Shunya / Maths Explainers

Shunya · Maths lab

Maths explainers

Six ideas that are easier to see than to read. Watch a fraction turn into a repeating decimal, drag a complex number around the Argand plane, count possibilities with a tree and Pascal's triangle, grow an arithmetic or geometric series, fill in a Venn diagram, and solve a linear programming problem by sliding a line across a feasible region.

Lessons

Pick your class, follow the steps in the lab, and check what you see against the working under the graph.

Class 9 — Rational and irrational numbers

Concept: a rational number is a fraction p/q; its decimal either stops (terminates) or repeats a block forever. An irrational number, such as √2, has a decimal that neither stops nor repeats.

Open Number systems and press 1/8: the remainder reaches 0 and the decimal stops (0.125).
Press 1/7: the remainders 1, 3, 2, 6, 4, 5 come round again, so 142857 repeats forever.
Try 1/6, 7/13 and 1/97. Which denominators terminate? Only those made of 2s and 5s.
Switch to √n on the number line and slide the zoom: √2 is always trapped between two decimals but never lands on one.
Press √16 (a perfect square): the trap collapses to the whole number 4.

Takeaway: p/q gives a terminating or repeating decimal; a square root of a non-perfect-square never does.

Class 9 — Repeating decimals as fractions and the square-root spiral

Concept: multiplying a repeating decimal by a power of 10 and subtracting removes the endless tail, which leaves a fraction. The square-root spiral shows how to mark √2, √3, √5 … on a line using right triangles.

Open Decimal → fraction and press 0.333…: 10x − x = 3 gives x = 1/3.
Press 0.1666… and follow the two shifts (10x and 100x) and the subtraction to reach 1/6.
Press 0.999… and see that it equals exactly 1.
Open Root spiral and raise the number of triangles: each hypotenuse is √k, and √4, √9, √16 land on whole numbers.
Check one triangle by hand: (√5)² + 1² = 6, so the next hypotenuse is √6.

Takeaway: every repeating decimal is a fraction, and Pythagoras builds every square root on a ruler.

Class 10 — Arithmetic progressions

Concept: an AP has a constant difference d between neighbouring terms. The nth term is a + (n − 1)d and the sum of n terms is n/2 · [2a + (n − 1)d].

Open Sequences & series, choose Arithmetic (AP) and press 2, 5, 8, ….
Read the nth term and the sum in the working, then drag the last dot to change n.
Tick the bars: the total area of the bars is the sum Sₙ.
Move Which term equals… to 20 and read term number 7; try 21 and see it is not in the progression.
Press 10, 8, 6, …: a negative difference gives a decreasing line.

Takeaway: the terms of an AP lie on a straight line, and Sₙ = n/2 · (first + last).

Class 11 — Complex numbers on the Argand plane

Concept: z = a + bi is the point (a, b). Its modulus |z| is its distance from the origin, its argument is its angle, and its conjugate z̄ is its mirror image in the real axis. Adding is the parallelogram rule; multiplying adds the angles and multiplies the moduli.

Open Complex numbers, press 3 + 4i and read |z| = 5 and the argument from the working.
Tick show the conjugate and check that z·z̄ = |z|² = 25.
Open Add & subtract, drag z and w and watch z + w as the diagonal of the parallelogram.
Open Multiply & divide and press multiply by i (w = i): z turns a quarter turn.
Press multiply by z̄ to see that the product is real.

Takeaway: modulus, argument and conjugate describe z; multiplying by w turns and stretches z.

Class 11 — Roots of complex numbers and De Moivre's theorem

Concept: zⁿ has modulus |z|ⁿ and argument n times the argument of z. The n nth roots of z lie at equal angles 360° ÷ n apart on a circle, forming a regular polygon.

Open Roots & powers and press roots of unity (z = 1) with n = 3: three points form an equilateral triangle.
Increase n to 6 and check that the roots still add up to 0.
Press roots of 16i, set n = 4 and read the four roots, 2 ∠ 22.5° and three more.
Switch to powers of z and drag z inside and outside the unit circle to see the powers spiral in or out.
Read the angle between neighbouring roots in the cards.

