Read each topic in order. Study the formula cards first, then work through the solved examples on paper with the solution covered. Finally, attempt the exam-style questions before looking at the answers. Questions marked ICSE Board are verified questions from CISCE board papers and are reused here exactly for practice.
Topic 1 – Linear Inequations
A. Learning Objectives
After studying this topic you should be able to:
- Distinguish between a linear equation and a linear inequation (inequality).
- Understand the replacement set and find the solution set of an inequation.
- Apply the four rules for solving inequations, including reversing the sign when multiplying or dividing by a negative number.
- Represent solution sets on the real number line using closed and open dots.
- Solve a chain (compound) inequation such as \(a < bx + c \le d\).
B. Basic Concepts and Definitions
A statement that relates two algebraic expressions using one of the symbols \(\le\) (less than or equal to), \(\ge\) (greater than or equal to), < (less than) or > (greater than). Unlike an equation, it almost always has a set of solutions, not just one value.
The replacement set is the set from which the variable can be chosen – typically \(x \in \mathbb{W}\) (whole numbers), \(x \in \mathbb{N}\), \(x \in \mathbb{I}\) (integers) or \(x \in \mathbb{R}\) (real numbers). The solution set is the set of all values from the replacement set that actually satisfy the inequation.
Example: for \(2x + 4 \le 14\) with \(x \in \mathbb{W}\), the solution set is \(\{0, 1, 2, 3, 4, 5\}\). With \(x \in \mathbb{R}\) the same inequation has infinitely many solutions: \(x \le 5\).
- The same number can be added to or subtracted from both sides.
- Both sides can be multiplied or divided by the same positive number – the sign stays the same.
- Both sides can be multiplied or divided by the same negative number – the sign must be reversed.
Also: you may not multiply or divide by zero, and you may not multiply or divide by an unknown whose sign you do not know.
Use a closed (filled) dot for \(\le\) or \(\ge\) (the endpoint is included) and an open (hollow) dot for < or > (the endpoint is excluded). Shade the line in the direction of the solution set.
C. Important Formulas
D. Step-by-Step Solved Examples
Level 1 Inequation with whole-number replacement set
Solve \(7x - 3 \ge 11\) where \(x \in \mathbb{W}\). Write down the solution set and show it on a number line.
Required: solution set of the inequation.
Level 2 Real-number solution and half-open chain
Solve \(-3 < 2x + 1 \le 9\) where \(x \in \mathbb{R}\). Represent the solution on the real number line.
Level 3 Chain with a negative coefficient (verified ICSE board question)
Solve the inequation \(-5(x - 9) \ge 17 - 9x > x + 2, \; x \in \mathbb{R}\). Write down the solution set and represent it on the real number line.
E. Exam-Style Questions
Solve the inequation \(2x + 4 \le 14\) where \(x \in \mathbb{W}\) (whole numbers). Write down the solution set.
Show answer
\(2x \le 10\) so \(x \le 5\). Solution set = \(\{0, 1, 2, 3, 4, 5\}\).
Solve \(3x + 2 \ge 2x + 9\) where \(x \in \mathbb{I}\) (integers).
Show answer
\(3x - 2x \ge 9 - 2\), so \(x \ge 7\). Solution set = \(\{7, 8, 9, \dots\}\).
Solve \(-2 \le 3x - 1 < 8\) where \(x \in \mathbb{R}\).
Show answer
Add 1: \(-1 \le 3x < 9\), divide by 3: \(-\frac{1}{3} \le x < 3\). Solution set = \(\{x : x \in \mathbb{R},\; -\frac{1}{3} \le x < 3\}\).
If \(x \in \mathbb{N}\), solve \(4x - 7 \le 13\) and state how many solutions there are.
Show answer
\(4x \le 20\) so \(x \le 5\). Since \(x \in \mathbb{N} = \{1, 2, 3, 4, 5\}\), the solution set is \(\{1, 2, 3, 4, 5\}\) – five solutions.
Solve \(\dfrac{2x - 1}{3} > 3\) for \(x \in \mathbb{R}\) and show the solution on a number line.
Show answer
Multiply by 3: \(2x - 1 > 9\), so \(2x > 10\) and \(x > 5\). Open dot at 5, shade right.
A solution of a linear inequation is every real number strictly between \(-4\) and 1, inclusive of \(-4\). Write the inequation chain and its solution set.
Show answer
Chain: \(-4 \le x < 1\). Solution set = \(\{x : x \in \mathbb{R},\; -4 \le x < 1\}\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Multiplying or dividing by a negative without reversing the inequality sign. | Whenever you multiply or divide both sides by a negative number, reverse the symbol (\(\le \leftrightarrow \ge\) and < \(\leftrightarrow\) >). |
| Giving a finite set of solutions when \(x \in \mathbb{R}\). | For \(x \in \mathbb{R}\), write the set-builder form such as \(\{x : x \in \mathbb{R},\; x > 5\}\) and show it on a number line. |
| Using an open dot for \(\le\) (or \(\ge\)). | Use a closed (filled) dot when the endpoint is included; an open dot only for strict < or >. |
| Cross-multiplying or cancelling an unknown factor in an inequation. | Never multiply or divide by a quantity whose sign you do not know – it may reverse the inequality. |
| Splitting a chain inequation but treating it as OR. | A chain like \(a < bx + c \le d\) means both parts (AND); take the overlapping interval. |
| Stopping before dividing, leaving \(2x \ge 10\) as the final answer. | Always isolate \(x\) completely: divide to get \(x \ge 5\). |
| Including values outside the replacement set in the solution set. | Filter the solved inequality using the replacement set: for \(x \in \mathbb{W}\), keep only whole numbers that satisfy it. |
| Writing the solution set with wrong notation, e.g. \(\{x < 5\}\) or \((5, \infty]\) but excluding the endpoint wrongly. | Use proper set-builder notation \(\{x : x \in \mathbb{R},\; x \le 5\}\); endpoint 5 included → closed, \((5, \infty)\) would wrongly exclude 5. |
G. Exam Tips
- Isolate the variable exactly as in an equation; the only extra step is the sign reversal rule.
