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Sequences and Series

Arithmetic Progressions

A sequence is an ordered list of numbers $a_1, a_2, a_3, \dots$ following some rule; a series is what you get when you add the terms of a sequence. An Arithmetic Progression (AP) is the simplest pattern of all: each term is obtained by adding a fixed number to the one before it. That fixed number is the common difference $d$, and the starting value is the first term $a$.

So an AP looks like $a,\ a+d,\ a+2d,\ a+3d, \dots$. The defining test is simple — if $a_{n} - a_{n-1}$ is the same constant for every $n$, the sequence is an AP. The general (nth) term follows directly from this:

$$a_n = a + (n-1)d$$

To add up the first $n$ terms you use the sum formula. There are two equivalent versions — use whichever fits the information you are given:

$$S_n = \dfrac{n}{2}\left[2a + (n-1)d\right] = \dfrac{n}{2}\left(a + \ell\right)$$

Here $\ell = a_n$ is the last term. The second form is Gauss's pairing idea — pair the first and last terms, the second and second-last, and so on; each pair sums to $a + \ell$, and there are $\tfrac{n}{2}$ such pairs.

The Arithmetic Mean (AM) of two numbers $a$ and $b$ is the number $A$ that sits exactly between them so that $a, A, b$ form an AP. Equating the common differences ($A - a = b - A$) gives:

$$A = \dfrac{a + b}{2}$$

More generally, to insert $k$ arithmetic means between $a$ and $b$ means to find $k$ numbers that, placed between them, make the whole list an AP of $k+2$ terms. Since $b$ is then the $(k+2)$th term, the common difference is $d = \dfrac{b-a}{k+1}$.

It helps to see the AP machinery beside its multiplicative cousin, the GP, which you meet in the next topic:

FeatureArithmetic ProgressionGeometric Progression
Rule between termsadd $d$multiply by $r$
nth term$a_n = a + (n-1)d$$a_n = ar^{n-1}$
Sum of $n$ terms$\dfrac{n}{2}[2a+(n-1)d]$$\dfrac{a(r^n-1)}{r-1}$
Mean of $a,b$$\dfrac{a+b}{2}$$\sqrt{ab}$

Deeper Insight — why the AP is the natural home of constant change: An AP is the discrete version of a straight line. Plot the terms against their position $n$ and you get equally spaced points lying on a line of slope $d$ — that is the whole geometric meaning of "common difference". This is why APs model anything growing by a fixed amount each step: simple interest, a salary that rises by a flat increment, seats increasing row by row in an auditorium. The single most useful problem-solving trick in this topic is choosing terms symmetrically about the middle. When three numbers are in AP, write them as $a-d,\ a,\ a+d$ so the $d$'s cancel when you add; for four terms use $a-3d, a-d, a+d, a+3d$ with common difference $2d$. This symmetry turns ugly simultaneous equations into one-line solutions, and it is the reason examiners can pose "three numbers in AP" puzzles that look hard but collapse instantly once you set them up the right way.

An arithmetic progression as equally spaced points climbing by the common difference d AP: equal steps of size d a a+d a+2d a+3d a+4d d Arithmetic mean A sits exactly halfway between a and b Arithmetic Mean: A = (a + b) / 2 a A b equal gap equal gap
Example 1: Find the 20th term of the AP $7, 11, 15, 19, \dots$
  1. Identify $a = 7$ and $d = 11 - 7 = 4$.
  2. Apply $a_n = a + (n-1)d$ with $n = 20$.
  3. $a_{20} = 7 + (20-1)\times 4 = 7 + 76 = 83$.

Answer: $a_{20} = 83$.

Example 2: How many terms of the AP $3, 8, 13, \dots$ are needed to give a sum of $1275$?
  1. Here $a = 3$, $d = 5$, and $S_n = 1275$.
  2. $S_n = \dfrac{n}{2}\left[2a + (n-1)d\right] \Rightarrow 1275 = \dfrac{n}{2}\left[6 + 5(n-1)\right]$.
  3. $2550 = n(5n + 1) \Rightarrow 5n^2 + n - 2550 = 0$.
  4. Solve: $n = \dfrac{-1 + \sqrt{1 + 51000}}{10} = \dfrac{-1 + 226}{10} = \dfrac{225}{10}$... checking the discriminant $\sqrt{51001}=225.8$ is not exact, so factor instead: $5n^2 + n - 2550 = (5n + 51)(n - 50) = 0$.
  5. The positive integer solution is $n = 50$.