Takeaway: the nth roots of z are equally spaced on a circle of radius |z|^(1/n).

Class 11 — Counting: the fundamental principle, permutations and combinations

Concept: if a job has stages that can be done in a, b and c ways, the whole job can be done in a × b × c ways. Choosing r items from n gives P(n, r) = n!/(n − r)! ordered arrangements or C(n, r) = P(n, r) ÷ r! unordered groups.

Open Counting and set 3 shirts, 2 trousers and 2 pairs of shoes: count the leaves of the tree (12).
Slide the path number to follow one outfit from root to leaf.
Open Permutations & combinations, set n = 5 and r = 3 and list all 60 permutations, then switch to the 10 combinations.
Check that P = C × r! in the cards.
Open Arrangements of a word, press BANANA and read 6! ÷ (3! 2!) = 60.

Takeaway: multiply the ways for each stage; divide by r! when order does not matter.

Class 11 — Pascal's triangle and the binomial theorem

Concept: row n of Pascal's triangle gives C(n, 0), C(n, 1), …, C(n, n), the coefficients of (a + b)ⁿ. Each entry is the sum of the two above it and every row adds to 2ⁿ.

Open Pascal's triangle and set the row n = 5: 1 5 10 10 5 1.
Move the entry slider and see the two entries above it in the row before highlighted in teal.
Read (a + b)⁵ in the working and check that its coefficients are the row.
Check that the row adds to 2⁵ = 32.
Raise the rows to 12 and look for the symmetry C(n, k) = C(n, n − k).

Takeaway: C(n, k) = C(n − 1, k − 1) + C(n − 1, k), and the row sum is 2ⁿ.

Class 11 & 12 — The binomial theorem: general term, middle term, term free of x

Concept: (a + b)ⁿ has n + 1 terms and the general term is T_r+1 = C(n, r) a^n−r b^r. Questions ask for one particular term without expanding the rest.

Open Counting → Binomial theorem. It opens on (x + 1/x)⁶: the seven terms are x⁶, 6x⁴, 15x², 20, 15x⁻², 6x⁻⁴, x⁻⁶. The power of x in T_r+1 is (6 − r) − r = 6 − 2r, which is 0 when r = 3, so the term free of x is C(6, 3) = 20.
Press (x² − 1/x)⁹: coefficient of x³. The power is 2(9 − r) − r = 18 − 3r. Setting 18 − 3r = 3 gives r = 5, so the coefficient is C(9, 5) · (−1)⁵ = −126. If the equation gave a fraction (try m = 4), no term has that power and the coefficient is 0.
Middle term. With 9 + 1 = 10 terms there are two middle terms, T₅ = 126x⁶ and T₆ = −126x³. For an even power n there is a single middle term T_n/2+1, as in (1 + x)¹⁰ where it is C(10, 5) = 252 x⁵.
Press (2x + 1/x²)⁶. The power of x is 6 − 3r; it is 0 when r = 2, so the term free of x is C(6, 2) · 2⁴ = 15 × 16 = 240, and the middle term is T₄ = C(6, 3) · 2³ = 160x⁻³.
Sum of coefficients: put x = 1. For (2x + 1/x²)⁶ it is (2 + 1)⁶ = 729. The sum of the binomial coefficients alone is 2ⁿ = 64.

Takeaway: write the power of x in T_r+1, set it equal to the target, solve for r, and check that r is a whole number from 0 to n.

Class 11 & 12 — Counting with conditions: together, round a table, stars and bars, derangements

Concept: JEE counting questions add conditions to the basic P(n, r) and C(n, r). Four tools cover most of them.