- Always state the replacement set in your final answer – ICSE marks are awarded for it.
- For number-line questions, draw dots first, then shade; a small open/closed dot error loses marks.
- If the variable disappears (e.g. \(0 \ge -5\)), state whether the statement is true for all reals or false for all reals.
- In chain inequations write \(x\) alone in the middle (e.g. \(-2 < x \le 4\)) in the final step.
H. Quick Revision Sheet – Linear Inequations
- \(\ge, \le, >, <\) – solution is usually a set, not a single value.
- Adding/subtracting: sign unchanged. Multiply/divide by positive: sign unchanged. Multiply/divide by negative: reverse sign.
- Closed dot \(\bullet\) for included; open dot \(\circ\) for excluded.
- Chain \(a < bx + c \le d\) → split and take the common part.
- Filter the final inequality using the replacement set.
Topic 2 – Quadratic Equations
A. Learning Objectives
After studying this topic you should be able to:
- Identify a quadratic equation in one variable and write it in the standard form \(ax^2 + bx + c = 0\).
- Solve quadratic equations by factorisation and by the quadratic formula.
- Use the discriminant \(b^2 - 4ac\) to determine the nature of the roots.
- Give roots to a stated degree of accuracy (one decimal place, two, or three significant figures) as required by the question.
- Solve word problems that reduce to quadratic equations.
B. Basic Concepts and Definitions
An equation of the form \(ax^2 + bx + c = 0\) where \(a \ne 0\). The values of \(x\) that satisfy it are called its roots (or solutions). A quadratic equation has at most two real roots.
Express \(ax^2 + bx + c\) as a product of two linear factors. If the product is zero, then one of the factors must be zero (the zero-product rule). For example \(x^2 - 5x + 6 = (x - 2)(x - 3) = 0\) gives \(x = 2\) or \(x = 3\).
For the equation \(px^2 - qx + r = 0\) (here \(a = p\), \(b = -q\), \(c = r\)), the roots are real and equal if and only if the discriminant is zero: \[b^2 - 4ac = (-q)^2 - 4pr = q^2 - 4pr = 0 \;\Rightarrow\; q^2 = 4pr.\]
For \(ax^2 + bx + c = 0\), the discriminant is \(D = b^2 - 4ac\):
- \(D > 0\) – two distinct real roots;
- \(D = 0\) – real and equal roots;
- \(D < 0\) – no real roots.
If the roots are required "correct to one decimal place" or "to 3 significant figures", use the quadratic formula and round only at the end.
C. Important Formulas
D. Step-by-Step Solved Examples
Level 1 Solving by factorisation
Solve \(x^2 - 5x + 6 = 0\).
Level 2 Quadratic formula to 3 significant figures (verified ICSE board question)
Solve the equation \(2x^2 - 10x + 5 = 0\), giving the roots correct to 3 significant figures.
Level 3 Quadratic formula to one decimal place (verified ICSE board question)
Solve the quadratic equation \(x^2 + 4x - 8 = 0\), correct to one decimal place.
E. Exam-Style Questions
If 3 is a root of the quadratic equation \(x^2 - px + 3 = 0\), find the value of \(p\).
Show answer
Substitute \(x = 3\): \(9 - 3p + 3 = 0\) → \(12 - 3p = 0\) → \(p = 4\).
The roots of the quadratic equation \(px^2 - qx + r = 0\) are real and equal. Show that \(q^2 = 4pr\).
Show answer
Real and equal roots require discriminant \(= 0\). Here \(a = p,\; b = -q,\; c = r\), so \(b^2 - 4ac = q^2 - 4pr = 0\), giving \(q^2 = 4pr\).
Solve \(x^2 - 8x + 15 = 0\) by factorisation.
Show answer
\((x - 3)(x - 5) = 0\), so \(x = 3\) or \(x = 5\).
Without solving, state the nature of the roots of \(2x^2 - 4x + 3 = 0\).
Show answer
\(D = (-4)^2 - 4(2)(3) = 16 - 24 = -8 < 0\). The equation has no real roots.
Solve \(x(x - 6) = 7\).
Show answer
\(x^2 - 6x - 7 = 0\) → \((x - 7)(x + 1) = 0\) → \(x = 7\) or \(x = -1\).
The product of two consecutive positive integers is 420. Find the integers.
Show answer
Let the integers be \(x\) and \(x + 1\). Then \(x(x + 1) = 420\) → \(x^2 + x - 420 = 0\) → \((x + 21)(x - 20) = 0\) → \(x = 20\) (positive). Integers are 20 and 21.