Answer: $n = 50$ terms.

Example 3: The 4th term of an AP is $11$ and the 9th term is $26$. Find the first term and the common difference.
  1. $a_4 = a + 3d = 11$ and $a_9 = a + 8d = 26$.
  2. Subtract: $(a + 8d) - (a + 3d) = 26 - 11 \Rightarrow 5d = 15 \Rightarrow d = 3$.
  3. Back-substitute: $a + 3(3) = 11 \Rightarrow a = 2$.

Answer: $a = 2$, $d = 3$.

Example 4: Find the sum of all natural numbers between $100$ and $200$ that are divisible by $4$.
  1. The relevant numbers form the AP $104, 108, 112, \dots, 196$ with $a = 104$, $d = 4$, $\ell = 196$.
  2. Number of terms: $\ell = a + (n-1)d \Rightarrow 196 = 104 + 4(n-1) \Rightarrow n - 1 = 23 \Rightarrow n = 24$.
  3. $S_n = \dfrac{n}{2}(a + \ell) = \dfrac{24}{2}(104 + 196) = 12 \times 300 = 3600$.

Answer: $3600$.

Example 5: Insert three arithmetic means between $3$ and $19$.
  1. Inserting $3$ means makes a $5$-term AP: $3, \_, \_, \_, 19$, so $19$ is the 5th term.
  2. $a_5 = a + 4d \Rightarrow 19 = 3 + 4d \Rightarrow d = 4$.
  3. The means are $3 + 4 = 7$, $7 + 4 = 11$, $11 + 4 = 15$.

Answer: The three arithmetic means are $7, 11, 15$.

Example 6: Three numbers are in AP and their sum is $24$. If the product of the first and the third is $63$, find the numbers.
  1. Take the symmetric form $a - d,\ a,\ a + d$.
  2. Sum: $(a - d) + a + (a + d) = 3a = 24 \Rightarrow a = 8$.
  3. Product of first and third: $(a - d)(a + d) = a^2 - d^2 = 63 \Rightarrow 64 - d^2 = 63 \Rightarrow d^2 = 1 \Rightarrow d = \pm 1$.
  4. With $a = 8$, $d = 1$ gives $7, 8, 9$ (and $d = -1$ gives the same set reversed).

Answer: The numbers are $7, 8, 9$.

Quick recap
  • An AP adds a fixed common difference $d$ each step; its nth term is $a_n = a + (n-1)d$.
  • Sum of $n$ terms: $S_n = \dfrac{n}{2}\left[2a + (n-1)d\right] = \dfrac{n}{2}(a + \ell)$, where $\ell$ is the last term.
  • The arithmetic mean of $a$ and $b$ is $A = \dfrac{a+b}{2}$.
  • To insert $k$ means between $a$ and $b$, use $d = \dfrac{b-a}{k+1}$ for the resulting $(k+2)$-term AP.
  • For three terms in AP take $a-d,\,a,\,a+d$; for four take $a-3d,\,a-d,\,a+d,\,a+3d$ — the symmetry simplifies everything.
✓ Quick check
The 7th term of the GP 1, 3, 9, 27, ... is:
T₇ = 3⁶ = 729.
The sum of the first n natural numbers is given by:
The sum of the sequence 1 + 2 + 3 + ... + n forms an AP with a=1, d=1. S = (n/2)[2(1) + (n − 1)1] = n(n + 1)/2.

Geometric Progressions

A Geometric Progression (GP) is built by multiplying, not adding. Each term is the previous one times a fixed number called the common ratio $r$. So a GP reads $a,\ ar,\ ar^2,\ ar^3, \dots$, where $a$ is the first term and $r \ne 0$. The test for a GP is that the ratio $\dfrac{a_n}{a_{n-1}}$ is the same for every $n$.

The general (nth) term records how many times you have multiplied by $r$:

$$a_n = ar^{\,n-1}$$

The sum of the first $n$ terms has two forms; pick the one that keeps the numerator positive:

$$S_n = \dfrac{a\left(r^{n} - 1\right)}{r - 1}\ \ (r > 1) \qquad S_n = \dfrac{a\left(1 - r^{n}\right)}{1 - r}\ \ (r < 1)$$

When $r = 1$ every term equals $a$, so $S_n = na$ — do not use the formula then, since it divides by zero.