People together. Open Counting → Together & circular with 5 people, 2 together, in a row: glue the two into one block, arrange 4 objects in 4! = 24 ways, and the block in 2! = 2 ways: 48 of the 120 arrangements. “Not together” is the complement: 120 − 48 = 72.
Round a table. Rotations are the same arrangement, so fix one person: 5 people give (5 − 1)! = 24. With two together it is (5 − 2)! × 2! = 12 (the chips list all 12). If clockwise and anticlockwise count as the same (a necklace), halve it: 12 for all 5.
Stars and bars. Open Stars & bars: sharing 10 identical sweets among 3 children is the number of solutions of x₁ + x₂ + x₃ = 10, which is C(12, 2) = 66. If every child must get at least one, give each one first and share the other 7: C(9, 2) = 36.
Derangements. Open Derangements: 4 letters in 4 envelopes, none in its own: D(4) = 9, found as 24 − 24 + 12 − 4 + 1 by inclusion–exclusion. The list gives D(1) to D(8) = 0, 1, 2, 9, 44, 265, 1854, 14833, and D(n) ÷ n! → 1/e ≈ 0.368.

Takeaway: block the “together” people, fix one person for a circle, count stars and bars for identical objects, and use inclusion–exclusion for “none in its own place”.

Class 11 & 12 — Locus of a complex number: circles, bisectors, ellipses and arcs

Concept: |z − z₁| is the distance from z to z₁, and arg(z − z₁) is the direction of the line from z₁ to z. Every rule about them is a locus problem in the Argand plane: write z = x + iy and turn the rule into an equation in x and y.

Open Complex numbers → Locus of z. |z − (1 + i)| = 2 is a circle with centre (1, 1) and radius 2: (x − 1)² + (y − 1)² = 4.
Press |z − 2| = |z + 2|. Squaring gives (x − 2)² + y² = (x + 2)² + y², so −8x = 0: the imaginary axis, the perpendicular bisector of the points 2 and −2.
Press |z − 1| = 2|z + 2|. After squaring and dividing by 1 − 4 the equation is x² + y² + 6x + 5 = 0: a circle with centre (−3, 0) and radius 2, the circle of Apollonius. If the ratio is 1 the locus is the bisector line instead.
Press |z − 3| + |z + 3| = 10: an ellipse with foci ±3, a = 5, b² = 25 − 9 = 16, e = 3/5. It exists only when the constant is greater than |z₁ − z₂| = 6 (the triangle inequality); with k = 6 it shrinks to the segment, with less there is nothing. For a constant difference 4 it is the hyperbola a = 2, b² = 9 − 4 = 5, e = 3/2, which needs k < 6.
Directions. arg(z − 1) = 45° is a ray from 1 at 45° (the point 1 itself is excluded). arg((z − 2) ÷ (z + 2)) = 90° is the semicircle on the segment from −2 to 2: the angle in a semicircle is a right angle. For another θ the locus is an arc of radius |z₁ − z₂| ÷ (2 sin θ); for θ = 30° and the same points R = 4 ÷ (2 × 0.5) = 4.
Drag z: the readout shows how far it is from obeying the rule, and Put z on the locus snaps it onto the curve.

Takeaway: equal distances give a line, a constant ratio gives a circle, a constant sum or difference gives an ellipse or hyperbola, and a constant argument gives a ray, or an arc when it is the angle at z between two fixed points.

Class 11 & 12 — Rotation in the Argand plane: multiplying by e^iθ

Concept: multiplying by e^iθ = cos θ + i sin θ turns a point through θ about the origin without changing its distance. To turn z about another point c use z′ − c = e^iθ(z − c).

Open Complex numbers → Rotation. The opening example turns z = 2 + i through 90° about the origin: i(2 + i) = 2i + i² = −1 + 2i. The distance √5 = 2.236 is unchanged.
Press 60°: c, z, z′ make an equilateral triangle. Rotating z about c by 60° gives z′ with |z′ − c| = |z − c| and an angle of 60° between them, so the triangle c z z′ has two equal sides with a 60° angle between them: it is equilateral.
Press 120°, 3 points to see the three points z, ωz, ω²z where ω = e^2πi/3: an equilateral triangle centred on c. This is why z₁ + ωz₂ + ω²z₃ = 0 characterises an equilateral triangle, and why 1 + ω + ω² = 0.
Press 90° about c (square): the vertices are c + (z − c), c + i(z − c), c − (z − c), c − i(z − c). If you know two adjacent vertices of a square you find the others by rotating through 90°.
Change c: to rotate z = 2 + i through 90° about c = 1 + i, z − c = 1, times i gives i, so z′ = c + i = 1 + 2i.

