

Then $a = 40 - 9(-40) = 400$ .

Answer: $a = 400$ , $d = -40$ .

Example 2 — Medium (JEE Main Level)

Find the sum: $1 + 2\cdot2 + 3\cdot4 + 4\cdot8 + \ldots$ up to $n$ terms.

The general term is $T_k = k \cdot 2^{k-1}$ . This is an AGP with $a = 1$ , $d = 1$ , $b = 1$ , $r = 2$ .

Let $S = 1\cdot1 + 2\cdot2 + 3\cdot4 + \ldots + n\cdot2^{n-1}$

Multiply by 2:

2S = 1\cdot2 + 2\cdot4 + 3\cdot8 + \ldots + (n-1)\cdot2^{n-1} + n\cdot2^n

Subtract ( $S - 2S$ ):

-S = (1 + 2 + 4 + \ldots + 2^{n-1}) - n\cdot2^n = (2^n - 1) - n\cdot2^n

S = (n-1)\cdot2^n + 1

Example 3 — Hard (JEE Advanced Level)

If $a_1, a_2, \ldots, a_n$ are in GP with common ratio $r$ , show that:

\sum_{k=1}^{n-1} \frac{1}{a_k \cdot a_{k+1}} = \frac{1 - r^{-(n-1)}}{a_1 a_2 (r-1)/r}

(Or more cleanly: find $\sum \frac{1}{a_k a_{k+1}}$ in a closed form.)

Since $a_k = a_1 r^{k-1}$ :

\frac{1}{a_k a_{k+1}} = \frac{1}{a_1 r^{k-1} \cdot a_1 r^k} = \frac{1}{a_1^2 r^{2k-1}}

So the sum becomes:

\sum_{k=1}^{n-1} \frac{1}{a_1^2 r^{2k-1}} = \frac{1}{a_1^2 r} \sum_{k=1}^{n-1} \left(\frac{1}{r^2}\right)^{k-1} = \frac{1}{a_1^2 r} \cdot \frac{1 - r^{-2(n-1)}}{1 - r^{-2}}

This simplifies — the key insight is recognizing $\{a_k a_{k+1}\}$ forms its own GP. That’s the “see the structure” skill JEE Advanced rewards.

Exam-Specific Tips

JEE Main Weightage: Sequences and Series consistently gives 2–3 questions per paper. The split is usually: 1 AP/GP theory question, 1 special sums question, 1 AGP or telescoping sum. Total marks: 12–16 per paper. This chapter appeared in JEE Main 2024 Session 1 (Jan), Session 2 (Apr), and both sessions of 2023.

For CBSE Class 11 Boards:

Sum formulas for AP and GP are directly asked (5-mark questions).
The proof of $S_\infty = \frac{a}{1-r}$ for infinite GP is a common 3-marker — know the derivation, not just the formula.
Marking scheme gives 1 mark for setting up the correct formula, 2 marks for correct working, 2 marks for answer. If you make a calculation error but the method is right, you still get 3/5.

For JEE Main:

Use the 3-term and 4-term substitution tricks — they convert 4-step algebra problems into 2-step ones.
When a question says “sum to infinity”, immediately check if $|r| < 1$ . If not, write “series diverges” — you won’t get such a question but it signals clear understanding.

For JEE Advanced:

Telescoping sums appear heavily. Practice writing $T_k = f(k+1) - f(k)$ decomposition.
AM-GM inequality combined with AP/GP conditions is a classic Advanced trap.

Common Mistakes to Avoid

Mistake 1 — Wrong $n$ -th term formula. Students write $a_n = a + nd$ instead of $a + (n-1)d$ . The first term corresponds to $n=1$ , so we need $(n-1)$ steps of $d$ , not $n$ . Check: for $n=1$ , $a_1 = a + (1-1)d = a$ . Correct.

Mistake 2 — Using infinite GP formula when $|r| \geq 1$ . If someone gives you a GP with $r = 2$ and asks for sum to infinity, the answer is “doesn’t exist” — the series diverges. Applying the formula anyway is a conceptual error that loses you marks in JEE.

Mistake 3 — Confusing $S_n$ and $a_n$ . The $n$ -th term of a series is $a_n = S_n - S_{n-1}$ for $n \geq 2$ , and $a_1 = S_1$ . Many students forget the special case for $n=1$ , which causes wrong answers in “find the general term given $S_n$ ” problems.