Something remarkable happens when the ratio is a proper fraction. If $|r| < 1$, the terms shrink toward zero and the running total settles on a finite limit even though the series never stops. This is the sum to infinity:

$$S_\infty = \dfrac{a}{1 - r}, \qquad |r| < 1$$

The Geometric Mean (GM) of two positive numbers $a$ and $b$ is the value $G$ that makes $a, G, b$ a GP. Equating ratios ($\tfrac{G}{a} = \tfrac{b}{G}$) gives $G^2 = ab$, so:

$$G = \sqrt{ab}$$

There is a clean and famous relationship between the two means of the same pair of positive numbers — the AM–GM inequality:

$$\text{AM} \ge \text{GM} \quad\Longleftrightarrow\quad \dfrac{a + b}{2} \ge \sqrt{ab}$$

Equality holds only when $a = b$. The proof is one line: $\left(\sqrt{a} - \sqrt{b}\right)^2 \ge 0$ expands to $a + b \ge 2\sqrt{ab}$.

QuantityFormulaCondition
nth term$a_n = ar^{n-1}$$r \ne 0$
Sum of $n$ terms$\dfrac{a(r^n - 1)}{r - 1}$$r \ne 1$
Sum to infinity$\dfrac{a}{1 - r}$$|r| < 1$
Geometric mean$\sqrt{ab}$$a, b > 0$

Deeper Insight — multiplication, growth, and the surprise of a finite infinite sum: Where the AP is the discrete cousin of a straight line, the GP is the discrete cousin of exponential growth and decay. A constant ratio means each term is a fixed percentage larger (or smaller) than the last, which is exactly how compound interest, population growth, radioactive decay, and bouncing balls behave in the real world. The deepest idea in this topic is the sum to infinity. It feels paradoxical that adding infinitely many positive numbers can give a finite answer, but the resolution is that when $|r| < 1$ the terms vanish so fast that the leftover "tail" you keep adding becomes negligibly small — the partial sums creep ever closer to $\tfrac{a}{1-r}$ without ever overshooting it. This single mechanism is what lets a recurring decimal like $0.\overline{3}$ equal the exact fraction $\tfrac{1}{3}$. Finally, the AM–GM inequality is one of the most powerful tools in all of mathematics for proving things are as large or as small as possible; remembering that the two means coincide only when the numbers are equal is the key that unlocks countless optimisation arguments.

A geometric progression shown as bars growing by a constant ratio r GP: each bar is r times the last (a=4, r=1.5) a4 ar6 ar²9 ar³13.5 ar⁴20.25 Sum to infinity of a shrinking geometric series approaches a finite total |r| < 1: the tail vanishes, sum → a / (1 − r) limit a ar ar²
Example 1: Find the 8th term of the GP $5, 10, 20, 40, \dots$
  1. Identify $a = 5$ and $r = \dfrac{10}{5} = 2$.
  2. Apply $a_n = ar^{n-1}$ with $n = 8$.
  3. $a_8 = 5 \times 2^{7} = 5 \times 128 = 640$.

Answer: $a_8 = 640$.

Example 2: Find the sum of the first $6$ terms of the GP $2, 6, 18, \dots$
  1. Here $a = 2$, $r = 3$ (so use the $r > 1$ form).
  2. $S_n = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{2(3^6 - 1)}{3 - 1}$.
  3. $3^6 = 729$, so $S_6 = \dfrac{2(729 - 1)}{2} = 729 - 1 = 728$.

Answer: $S_6 = 728$.

Example 3: Find the sum to infinity of $9 + 3 + 1 + \dfrac{1}{3} + \dots$
  1. First term $a = 9$ and ratio $r = \dfrac{3}{9} = \dfrac{1}{3}$, with $|r| < 1$ so the sum exists.
  2. $S_\infty = \dfrac{a}{1 - r} = \dfrac{9}{1 - \tfrac{1}{3}} = \dfrac{9}{\tfrac{2}{3}}$.
  3. $= 9 \times \dfrac{3}{2} = \dfrac{27}{2} = 13.5$.

Answer: $S_\infty = \dfrac{27}{2}$.

Example 4: Express the recurring decimal $0.\overline{4} = 0.4444\dots$ as a fraction using a GP.
  1. Write it as the series $\dfrac{4}{10} + \dfrac{4}{100} + \dfrac{4}{1000} + \dots$
  2. This is a GP with $a = \dfrac{4}{10}$ and $r = \dfrac{1}{10}$, and $|r| < 1$.
  3. $S_\infty = \dfrac{a}{1 - r} = \dfrac{4/10}{1 - 1/10} = \dfrac{4/10}{9/10} = \dfrac{4}{9}$.