Takeaway: multiplying by a unit complex number rotates; to rotate about a point, subtract it, multiply, and add it back.

Class 11 & 12 — Mathematical reasoning: statements, connectives and truth tables

Concept: a statement is a sentence that is either true or false. Statements are joined with not (¬), and (∧), or (∨, inclusive), if … then (→) and if and only if (↔). A truth table lists the result for every combination of true (T) and false (F).

Open Mathematical reasoning → Truth tables. It opens on A = p → q and B = ¬q → ¬p. With two letters there are 2² = 4 rows, and p → q is false in exactly one of them (p true, q false): it is true in 3 of 4 rows.
The two columns A and B match in every row, so the statement and its contrapositive are equivalent. Press p → q and its converse q → p: the columns differ in rows 2 and 3 (p = T, q = F and p = F, q = T), so a conditional and its converse are not equivalent.
De Morgan's laws. Press the De Morgan examples: ¬(p ∧ q) ≡ ¬p ∨ ¬q and ¬(p ∨ q) ≡ ¬p ∧ ¬q. Also p → q ≡ ¬p ∨ q, and the negation of a conditional is ¬(p → q) ≡ p ∧ ¬q (the lab confirms the columns match).
Tautology and contradiction. p ∨ ¬p is true in every row (a tautology) and p ∧ ¬p is false in every row (a contradiction). The longer statement ((p → q) ∧ (q → r)) → (p → r) is true in all 8 rows: this is the rule of reasoning “if p implies q and q implies r, then p implies r”.
Type your own statement, for example (p ∨ q) ∧ ¬p, and read off the class: it is true in 1 of 4 rows (p false, q true), so it is neither a tautology nor a contradiction.

Takeaway: p → q is false only when p is true and q is false; two statements are equivalent when their truth-table columns agree; ¬(p ∧ q) ≡ ¬p ∨ ¬q and ¬(p ∨ q) ≡ ¬p ∧ ¬q.

Class 11 & 12 — Converse, contrapositive, quantifiers and methods of proof

Concept: from “if p, then q” we build the converse (if q, then p), the inverse (if not p, then not q) and the contrapositive (if not q, then not p). A conditional is disproved by one counter-example: a case where p holds and q fails.

Open If–then family with “If a number is divisible by 4, then it is even.” The original is true for every n from 1 to 100. Its contrapositive (if a number is odd, it is not divisible by 4) is also true. The converse (if even, then divisible by 4) fails for n = 2, and so does the inverse.
Choose “If x² = 9, then x = 3.” It is false: x = −3 has x² = 9 but x ≠ 3. The contrapositive is false for the same reason, while the converse (if x = 3 then x² = 9) and the inverse are true. The original and the contrapositive always agree; the converse and the inverse always agree with each other.
Choose “If n is a multiple of 6, then n is a multiple of both 2 and 3.” The converse is also true, so we may write n is a multiple of 6 if and only if it is a multiple of both 2 and 3 (p ⇔ q).
Quantifiers. Open Quantifiers & negation. “For every x in {1, …, 10}, x² > x” is false: the counter-example x = 1 (1 > 1 fails). Its negation, “there is an x with x² ≤ x”, is true. “There is an x with x² = 49” is true with the witness x = 7 once N ≥ 7, and its negation (for every x, x² ≠ 49) is then false.
The rule: ¬(for every x, P(x)) is “there is an x with not P(x)”, and ¬(there is an x with P(x)) is “for every x, not P(x)”. A statement and its negation always have opposite truth values.
Methods of proof. Direct: assume p and deduce q. Contrapositive: assume not q and deduce not p (valid because it is equivalent). Contradiction: assume the statement is false and reach an impossible result. A single counter-example is enough to disprove a “for every” statement.