Mistake 4 — Wrong substitution for 3 terms in GP. Using $a, ar, ar^2$ instead of $\frac{a}{r}, a, ar$ means you have two unknowns ( $a$ and $r$ ) instead of being able to use the product condition $a^3 =$ product directly. Always use the symmetric form.

Mistake 5 — Sign errors in the vS trick for AGP. When you compute $S - rS$ , be careful with the last term. The final term of $rS$ is $n \cdot br^n$ , not what you had in $S$ . Write out the subtraction term by term for at least the first two problems until the pattern becomes muscle memory.

Practice Questions

Q1. If the $p$ -th, $q$ -th, and $r$ -th terms of an AP are $a$ , $b$ , $c$ respectively, prove that $a(q-r) + b(r-p) + c(p-q) = 0$ .

Let first term be $A$ and common difference $D$ . Then $a = A + (p-1)D$ , $b = A + (q-1)D$ , $c = A + (r-1)D$ .

a(q-r) + b(r-p) + c(p-q)

= [A+(p-1)D](q-r) + [A+(q-1)D](r-p) + [A+(r-1)D](p-q)

Grouping the $A$ terms: $A[(q-r)+(r-p)+(p-q)] = A \cdot 0 = 0$

Grouping the $D$ terms: $D[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]$

Expand: $pq - pr - q + r + qr - pq - r + p + rp - rq - p + q = 0$

Both groups vanish. QED.

Q2. Find the sum: $3 + 33 + 333 + 3333 + \ldots$ to $n$ terms.

Factor out 3: $S = 3(1 + 11 + 111 + \ldots)$

Multiply and divide by 9: $S = \frac{3}{9}(9 + 99 + 999 + \ldots) = \frac{1}{3}[(10-1) + (100-1) + (1000-1) + \ldots]$

= \frac{1}{3}\left[(10 + 100 + \ldots + 10^n) - n\right] = \frac{1}{3}\left[\frac{10(10^n - 1)}{9} - n\right]

\boxed{S = \frac{10(10^n - 1)}{27} - \frac{n}{3}}

Q3. The sum of an infinite GP is 4 and the sum of the cubes of its terms is 192. Find the GP.

Let first term $= a$ , common ratio $= r$ .

$\frac{a}{1-r} = 4 \quad \ldots(1)$

The GP of cubes has first term $a^3$ and ratio $r^3$ : $\frac{a^3}{1-r^3} = 192 \quad \ldots(2)$

Cube equation (1): $\frac{a^3}{(1-r)^3} = 64$

Divide (2) by this: $\frac{(1-r)^3}{1-r^3} = \frac{192}{64} = 3$

Now $1 - r^3 = (1-r)(1+r+r^2)$ , so:

$\frac{(1-r)^2}{1+r+r^2} = 3 \implies (1-r)^2 = 3(1+r+r^2)$

$1 - 2r + r^2 = 3 + 3r + 3r^2 \implies 2r^2 + 5r + 2 = 0$

$r = -2$ or $r = -\frac{1}{2}$ . Since $|r| < 1$ , we need $r = -\frac{1}{2}$ .

From (1): $a = 4(1 - (-\frac{1}{2})) = 6$ .

GP: $6, -3, \frac{3}{2}, -\frac{3}{4}, \ldots$

Q4. Find the sum of $n$ terms: $\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \ldots$

Use partial fractions: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$

This is a telescoping series:

S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \ldots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}

Q5. If $a, b, c$ are in GP and $a^x = b^y = c^z$ , show that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in AP.

Let $a^x = b^y = c^z = k$ . Then $a = k^{1/x}$ , $b = k^{1/y}$ , $c = k^{1/z}$ .

Since $a, b, c$ are in GP: $b^2 = ac$

$k^{2/y} = k^{1/x} \cdot k^{1/z} = k^{1/x + 1/z}$

Equating exponents: $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$

This is exactly the condition that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in AP. QED.

Q6. Evaluate $\sum_{k=1}^{n} k(k+1)(k+2)$ .