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Writing the quadratic formula with a wrong sign, e.g. \(x = -b \pm \sqrt{b^2 - 4ac}\). | The minus in front of \(b\) comes from the formula itself: \(x = \dfrac{-b \pm \sqrt{D}}{2a}\); keep \(-b\) over the denominator. |
| Forgetting the denominator \(2a\) and dividing by 2 only. | Put the whole numerator over \(2a\); when \(a \ne 1\), the denominator is not 2. |
| Rounding intermediate values while using the formula. | Keep the exact surd form as long as possible; round only the final answer to the asked accuracy. |
| Cancelling a factor of \(x\) from both sides (e.g. \(x^2 = x \Rightarrow x = 1\)). | Bring all terms to one side and factorise: \(x^2 - x = 0 \Rightarrow x(x - 1) = 0 \Rightarrow x = 0\) or \(x = 1\). |
| Discarding a negative root even when nothing in the question rules it out. | Discard a root only if it is impossible in the context (e.g. a negative length or time). |
| Confusing the roles of \(a, b, c\) when the equation is written as \(px^2 - qx + r = 0\). | First write the equation in standard form \(ax^2 + bx + c = 0\); then \(b = -q\), not \(q\). |
| Saying "no solution" for a negative discriminant in a word problem. | State "no real roots" – the equation still has complex roots, but for the ICSE syllabus you conclude no real solution. |
| Not checking roots are the answer the question wanted (e.g. giving both roots when only positive length is needed). | Re-read the question; for applied problems, write the meaningful value with its units. |
G. Exam Tips
- Try factorisation mentally for simple equations; if the middle term is awkward, go straight to the formula.
- Show \(a = \ldots,\; b = \ldots,\; c = \ldots\) and the discriminant before the answer – partial marks are awarded.
- When roots are asked "correct to one decimal place" or "to 3 significant figures", write the exact surd form first and round at the end.
- For nature-of-roots questions, quote the rule (\(D > 0\), \(D = 0\), \(D < 0\)) after computing \(D\).
- Verify by substituting the roots back into the original equation.
H. Quick Revision Sheet – Quadratic Equations
- Standard form: \(ax^2 + bx + c = 0\), \(a \ne 0\).
- Factorise → zero-product rule; otherwise quadratic formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- \(D = b^2 - 4ac\): >0 two roots, =0 equal roots, <0 no real roots.
- Sum of roots \(= -b/a\); product of roots \(= c/a\).
- Round to the asked accuracy only at the final step.
Topic 3 – Ratio and Proportion
A. Learning Objectives
After studying this topic you should be able to:
- Express a ratio and write equivalent ratios in simplest form.
- Check whether four quantities are in proportion and find the missing term.
- Find the mean proportional, third proportional and fourth proportional.
- Apply the properties \(a : b = c : d \Leftrightarrow ad = bc\).
- Apply componendo and dividendo to solve problems and equations.
B. Basic Concepts and Definitions
The ratio \(a : b\) is the comparison of two quantities of the same kind, written as a fraction \(\frac{a}{b}\) in its lowest terms. Example: the ratio 12 : 8 equals 3 : 2.
Four numbers \(a, b, c, d\) are in proportion when \(a : b = c : d\), written \(a : b :: c : d\). The key property is the cross product:
\[a : b = c : d \quad \Leftrightarrow \quad ad = bc.\]
- Mean proportional between \(a\) and \(b\): a number \(x\) such that \(a : x = x : b\), i.e. \(x^2 = ab\), so \(x = \sqrt{ab}\).
- Third proportional to \(a\) and \(b\): a number \(x\) such that \(a : b = b : x\), so \(x = \frac{b^2}{a}\).
- Fourth proportional to \(a, b, c\): a number \(x\) such that \(a : b = c : x\), so \(x = \frac{bc}{a}\).
If \(a, b, c\) are in continued proportion, then \(b^2 = ac\).
If \(\dfrac{a}{b} = \dfrac{c}{d}\), then \(\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}\). This rule lets you "compose" and "divide" numerators and denominators together and is very useful for solving equations with surds.
C. Important Formulas
D. Step-by-Step Solved Examples
Level 1 Mean proportional (verified ICSE board question)
Find the mean proportional between 4 and 9.
Level 2 Ratio from a cube ratio (verified ICSE board question)
If \(\dfrac{(a + b)^3}{(a - b)^3} = \dfrac{64}{27}\), find the ratio \(a : b\).
Level 3 Componendo and dividendo (verified ICSE board question)
Using componendo and dividendo, solve for \(x\): \[\frac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3.\]
E. Exam-Style Questions
Find the fourth proportional to 5, 8 and 15.
Show answer
\(5 : 8 = 15 : x\) → \(5x = 120\) → \(x = 24\).
If \(a : b = 3 : 4\) and \(b : c = 2 : 5\), find \(a : c\).
Show answer
Make \(b\) the same in both ratios: \(b : c = 2 : 5 = 4 : 10\). So \(a : b : c = 3 : 4 : 10\), hence \(a : c = 3 : 10\).
Find the mean proportional between 2 and 18.
Show answer
\(x^2 = 2 \times 18 = 36\), so \(x = 6\).