Takeaway: original ≡ contrapositive, converse ≡ inverse, and a conditional is disproved by one counter-example; negating a quantified statement flips ∀ and ∃ and negates the property.

Class 11 — Geometric progressions, AM–GM–HM and standard sums

Concept: in a GP each term is r times the one before, so aₙ = a·rⁿ⁻¹. If |r| < 1 the terms shrink and the sum settles on a/(1 − r). For two positive numbers, AM ≥ GM ≥ HM.

Choose Geometric (GP) and press 8, 4, 2, … (r = ½): the teal partial sums climb to the dashed line at 16.
Press 2, 4, 8, … (r = 2): the partial sums have no limit.
Open AM, GM, HM and drag the meeting point: the GM (height) never exceeds the AM (radius).
Set a = b and see all three means equal.
Open Standard sums, compare 1 + 2 + … + n with n(n + 1)/2, and see the cubes add up to the square of the sum.

Takeaway: a geometric series converges only when |r| < 1, and AM ≥ GM ≥ HM with equality when the numbers are equal.

Class 11 — Sets and Venn diagrams

Concept: a set is a collection of distinct members. |A ∪ B| = |A| + |B| − |A ∩ B|, because the overlap is counted twice. De Morgan's laws say (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.

Open Sets & Venn diagrams and press Tea and coffee: 28 + 22 − 9 = 41 people drink at least one.
Use Shade this region to highlight A ∪ B, A ∩ B and A − B and read how many are in each.
Open Three sets, press Survey of 100 students and read the region counts.
Press An impossible one and see the red region with a negative count.
Open Sets of elements, press overlap and check De Morgan's laws in the working.

Takeaway: count the overlap once, and a set of n members has 2ⁿ subsets.

Class 12 — Linear programming

Concept: maximise or minimise a linear objective Z = ax + by subject to linear constraints. The constraints bound a feasible region, and the best value of Z occurs at one of its corner points.

Open Linear programming and press Furniture: max 5x + 4y to see four constraints and the teal feasible region.
Read the corner points and the value of Z at each in the working; the largest is the answer.
Slide the objective line from the worst corner to the best: the last corner it touches is the optimum.
Press Diet: min 4x + 3y and switch between maximise and minimise on the same region.
Press Unbounded and Infeasible to see the two special cases, and Two best points for many optimal solutions.

Takeaway: evaluate Z at every corner point; for an unbounded region check that the best value really exists.

The maths behind it

What each explainer covers

Number systems — a fraction turned into a decimal by long division (and why it terminates or repeats), a repeating decimal turned back into a fraction, √n squeezed between decimals on the number line, and the square-root spiral. Classes 9 and 10.

Complex numbers — a draggable z on the Argand plane with its conjugate, modulus and argument; sum and difference as a parallelogram; product and quotient as a turn and a stretch; nth roots and powers (De Moivre). Class 11.

Counting — the multiplication principle as a tree you can follow, permutations and combinations listed out, Pascal's triangle and the binomial coefficients, and the arrangements of the letters of a word. Class 11.

Sequences & series — arithmetic progressions with draggable terms and bars for the sum, geometric progressions with partial sums closing in on the sum to infinity, AM ≥ GM ≥ HM on a semicircle, and the standard sums and Fibonacci numbers. Classes 10 and 11.

Sets & Venn diagrams — two and three sets by counts (inclusion–exclusion, impossible numbers flagged) and by typed elements (union, intersection, difference, complement, subsets, De Morgan). Class 11.

Linear programming — type an objective and up to six constraints, see the feasible region and its corner points, slide the objective line to the best corner, and meet unbounded, infeasible and many-optima cases. Class 12.

Part of Shunya. See the syllabus map for the chapters these explainers map to. Complex numbers and counting can be checked against the Function & Graph Explorer and the Statistics & Probability lab.