Answer: $0.\overline{4} = \dfrac{4}{9}$.

Example 5: Find the geometric mean of $4$ and $25$.
  1. The GM of two positive numbers is $G = \sqrt{ab}$.
  2. $G = \sqrt{4 \times 25} = \sqrt{100}$.
  3. $= 10$.

Answer: $G = 10$.

Example 6: The AM of two positive numbers is $10$ and their GM is $8$. Find the numbers.
  1. $\dfrac{a + b}{2} = 10 \Rightarrow a + b = 20$, and $\sqrt{ab} = 8 \Rightarrow ab = 64$.
  2. The numbers are roots of $t^2 - (a+b)t + ab = 0$, i.e. $t^2 - 20t + 64 = 0$.
  3. Factorise: $(t - 4)(t - 16) = 0 \Rightarrow t = 4$ or $t = 16$.
  4. Check: $\text{AM} = 10 \ge \text{GM} = 8$, consistent with AM–GM.

Answer: The numbers are $4$ and $16$.

Quick recap
  • A GP multiplies by a fixed common ratio $r$; its nth term is $a_n = ar^{n-1}$.
  • Sum of $n$ terms: $S_n = \dfrac{a(r^n - 1)}{r - 1}$ for $r \ne 1$ (use $S_n = na$ when $r = 1$).
  • When $|r| < 1$, the sum to infinity is $S_\infty = \dfrac{a}{1 - r}$.
  • The geometric mean of positive $a, b$ is $G = \sqrt{ab}$.
  • AM–GM inequality: $\dfrac{a+b}{2} \ge \sqrt{ab}$, with equality only when $a = b$.
✓ Quick check
If the sum of n terms of an AP is Pn + Qn², where P and Q are constants, the common difference is:
The sum Sⁿ = Qn² + Pn. The nth term tⁿ = Sⁿ − Sⁿ⁻¹ = 2Qn + (P − Q). This is a linear expression in n, and the coefficient of n is the common difference, which is 2Q.
The Arithmetic Mean (AM) between 13 and 19 is:
The AM between two numbers a and b is (a + b)/2. So, (13 + 19)/2 = 32/2 = 16.

Means and Special Series

Beyond APs and GPs, three particular sums appear so often that their closed formulas are worth memorising. They are the sums of the first $n$ natural numbers, their squares, and their cubes. A special series is one whose terms follow a simple power pattern, letting us replace a long addition with a single tidy expression.

Sum of the first $n$ natural numbers — this is just an AP with $a = 1$, $d = 1$, so Gauss's pairing gives:

$$\sum_{k=1}^{n} k = 1 + 2 + 3 + \dots + n = \dfrac{n(n+1)}{2}$$

Sum of the first $n$ squares:

$$\sum_{k=1}^{n} k^2 = 1^2 + 2^2 + \dots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$$

Sum of the first $n$ cubes — strikingly, this is exactly the square of the first formula:

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

These three are the toolkit; their real power shows when you combine them. Because summation is linear, you can split and scale any polynomial term and apply each formula separately:

$$\sum_{k=1}^{n}\left(\alpha k^2 + \beta k + \gamma\right) = \alpha\sum k^2 + \beta\sum k + \gamma\, n$$

Here is the full reference set side by side:

SeriesClosed form$n = 4$ check
$\sum k$$\dfrac{n(n+1)}{2}$$1+2+3+4 = 10$
$\sum k^2$$\dfrac{n(n+1)(2n+1)}{6}$$1+4+9+16 = 30$
$\sum k^3$$\left[\dfrac{n(n+1)}{2}\right]^2$$1+8+27+64 = 100$

The way to use these on an unfamiliar series is the nth-term method: find a formula $a_k$ for the general term, then sum it. For instance, the series $1\cdot 2 + 2\cdot 3 + 3\cdot 4 + \dots$ has $a_k = k(k+1) = k^2 + k$, so its sum is $\sum k^2 + \sum k$, which you can evaluate immediately from the table.

Deeper Insight — find the term, then the formulas do the rest: The single most transferable skill in this topic is learning to read a series, spot the rule generating its general term $a_k$, and rewrite that term as a combination of powers of $k$. Once a sum is expressed as $\sum(\alpha k^2 + \beta k + \gamma)$, the three closed forms turn an apparently hard problem into routine arithmetic — no clever trick required, just bookkeeping. The cube identity $\sum k^3 = (\sum k)^2$ deserves a moment of wonder: it says the sum of the first $n$ cubes is a perfect square, and not just any square but the square of the triangular number $\tfrac{n(n+1)}{2}$. This is the kind of unexpected harmony that makes the subject worth studying, and it has a beautiful geometric proof involving nested square frames. Mastering the nth-term method here pays off far beyond this chapter: the same idea underlies the method of differences, telescoping sums, and ultimately the definite integral, where summing infinitely many tiny terms is exactly what you will be doing in Class 12.