If 2, \(x\) and 18 are in continued proportion, find \(x\).
Show answer
Continued proportion: \(2 : x = x : 18\) → \(x^2 = 36\) → \(x = 6\).
If \(\dfrac{x}{y} = \dfrac{7}{9}\), find the value of \(\dfrac{x + y}{x - y}\).
Show answer
By componendo and dividendo: \(\dfrac{x + y}{x - y} = \dfrac{7 + 9}{7 - 9} = \dfrac{16}{-2} = -8\).
Find the third proportional to 4 and 8.
Show answer
\(4 : 8 = 8 : x\) → \(4x = 64\) → \(x = 16\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Comparing quantities of different units in a ratio. | Convert to the same unit first (e.g. 1 kg : 500 g becomes 1000 : 500 = 2 : 1). |
| Taking only the positive or negative mean proportional without context. | Mean proportional \(= \pm\sqrt{ab}\) in general; for lengths/positive quantities take the positive value. |
| Writing \(a : b = c : d\) as \(ac = bd\). | The correct identity is \(ad = bc\) (extremes × extremes = means × means). |
| Confusing third, fourth and mean proportional definitions. | Mean: \(a : x = x : b\). Third: \(a : b = b : x\). Fourth: \(a : b = c : x\). Memorise the three patterns. |
| In componendo and dividendo, applying it when the ratio is not in the form \(\frac{a}{b} = \frac{c}{d}\). | The rule applies only to equal ratios; if the RHS is a plain number \(k\), write it as \(\frac{k}{1}\) first. |
| Forgetting to reduce a ratio to lowest terms. | Always write the final ratio in simplest form, e.g. 12 : 8 = 3 : 2. |
| Assuming \(a : b = 3 : 4\) means \(a = 3\) and \(b = 4\). | Write \(a = 3k, b = 4k\); then apply the other given condition to find \(k\). |
| Squaring a surd equation and not checking for extraneous roots. | After squaring, substitute the value back into the original equation and discard any root that does not satisfy it. |
G. Exam Tips
- State the cross product step \(ad = bc\) explicitly – it is a mark-bearing step.
- For "continued proportion", always write \(a : x = x : b\) before squaring.
- With componendo and dividendo, show the two fractions \((\text{sum}, \text{difference})\) explicitly.
- Check whether the answer needs to be positive (lengths, ages, money).
- In ratio problems introduce \(k\): \(a : b = p : q \Rightarrow a = pk, b = qk\).
H. Quick Revision Sheet – Ratio and Proportion
- \(a : b = c : d \Leftrightarrow ad = bc\).
- Mean proportional \(= \sqrt{ab}\); third \(= \frac{b^2}{a}\); fourth \(= \frac{bc}{a}\).
- Continued proportion: \(b^2 = ac\).
- Componendo and dividendo: \(\frac{a}{b} = \frac{c}{d} \Rightarrow \frac{a+b}{a-b} = \frac{c+d}{c-d}\).
- Use \(k\) for ratios: \(a : b = p : q \Rightarrow a = pk, b = qk\).
Topic 4 – Factorisation of Polynomials
A. Learning Objectives
After studying this topic you should be able to:
- State and apply the remainder theorem to find remainders without long division.
- State and apply the factor theorem to test whether a linear expression is a factor.
- Find an unknown coefficient when a given linear expression is a factor.
- Factorise a cubic polynomial completely using the factor theorem and polynomial division.
B. Basic Concepts and Definitions
If a polynomial \(P(x)\) is divided by \(x - a\), the remainder is \(P(a)\). You never need to divide: just substitute \(x = a\) into the polynomial.
\(x - a\) is a factor of \(P(x)\) if and only if \(P(a) = 0\). This follows from the remainder theorem when the remainder is zero.
If \(x - a\) is a factor of a polynomial \(P(x)\) that contains an unknown coefficient \(k\), substitute \(x = a\), set \(P(a) = 0\), and solve the resulting linear equation for \(k\).
Test small integer values (\(x = \pm 1, \pm 2, \pm 3, \ldots\)) to find one linear factor. Divide to obtain a quadratic, then factorise the quadratic (or use the quadratic formula) to get all three factors.
C. Important Formulas
D. Step-by-Step Solved Examples
Level 1 Unknown coefficient (verified ICSE board question)
Find the value of \(a\) if \(x - a\) is a factor of \(3x^3 + x^2 - ax - 81\).
Level 2 Find the coefficient, then factorise completely (verified ICSE board question)
Find the value of \(k\) if \(3x^3 + 8x^2 - 15x + k\) has a factor \((x - 1)\). Hence factorise the polynomial completely.
Level 3 Factorise a cubic completely (verified ICSE board question)
Using factor theorem, factorise completely: \(2x^3 - x^2 - 13x - 6\).
E. Exam-Style Questions
If \(x - 2\) is a factor of \(x^3 - kx - 12\), find the value of \(k\).
Show answer
\(P(2) = 8 - 2k - 12 = -4 - 2k = 0\) → \(k = -2\).
Find the remainder when \(x^3 - 2x^2 + x + 5\) is divided by \(x - 1\).
Show answer
Remainder \(= P(1) = 1 - 2 + 1 + 5 = 5\).