$k(k+1)(k+2) = k^3 + 3k^2 + 2k$

\sum k^3 + 3\sum k^2 + 2\sum k = \frac{n^2(n+1)^2}{4} + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}

= \frac{n(n+1)}{4}\left[n(n+1) + 2(2n+1) + 4\right] = \frac{n(n+1)(n+2)(n+3)}{4}

This is the “falling factorial” pattern — the answer is always $\frac{n(n+1)\cdots(n+r)}{r+1}$ for product of $r$ consecutive integers starting at $k$ . Useful shortcut for JEE.

Q7. Find the sum of the series $1 + 3x + 6x^2 + 10x^3 + \ldots$ to infinity, where $|x| < 1$ .

The coefficients $1, 3, 6, 10, \ldots$ are triangular numbers: $T_k = \frac{k(k+1)}{2}$ .

Let $S = \sum_{k=1}^{\infty} \frac{k(k+1)}{2} x^{k-1}$

We know $\sum x^k = \frac{1}{1-x}$ . Differentiating: $\sum k x^{k-1} = \frac{1}{(1-x)^2}$ .

Differentiating again: $\sum k(k-1)x^{k-2} = \frac{2}{(1-x)^3}$ , so $\sum k(k-1)x^{k-1} = \frac{2x}{(1-x)^3}$ .

Note $\frac{k(k+1)}{2} = \frac{k(k-1)}{2} + k$ .

S = \frac{1}{2} \cdot \frac{2x}{(1-x)^3} \cdot \frac{1}{x} \cdot x + \frac{1}{(1-x)^2} = \frac{1}{(1-x)^3}

Wait — more cleanly: this is the coefficient of $x^{k-1}$ for the series $\frac{1}{(1-x)^3}$ .

\boxed{S = \frac{1}{(1-x)^3}}

Q8. Insert 5 arithmetic means between 8 and 26.

With 5 means inserted, we have 7 terms total: 8, $a_1$ , $a_2$ , $a_3$ , $a_4$ , $a_5$ , 26.

$a_7 = 26$ : $8 + 6d = 26 \implies d = 3$ .

Means: $11, 14, 17, 20, 23$ .

Frequently Asked Questions

What’s the difference between a sequence and a series?

A sequence is a list: $2, 5, 8, 11, \ldots$ A series is its sum: $2 + 5 + 8 + 11 + \ldots$ In board exams, marks are specifically deducted if you confuse these two. The notation $S_n$ always refers to a series (sum), never a sequence.

Can a sequence be both AP and GP simultaneously?

Yes — a constant sequence like $3, 3, 3, 3, \ldots$ is both an AP (with $d = 0$ ) and a GP (with $r = 1$ ). This is the only such sequence (ignoring the trivial all-zero sequence). JEE sometimes uses this as a trick option in multiple-correct problems.

When does an infinite GP converge?

When $|r| < 1$ . The formula $S_\infty = \frac{a}{1-r}$ is only valid then. If $|r| \geq 1$ , the series either diverges to infinity (if $r \geq 1$ ) or oscillates without converging (if $r \leq -1$ ).

How do I identify an AGP in an exam question?

Look at the general term. If it’s of the form (linear in $k$ ) × (exponential in $k$ ), it’s an AGP. For example, $T_k = (3k-1) \cdot 2^{k-1}$ — linear factor $(3k-1)$ times exponential $2^{k-1}$ . Apply the vS trick.

What’s the AM-GM inequality and why does it matter here?

For positive numbers: $\frac{a+b}{2} \geq \sqrt{ab}$ , with equality iff $a = b$ . This connects AP and GP directly — the arithmetic mean of two numbers is always $\geq$ their geometric mean. JEE Advanced uses this to set minimum/maximum problems where AP or GP conditions are constraints.

How are special sums ( $\sum k$ , $\sum k^2$ , $\sum k^3$ ) derived?

$\sum k = \frac{n(n+1)}{2}$ comes from pairing terms (Gauss’s method). $\sum k^2$ comes from the telescoping identity $(k+1)^3 - k^3 = 3k^2 + 3k + 1$ . $\sum k^3 = \left(\sum k\right)^2$ can be proved by induction. CBSE sometimes asks for these derivations — knowing them is non-negotiable.

In JEE, should I memorize the AGP sum formula or derive it each time?

Derive it. The vS method takes 3–4 lines and is completely reliable. Memorized formulas get garbled under exam pressure and there’s no way to verify them quickly. Practice the derivation 10 times so it becomes automatic.