Sum of the first n natural numbers as a triangular staircase of unit blocks Σ k = n(n+1) / 2 (triangular staircase, n = 5) 12345 = 15 The three special series formulas grouped together Three formulas to memorise Σ kn(n+1) / 2 Σ k²n(n+1)(2n+1) / 6 Σ k³[ n(n+1) / 2 ]²
Example 1: Find the sum of the first $50$ natural numbers.
  1. Use $\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ with $n = 50$.
  2. $= \dfrac{50 \times 51}{2} = \dfrac{2550}{2}$.
  3. $= 1275$.

Answer: $1275$.

Example 2: Find $\sum_{k=1}^{10} k^2$, the sum of the squares of the first $10$ natural numbers.
  1. Use $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$ with $n = 10$.
  2. $= \dfrac{10 \times 11 \times 21}{6} = \dfrac{2310}{6}$.
  3. $= 385$.

Answer: $385$.

Example 3: Find the sum of the cubes of the first $6$ natural numbers.
  1. Use $\sum_{k=1}^{n} k^3 = \left[\dfrac{n(n+1)}{2}\right]^2$ with $n = 6$.
  2. $\dfrac{6 \times 7}{2} = 21$, so the sum is $21^2$.
  3. $= 441$.

Answer: $441$.

Example 4: Find the sum of the series $1\cdot 2 + 2\cdot 3 + 3\cdot 4 + \dots$ up to $n$ terms.
  1. The general term is $a_k = k(k+1) = k^2 + k$.
  2. $\sum_{k=1}^{n} a_k = \sum k^2 + \sum k = \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n+1)}{2}$.
  3. Take out $\dfrac{n(n+1)}{6}$: $= \dfrac{n(n+1)}{6}\left[(2n+1) + 3\right] = \dfrac{n(n+1)(2n+4)}{6}$.
  4. $= \dfrac{n(n+1)(n+2)}{3}$.

Answer: $\dfrac{n(n+1)(n+2)}{3}$.

Example 5: Evaluate $\sum_{k=1}^{20} (3k^2 - 2k + 1)$.
  1. Split using linearity: $3\sum k^2 - 2\sum k + \sum 1$ for $n = 20$.
  2. $\sum k^2 = \dfrac{20 \times 21 \times 41}{6} = 2870$; $\sum k = \dfrac{20 \times 21}{2} = 210$; $\sum_{k=1}^{20} 1 = 20$.
  3. $3(2870) - 2(210) + 20 = 8610 - 420 + 20$.
  4. $= 8210$.

Answer: $8210$.

Example 6: Find the sum of the series $2^2 + 4^2 + 6^2 + \dots + (2n)^2$.
  1. The general term is $(2k)^2 = 4k^2$ for $k = 1$ to $n$.
  2. $\sum_{k=1}^{n} 4k^2 = 4\sum k^2 = 4 \times \dfrac{n(n+1)(2n+1)}{6}$.
  3. $= \dfrac{2n(n+1)(2n+1)}{3}$.

Answer: $\dfrac{2n(n+1)(2n+1)}{3}$.

Quick recap
  • $\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ — the first $n$ natural numbers.
  • $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$ — the first $n$ squares.
  • $\sum_{k=1}^{n} k^3 = \left[\dfrac{n(n+1)}{2}\right]^2$ — the first $n$ cubes, equal to $\left(\sum k\right)^2$.
  • Summation is linear: split $\sum(\alpha k^2 + \beta k + \gamma)$ and apply each formula separately.
  • For an unfamiliar series, find the nth term $a_k$ first, rewrite it in powers of $k$, then sum.
✓ Quick check
A bus fare increases by ₹5 every stage. If the first fare is ₹15, the fare at the 10th stage is:
T₁₀ = 15 + 9×5 = ₹60.
Ramesh's initial salary is ₹ 10,000 per month, and he receives an annual increment of ₹ 500. In which year will his salary reach ₹ 15,000?
This forms an AP with a = 10000, d = 500, tⁿ = 15000. 15000 = 10000 + (n − 1)500 → 5000 = 500(n − 1) → n − 1 = 10 → n = 11.
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