Show that \(x + 1\) is a factor of \(x^3 + 2x^2 - 1\).
Show answer
\(x + 1 = x - (-1)\), so test \(P(-1) = -1 + 2 - 1 = 0\). Since the value is zero, \(x + 1\) is a factor.
Find the value of \(b\) if \(x - 2\) is a factor of \(2x^3 - 3x^2 + bx - 6\).
Show answer
\(P(2) = 16 - 12 + 2b - 6 = -2 + 2b = 0\) → \(b = 1\).
Factorise completely: \(x^3 - 7x + 6\).
Show answer
\(P(1) = 1 - 7 + 6 = 0\), so \((x - 1)\) is a factor. Quotient = \(x^2 + x - 6 = (x + 3)(x - 2)\). Complete factorisation: \((x - 1)(x - 2)(x + 3)\).
Find \(p\) if the polynomial \(px^3 + x^2 - 11x - 6\) leaves a remainder of 4 when divided by \(x + 1\).
Show answer
Remainder \(= P(-1) = -p + 1 + 11 - 6 = -p + 6 = 4\), so \(p = 2\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Using the divisor value instead of its root, e.g. substituting \(x = a\) for divisor \(x + a\). | For \(x + a = x - (-a)\), substitute \(x = -a\). Remember the divisor is \(x - (\text{root})\). |
| Substituting the root but forgetting the sign, e.g. \((x - 2)\) then testing \(x = +2\) as \(P(-2)\). | For \(x - 2\), test \(x = +2\); for \(x + 3\), test \(x = -3\). |
| Stopping after finding one factor of a cubic. | Continue to divide and factorise the quadratic – a complete factorisation has three linear factors where possible. |
| Adding instead of subtracting when setting \(P(\text{root}) = 0\). | Factor means remainder zero: compute \(P(\text{root})\), set explicitly equal to 0, solve. |
| Forgetting that the remainder theorem works for linear divisors only. | The theorem applies when dividing by a linear expression \((x - a)\); it is not for quadratic divisors. |
| Division errors in the cubic step (wrong quotient). | Synthetic or long division carefully; verify by multiplying the quotient back by the divisor. |
| Not testing enough candidate roots for a cubic. | Test \(\pm 1, \pm 2, \pm 3, \ldots\); for the factor to exist, one of these small integer values normally works. |
| Writing the quotient factors with wrong signs, e.g. \((2x - 1)\) instead of \((2x + 1)\). | Expand your final factorisation to check it returns the original polynomial. |
G. Exam Tips
- Write \(P(x) = \ldots\) and then \(P(\text{root}) = \ldots\) in full – examiners award marks at each substitution.
- Set the bracket for the divisor from the sign in \(x - a\): \(x - 2\) means root 2; \(x + 1\) means root \(-1\).
- "Factorise completely" needs three linear factors for a cubic; always finish the quadratic.
- Check your final factorisation by expanding and matching coefficients.
- Remainder-theorem questions never require doing the long division – substitute only.
H. Quick Revision Sheet – Factorisation
- Remainder when \(P(x) \div (x - a)\) is \(P(a)\).
- \(x - a\) is a factor \(\Leftrightarrow P(a) = 0\).
- For divisor \(x + a\), use root \(x = -a\).
- Cubic: find one linear factor, divide, then factorise the quadratic.
- Always verify by expanding the final factors.
Topic 5 – Matrices
A. Learning Objectives
After studying this topic you should be able to:
- State the order of a matrix and write individual elements.
- Use matrix equality to find unknown elements.
- Add, subtract and multiply a matrix by a scalar.
- Multiply two matrices where the orders allow it.
- Use the identity matrix \(I\) and the algebra \(A(B + C) = AB + AC\).
B. Basic Concepts and Definitions
A matrix is a rectangular arrangement of numbers in rows and columns. Its order is \(m \times n\) (read "\(m\) by \(n\)"), where \(m\) = number of rows and \(n\) = number of columns. A matrix with one column is a column matrix; with one row, a row matrix.
Two matrices are equal when they have the same order and every corresponding element is equal. This allows you to compare entries and solve for unknowns. Note that \(\begin{bmatrix}1 & 2\end{bmatrix} \ne \begin{bmatrix}1 \\ 2\end{bmatrix}\) because the orders differ.
Matrices of the same order are added (or subtracted) element by element. In scalar multiplication, every element is multiplied by the scalar: \(k\begin{bmatrix}a & b \\ c & d\end{bmatrix} = \begin{bmatrix}ka & kb \\ kc & kd\end{bmatrix}\).
\(AB\) exists only when the number of columns of \(A\) equals the number of rows of \(B\). If \(A\) is \(m \times q\) and \(B\) is \(q \times n\), then \(AB\) is \(m \times n\), with each entry the "row × column" dot product. Matrix multiplication is not commutative in general.
The identity matrix is a square matrix with 1 on the main diagonal and 0 elsewhere, e.g. \(I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\). For any compatible matrix \(A\), \(AI = IA = A\).
C. Important Formulas
D. Step-by-Step Solved Examples
Level 1 Solving a matrix equation (verified ICSE board question)
If \(\begin{bmatrix}2 & 0 \\ 0 & 4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}2 \\ -8\end{bmatrix}\), find the values of \(x\) and \(y\).