Formulas used in this lab

Number systems Class 9

A rational number: $\dfrac{p}{q},\qquad p, q \text{ integers},\ q \ne 0$
Terminating decimal: $q = 2^m \cdot 5^n \ \Rightarrow\ \dfrac{p}{q} \text{ terminates}$
Repeating decimal to a fraction: $0.\overline{ab} = \dfrac{ab}{99},\qquad 0.\overline{a} = \dfrac{a}{9}$
Rationalising: $\dfrac{1}{a + b\sqrt{c}} = \dfrac{a - b\sqrt{c}}{a^2 - b^2 c}$
Laws of surds: $\sqrt{a}\sqrt{b} = \sqrt{ab},\qquad (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$

Arithmetic progressions Class 10

nth term: $a_n = a + (n - 1)d$
Sum of n terms: $S_n = \dfrac{n}{2}\,[2a + (n - 1)d]$
Sum from the last term l: $S_n = \dfrac{n}{2}\,(a + l)$
A term from the sums: $a_n = S_n - S_{n-1}$
Three numbers in AP: $2b = a + c$

Complex numbers Class 11

The imaginary unit: $i^2 = -1,\qquad i^4 = 1$
Modulus and conjugate: $|z| = \sqrt{a^2 + b^2},\qquad z\bar{z} = |z|^2$
Polar form: $z = r(\cos\theta + i\sin\theta),\qquad r = |z|$
Product and quotient: $|z_1z_2| = |z_1||z_2|,\qquad \arg(z_1z_2) = \arg z_1 + \arg z_2$
Quotient by the conjugate: $\dfrac{z_1}{z_2} = \dfrac{z_1\bar{z_2}}{|z_2|^2}$
Triangle inequality: $|z_1 + z_2| \le |z_1| + |z_2|$

Roots of complex numbers Class 11

De Moivre's theorem: $[r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)$
The nth roots of z (k = 0, 1, …, n − 1): modulus and angle: $r^{1/n},\quad \dfrac{\theta + 2\pi k}{n}$
Roots of unity: $\omega = e^{2\pi i/n},\qquad 1 + \omega + \dots + \omega^{n-1} = 0$
Cube roots of unity: $\omega^3 = 1,\qquad 1 + \omega + \omega^2 = 0$

Locus and rotation of complex numbers Class 11 & 12

Circle: $|z - z_1| = r$
Perpendicular bisector: $|z - z_1| = |z - z_2|$
Circle of Apollonius: $|z - z_1| = k\,|z - z_2|,\ k \ne 1$
Ellipse and hyperbola: $|z - z_1| + |z - z_2| = k \ (k > |z_1 - z_2|),\quad \bigl||z - z_1| - |z - z_2|\bigr| = k \ (k < |z_1 - z_2|)$
Ray and arc: $\arg(z - z_1) = \theta,\qquad \arg\dfrac{z - z_1}{z - z_2} = \theta$
Rotation about c: $z' - c = e^{i\theta}(z - c)$
Equilateral triangle: $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$

Counting Class 11

Multiplication principle: $m \times n \times p \text{ ways}$
Permutations: ${}^nP_r = \dfrac{n!}{(n - r)!}$
Combinations: ${}^nC_r = \dfrac{n!}{r!\,(n - r)!}$
Link between them: ${}^nP_r = {}^nC_r \cdot r!,\qquad {}^nC_r = {}^nC_{n-r}$
Pascal's rule: ${}^nC_r = {}^{n-1}C_{r-1} + {}^{n-1}C_r$
Arrangements with repeats: $\dfrac{n!}{p!\,q!\,r!}$
In a circle: $(n - 1)!$

Binomial theorem Class 11

Expansion: $(a + b)^n = \sum_{r=0}^{n} {}^nC_r\, a^{n-r}b^r$
General term: $T_{r+1} = {}^nC_r\, a^{n-r}b^r$
Number of terms: $n + 1$
Sum of the coefficients: ${}^nC_0 + {}^nC_1 + \dots + {}^nC_n = 2^n$
Alternating sum: ${}^nC_0 - {}^nC_1 + {}^nC_2 - \dots = 0$
General term of (ax^p + bx^q)ⁿ: $T_{r+1} = {}^nC_r\, a^{n-r}b^r\, x^{p(n-r)+qr}$
Coefficient of x^m: $p(n-r)+qr = m \Rightarrow r = \dfrac{pn-m}{p-q}\ (\text{a whole number, } 0 \le r \le n)$
Middle term: $n \text{ even: } T_{n/2+1};\quad n \text{ odd: } T_{(n+1)/2} \text{ and } T_{(n+3)/2}$
Sum of the coefficients: $\text{put } x = 1:\ (a+b)^n$