Level 2 Squaring a matrix (verified ICSE board question)
If \(A = \begin{bmatrix}2 & 2 \\ 0 & 2\end{bmatrix}\) and \(A^2 = \begin{bmatrix}4 & x \\ 0 & 4\end{bmatrix}\), find the value of \(x\).
Level 3 Matrix product with identity (verified ICSE board question)
Given \(A = \begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix}\), \(B = \begin{bmatrix}1 & 2 \\ 2 & 4\end{bmatrix}\), \(C = \begin{bmatrix}4 & 1 \\ 1 & 5\end{bmatrix}\) and \(I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\), find \(A(B + C) - 14I\).
E. Exam-Style Questions
State the order of the matrix \(\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix}\).
Show answer
It has 2 rows and 3 columns, so its order is \(2 \times 3\).
If \(\begin{bmatrix}x + 2 & 3 \\ 4 & 2y\end{bmatrix} = \begin{bmatrix}5 & 3 \\ 4 & 10\end{bmatrix}\), find \(x\) and \(y\).
Show answer
\(x + 2 = 5\) → \(x = 3\). \(2y = 10\) → \(y = 5\).
If \(A = \begin{bmatrix}2 & 3 \\ 4 & 5\end{bmatrix}\), find \(2A - 3I\).
Show answer
\(2A = \begin{bmatrix}4 & 6 \\ 8 & 10\end{bmatrix}\), \(3I = \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}\). \(2A - 3I = \begin{bmatrix}1 & 6 \\ 8 & 7\end{bmatrix}\).
If \(A = \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}\), find \(A^2\).
Show answer
\(A^2 = \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 4 \\ 0 & 1\end{bmatrix}\).
If \(A = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\), find \(A^2\).
Show answer
\(A^2 = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\). Entry (1,2) \(= 0\cdot1 + 1\cdot0 = 0\), so \(A^2 = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\).
Find \(x\) such that \(\begin{bmatrix}1 & x \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 4 \\ 0 & 1\end{bmatrix}\).
Show answer
LHS entry (1,2) \(= 1\cdot2 + x\cdot1 = 2 + x\). Equate to 4: \(x = 2\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Multiplying matrices whose orders are incompatible. | Check that columns of the first = rows of the second; only then define the product. |
| Swapping the order of multiplication \(AB\) and \(BA\). | Matrix multiplication is not commutative; always multiply in the order given. For identity: \(AI = IA = A\). |
| Treating a scalar times a matrix as multiplying only one row. | Scalar multiplication multiplies every element. |
| Confusing the order when reading \(m \times n\). | Rows × columns: \(2 \times 3\) means 2 rows, 3 columns \(-\) write \(m \times n\) with rows first and columns second. |
| Equating elements of matrices with different orders. | Equality requires identical order first; only then compare entries position by position. |
| Computing a row × row instead of row × column in a product. | Dot the row of the first matrix against a column of the second (sum of paired products). |
| Forgetting to subtract element-by-element in \(A(B + C) - 14I\). | Handle each matrix operation separately and write the resulting matrices before combining. |
| Thinking \((A + B)^2 = A^2 + 2AB + B^2\). | Since \(AB \ne BA\), \((A + B)^2 = A^2 + AB + BA + B^2\). Avoid the binomial shortcut in matrix work. |
G. Exam Tips
- Write the order compatibility check as a one-line statement before multiplying.
- Show at least one computed entry of a product so the examiner can see your row × column method.
- For matrix equations, always spell out the multiplied matrices before equating elements.
- Write \(I\) explicitly when used; it is \(2 \times 2\) for 2×2 matrices.
- Verify \(AI = A\) on small examples to confirm your multiplication.
H. Quick Revision Sheet – Matrices
- Order = rows × columns; equality requires same order and matching entries.
- Add/subtract element-wise; scalar multiplies every element.
- \(AB\) exists iff columns(\(A\)) = rows(\(B\)); result order is rows(\(A\)) × columns(\(B\)).
- Row of \(A\) × column of \(B\): sum of pairwise products.
- \(AB \ne BA\) in general; \(A(B + C) = AB + AC\); \(AI = IA = A\).
Topic 6 – Arithmetic Progression (AP)
A. Learning Objectives
After studying this topic you should be able to:
- Recognise an arithmetic progression and find its common difference.
- Write the nth term \(T_n = a + (n-1)d\).
- Find the first term, common difference or term number when other information is given.
- Use the sum of n terms formula, including the last-term form \(S_n = \frac{n}{2}(a + l)\).
- Solve problems that give two terms and require \(a\), \(d\) and a sum.
B. Basic Concepts and Definitions
A sequence in which each term after the first differs from the previous term by a fixed number called the common difference \(d\). Example: \(15, 30, 45, 60, \ldots\) is an AP with \(a = 15\) and \(d = 15\).
\(T_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference. To find a middle term given two other terms, set up equations in \(a\) and \(d\).
\(S_n = \dfrac{n}{2}\big[2a + (n-1)d\big]\), or \(S_n = \dfrac{n}{2}(a + l)\) where \(l = T_n\) is the last term. Both give the same total; use the second form when the last term is known.
If \(a, b, c\) are in AP, then \(2b = a + c\) (the middle term is the average of its neighbours). This is a quick checking tool.
C. Important Formulas
D. Step-by-Step Solved Examples
Level 1 Term directly from the nth term (verified ICSE board question)
The nth term of an AP is \(2n + 5\). Find its 10th term (and the common difference).
Level 2 First term and common difference from two terms (verified ICSE board question)
The 5th and 9th terms of an AP are 4 and \(-12\). Find (a) the first term, (b) the common difference, (c) the sum of the first 16 terms.
Level 3 Term number and sum (verified ICSE board question)
Which term of the AP \(15, 30, 45, 60, \ldots\) is 300? Hence find the sum of all the terms from the first term up to this term.
E. Exam-Style Questions
The terms of an AP are given by \(\dfrac{1}{a},\; \dfrac{1}{a} + 1,\; \dfrac{1}{a} + 2, \ldots\) Write the 7th term.
Show answer
First term \(= \frac{1}{a}\), \(d = 1\). \(T_7 = \frac{1}{a} + 6\).
The nth term of an AP is \(T_n = 6(7 - n)\). Find the first term \(a\), the common difference \(d\), and the sum of the first 25 terms.
Show answer
\(a = T_1 = 6(6) = 36\); \(T_2 = 6(5) = 30\) so \(d = -6\). \(S_{25} = \frac{25}{2}\big[2(36) + 24(-6)\big] = \frac{25}{2}(72 - 144) = \frac{25}{2}(-72) = -900\).
Find the 20th term of the AP \(5, 9, 13, \ldots\).
Show answer
\(a = 5\), \(d = 4\). \(T_{20} = 5 + 19(4) = 81\).
Find the sum \(1 + 2 + 3 + \cdots + 50\).
Show answer
\(S = \frac{50}{2}(1 + 50) = 25 \times 51 = 1275\).
If the nth term of an AP is \(2n + 1\), find the sum of its first 10 terms.
Show answer
\(a = 3\), \(d = 2\). \(S_{10} = \frac{10}{2}\big[2(3) + 9(2)\big] = 5(6 + 18) = 120\).
Find the sum of the first 15 terms of an AP whose first term is 5 and common difference is 7.
Show answer
\(S_{15} = \frac{15}{2}\big[2(5) + 14(7)\big] = \frac{15}{2}(10 + 98) = \frac{15}{2} \times 108 = 810\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Using \(a + nd\) instead of \(a + (n-1)d\). | The first term already accounts for \(d = 0\) at \(n = 1\); always write \((n - 1)\). |
| Computing the common difference from the first two terms incorrectly, e.g. \(T_1 - T_2\). | \(d = T_{k+1} - T_k\), e.g. \(T_2 - T_1\) (later minus earlier). |
| Writing the sum formula without halving: \(n(2a + (n-1)d)\). | The sum is \(\frac{n}{2}[2a + (n-1)d]\); the \(\frac{n}{2}\) factor is essential. |
| Using \(\frac{n}{2}(a + l)\) but with the wrong last term. | \(l\) must be the actual last term included in the sum, e.g. the 20th term for \(S_{20}\). |
| Mixing up two given terms: 5th term means \(a + 4d\), not \(a + 5d\). | Term \(k\) has \(k - 1\) differences; write \(T_5 = a + 4d\) carefully. |
| Forgetting the sign when \(d\) is negative. | Keep brackets: \(2a + (n-1)d\) with \(d = -4\) becomes \(2a - 4(n-1)\). |
| Not checking the arithmetic of long sums (e.g. \(\frac{25}{2} \times (-72)\)). | Split: \(25 \times (-36) = -900\); verify stepwise. |
| Reading "sum of all terms" but only giving the last term or term number. | The question asks for the sum \(S_n\) – compute it after finding \(n\). |
G. Exam Tips
- When two terms are given, always write out both equations in \(a\) and \(d\) before solving.
- For \(S_n\), substitute the values immediately: \(\frac{n}{2}\) then the bracket.
- State \(a = \ldots,\; d = \ldots\) explicitly – the examiner checks these first.
- Use \(\frac{n}{2}(a + l)\) whenever the last term is known or easy to get.
- Check totals: an AP with negative \(d\) gives a sum that eventually decreases.
H. Quick Revision Sheet – Arithmetic Progression
- \(d = T_{k+1} - T_k\).
- \(T_n = a + (n-1)d\).
- \(S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)\).
- \(a, b, c\) in AP \(\Rightarrow 2b = a + c\).
- Term \(k\) uses \(k - 1\) differences in its formula.
Topic 7 – Geometric Progression (GP)
A. Learning Objectives
After studying this topic you should be able to:
- Recognise a geometric progression and find its common ratio \(r\).
- Write the nth term \(T_n = ar^{n-1}\).
- Use the sum of n terms formula for a GP.
- Find a term number from a given term value (using powers).
- Distinguish AP situations from GP situations in problems.
B. Basic Concepts and Definitions
A sequence in which each term after the first is obtained by multiplying the previous term by a fixed number called the common ratio \(r\). Example: \(15, 30, 60, 120, \ldots\) is a GP with \(a = 15\), \(r = 2\).
\(T_n = ar^{n-1}\). The common ratio \(r = \frac{T_{k+1}}{T_k}\). A ratio \(r\) may be negative or fractional (e.g. \(1, -\frac12, \frac14, \ldots\)).