Counting with conditions Class 11 & 12

k particular objects together (row): $(n-k+1)!\; k!$
Round a table: $(n-1)!\quad\text{(necklace, both ways the same: } \tfrac{(n-1)!}{2}\text{)}$
k together, round a table: $(n-k)!\; k!$
Not together: $\text{all} - \text{together}$
x₁ + … + x_k = n, x_i ≥ 0: ${}^{n+k-1}C_{k-1}$
Same, every x_i ≥ 1: ${}^{n-1}C_{k-1}$
Derangements: $D_n = n!\left(1-\tfrac1{1!}+\tfrac1{2!}-\dots+\tfrac{(-1)^n}{n!}\right),\quad D_n=(n-1)(D_{n-1}+D_{n-2})$

Geometric progressions and means Class 11

nth term: $a_n = ar^{n-1}$
Sum of n terms: $S_n = \dfrac{a(1 - r^n)}{1 - r},\qquad r \ne 1$
Sum to infinity: $S_\infty = \dfrac{a}{1 - r},\qquad |r| \lt 1$
The three means: $AM = \dfrac{a + b}{2},\quad GM = \sqrt{ab},\quad HM = \dfrac{2ab}{a + b}$
Order and link: $AM \ge GM \ge HM,\qquad GM^2 = AM \cdot HM$

Standard sums Class 11

Natural numbers: $1 + 2 + \dots + n = \dfrac{n(n + 1)}{2}$
Squares: $1^2 + 2^2 + \dots + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$
Cubes: $1^3 + 2^3 + \dots + n^3 = \left[\dfrac{n(n + 1)}{2}\right]^2$
Fibonacci numbers: $F_n = F_{n-1} + F_{n-2},\qquad F_1 + \dots + F_n = F_{n+2} - 1$

Sets Class 11

Two sets: $|A \cup B| = |A| + |B| - |A \cap B|$
Three sets: $|A \cup B \cup C| = \sum |A| - \sum |A \cap B| + |A \cap B \cap C|$
De Morgan's laws: $(A \cup B)' = A' \cap B',\qquad (A \cap B)' = A' \cup B'$
Difference and complement: $A - B = A \cap B',\qquad A' = U - A$
Number of subsets: $|P(A)| = 2^n$

Linear programming Class 12

Objective function: $Z = ax + by$
Constraints: $a_1x + b_1y \le c_1,\quad a_2x + b_2y \ge c_2,\quad x, y \ge 0$
Corner point theorem: the best Z is at: $\text{a corner point}$
Two optimal corners: the optimum is: $\text{the whole edge}$
Unbounded region: check that: $Z \gt M \text{ meets it}$

Mathematical reasoning Class 11 & 12

Implication as or: $p \to q \equiv \lnot p \lor q$
Contrapositive, converse, inverse: $p\to q \equiv \lnot q\to\lnot p;\quad \text{converse } q\to p \equiv \text{inverse } \lnot p\to\lnot q$
De Morgan's laws: $\lnot(p\land q)\equiv \lnot p\lor\lnot q,\qquad \lnot(p\lor q)\equiv\lnot p\land\lnot q$
Negation of a conditional: $\lnot(p\to q)\equiv p\land\lnot q$
Biconditional: $p\leftrightarrow q \equiv (p\to q)\land(q\to p)$
Negating quantifiers: $\lnot\forall x\,P(x)\equiv \exists x\,\lnot P(x),\qquad \lnot\exists x\,P(x)\equiv\forall x\,\lnot P(x)$
Tautology and contradiction: $p\lor\lnot p \ \text{(always true)},\qquad p\land\lnot p\ \text{(always false)}$