\(S_n = \dfrac{a(r^n - 1)}{r - 1}\). For \(|r| < 1\) the same formula is often written \(S_n = \dfrac{a(1 - r^n)}{1 - r}\), which keeps the denominator positive. Both are identical.
\(a, b, c\) are in GP if and only if \(b^2 = ac\) (the middle term squared equals the product of its neighbours).
C. Important Formulas
D. Step-by-Step Solved Examples
Level 1 Finding a specific term
Find the 5th term of the GP \(3, 6, 12, \ldots\).
Level 2 nth term and term count from a sum (verified ICSE board question)
The first four terms of a GP are \(15, 30, 60, 120, \ldots\) (a) Write the nth term. (b) Find the value of \(n\) for which the sum of the first \(n\) terms is 945.
Level 3 Sum with a fractional ratio
Find the sum of the first 6 terms of the GP \(1, \frac12, \frac14, \ldots\).
E. Exam-Style Questions
Which term of the GP \(2, 4, 8, \ldots\) is 512?
Show answer
\(a = 2\), \(r = 2\). \(T_n = 2 \times 2^{n-1} = 2^n = 512 = 2^9\), so \(n = 9\).
Find the 6th term of the GP \(81, 27, 9, \ldots\).
Show answer
\(a = 81\), \(r = \frac13\). \(T_6 = 81 \times (\frac13)^5 = 81 \times \frac{1}{243} = \frac13\).
Find the sum \(1 + 2 + 4 + 8 + 16\).
Show answer
\(a=1\), \(r=2\), \(n=5\): \(S = \frac{1(2^5 - 1)}{2 - 1} = \frac{32 - 1}{1} = 31\).
Show that \(5, 10, 20\) are in GP, and write the common ratio.
Show answer
\(b^2 = 10^2 = 100\) and \(a \cdot c = 5 \cdot 20 = 100\); since \(b^2 = ac\), they are in GP with \(r = 2\).
The product of the first three terms of a GP with \(a = 2\) and \(r = 3\) equals the product of which three numbers?
Show answer
Terms are \(2, 6, 18\); product \(= 2 \times 6 \times 18 = 216\).
Find the sum of the GP \(3 + 6 + 12 + \cdots + 192\).
Show answer
\(a = 3\), \(r = 2\). Last term \(192 = 3 \times 2^{n-1}\) → \(2^{n-1} = 64\) → \(n = 7\). \(S_7 = \frac{3(2^7 - 1)}{2 - 1} = 3 \times 127 = 381\).
F. Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Using an AP formula for a GP problem (adding instead of multiplying). | AP adds a common difference; GP multiplies by a common ratio. Decide first which one is being described. |
| Writing the exponent as \(n\) instead of \(n - 1\) in \(T_n\). | The first term is \(ar^0\), so the \(n\)th term uses \(n - 1\) multiplications by \(r\). |
| Computing \(r\) backwards: \(r = \frac{T_k}{T_{k+1}}\). | \(r = \frac{T_{k+1}}{T_k}\), i.e. later term divided by earlier term. |
| Using \(S_n = \frac{a(r^n - 1)}{r - 1}\) with a fractional \(r < 1\) and getting a wrong sign. | For \(|r| < 1\) use \(\frac{a(1 - r^n)}{1 - r}\) so the denominator is positive. |
| Forgetting the term \(a\) multiplies the whole bracket in \(S_n\). | \(S_n = a \cdot \frac{r^n - 1}{r - 1}\); the leading \(a\) multiplies the entire fraction. |
| Treating negative \(r\) with an even power carelessly (sign errors). | With \(r = -2\), powers alternate sign: \(r^3 = -8\), \(r^4 = 16\). Decide the sign first. |
| Not recognising \(2^n = 64\) for \(n = 6\) and guessing. | Learn powers of 2 up to \(2^{10}\) and other common powers; then the term-number step is instant. |
| Misreading \(1 + \frac12 + \frac14 + \cdots\) as an AP. | The ratio \(\frac12\) (multiply) means it is a GP with \(|r| < 1\) – use the fraction form of the sum. |
G. Exam Tips
- Write \(a = \ldots,\; r = \ldots\) at the top; they are the first two marks.
- For the sum, show \(S_n = a\frac{r^n - 1}{r - 1}\) and substitute in one neat line.
- When a term value is given, work with powers: \(T_n = 512 \Rightarrow 2^{n-1} \times 2 = 2^n\).
- For \(|r| < 1\) choose the form with \(1 - r\) in the denominator to avoid negative signs.
- Practise the powers \(2^1\) to \(2^{10}\) and \(3^1\) to \(3^5\) for fast term-count work.
H. Quick Revision Sheet – Geometric Progression
- \(r = \frac{T_{k+1}}{T_k}\); GP multiplies by \(r\), AP adds by \(d\).
- \(T_n = ar^{n-1}\).
- \(S_n = \frac{a(r^n - 1)}{r - 1}\) (use \(\frac{a(1 - r^n)}{1 - r}\) for \(|r| < 1\)).
- \(a, b, c\) in GP \(\Rightarrow b^2 = ac\).
- Match powers to find the term number \(n\).
Now that you have revised all seven topics, attempt the topic quiz for instant feedback, then the practice paper and finally the timed chapter test.