Important Derivations

Steps

1. 1

   Let the first term be 'a' and common difference be 'd'

2. 2

   a₁ = a

3. 3

   a₂ = a₁ + d = a + d

4. 4

   a₃ = a₂ + d = a + 2d

5. 5

   a₄ = a + 3d, and by continuing the pattern …

6. 6

   The nth term: aₙ = a + (n − 1)d

Steps

1. 1

   Let l = last term = a + (n − 1)d

2. 2

   Write sum forward: Sₙ = a + (a+d) + (a+2d) + … + (l−d) + l …(i)

3. 3

   Write sum backward: Sₙ = l + (l−d) + (l−2d) + … + (a+d) + a …(ii)

4. 4

   Add (i) and (ii): 2Sₙ = (a+l) + (a+l) + … + (a+l) [n terms] = n(a+l)

5. 5

   Therefore Sₙ = n/2 × (a + l)

6. 6

   Substituting l = a + (n−1)d: Sₙ = n/2 [2a + (n−1)d]

Steps

1. 1

   Given: Triangle ABC with DE ∥ BC (D on AB, E on AC). To prove: AD/DB = AE/EC.

2. 2

   Join BE and CD. Draw EF ⊥ AB and DG ⊥ AC (altitudes for area calculation).

3. 3

   Area(△ADE) = ½ × AD × EF; Area(△BDE) = ½ × BD × EF

4. 4

   ∴ Area(△ADE)/Area(△BDE) = AD/BD …(i)

5. 5

   Area(△ADE) = ½ × AE × DG; Area(△CDE) = ½ × CE × DG

6. 6

   ∴ Area(△ADE)/Area(△CDE) = AE/CE …(ii)

7. 7

   △BDE and △CDE have the same base DE and lie between the same parallels DE ∥ BC ∴ Area(△BDE) = Area(△CDE) …(iii)

8. 8

   From (i), (ii), (iii): AD/BD = AE/CE → AD/DB = AE/EC (proved)

Steps

1. 1

   Given: Triangle ABC, right-angled at A. Draw AD ⊥ BC (D lies on BC). To prove: BC² = AB² + AC².

2. 2

   In △ABC and △DBA: ∠BAC = ∠BDA = 90°; ∠B is common ∴ △ABC ~ △DBA (AA similarity)

3. 3

   From similarity: AB/DB = BC/AB → AB² = DB × BC …(i)

4. 4

   In △ABC and △DAC: ∠BAC = ∠ADC = 90°; ∠C is common ∴ △ABC ~ △DAC (AA similarity)

5. 5

   From similarity: AC/DC = BC/AC → AC² = DC × BC …(ii)

6. 6

   Adding (i) and (ii): AB² + AC² = DB × BC + DC × BC = BC(DB + DC) = BC × BC = BC²

7. 7

   Therefore BC² = AB² + AC² (proved)

Steps

1. 1

   Given: Triangle ABC where AB² + AC² = BC². To prove: ∠BAC = 90°.

2. 2

   Construct a separate triangle PQR with PQ = AB, PR = AC, and ∠QPR = 90°

3. 3

   By the Pythagorean theorem applied to △PQR (right-angled at P): QR² = PQ² + PR² = AB² + AC²

4. 4

   But given AB² + AC² = BC² ∴ QR² = BC² → QR = BC

5. 5

   In △ABC and △PQR: AB = PQ, AC = PR, BC = QR (all sides equal) ∴ △ABC ≅ △PQR (SSS)

6. 6

   Therefore ∠BAC = ∠QPR = 90° (proved)

Steps

1. 1

   Given: Circle with centre O and radius OP; AB is a tangent to the circle at point P. To prove: OP ⊥ AB.

2. 2

   Proof by contradiction: Assume OP is NOT perpendicular to AB.

3. 3

   Then draw OQ ⊥ AB where Q is the foot of perpendicular (Q ≠ P, Q lies on AB).

4. 4

   In right triangle OQP: OQ < OP (perpendicular is the shortest distance from a point to a line).

5. 5

   But OP = radius ∴ OQ < radius → Q lies inside the circle.

6. 6

   This means the tangent AB passes through Q which is inside the circle — impossible (a tangent can touch the circle at only one external point).

7. 7

   This is a contradiction ∴ our assumption is wrong. Hence OP ⊥ AB. (proved)

Steps

1. 1

   Given: Circle with centre O; P is an external point; PA and PB are tangents where A and B are points of tangency. To prove: PA = PB.

2. 2

   Join OA, OB, and OP.

3. 3

   In triangles OAP and OBP:

4. 4

   OA = OB (radii of the same circle)

5. 5

   ∠OAP = ∠OBP = 90° (radius ⊥ tangent at point of contact — proved above)

6. 6

   OP = OP (common hypotenuse)

7. 7

   ∴ △OAP ≅ △OBP (RHS congruence criterion)

8. 8

   Therefore PA = PB (CPCT — proved)

Steps

1. 1

   Draw a concave mirror with pole P, focus F, and centre of curvature C

2. 2

   Place object AB beyond C; draw two rays — one parallel to principal axis (reflects through F) and one through C (reflects back on itself)

3. 3

   Mark image A'B' at intersection of reflected rays

4. 4

   Using similar triangles ABP and A'B'P: A'B'/AB = PA'/PA → m = −v/u

5. 5

   Using similar triangles ABF and A'B'F: (PA−PF)/PF = PA'/PA → (u−f)/f = v/u

6. 6

   Rearrange: uv = uf + fv → divide both sides by uvf → 1/f = 1/v + 1/u

Steps

1. 1

   Draw object AB and image A'B' for a concave mirror

2. 2

   Triangles ABP and A'B'P are similar (AA: right angle at base, common angle at P)

3. 3

   A'B'/AB = PA'/PA

4. 4

   With sign convention (distances from pole, real positive for image): h'/h = −v/u

5. 5

   Therefore m = h'/h = −v/u

Steps

1. 1

   Draw a convex lens with optical centre O; place object AB on left side

2. 2

   Draw two rays: one parallel to principal axis (refracts through F₂) and one through optical centre (passes straight)

3. 3

   Mark image A'B' at intersection of refracted rays

4. 4

   Using similar triangles OAB and OA'B': A'B'/AB = OA'/OA → h'/h = v/u

5. 5

   Using similar triangles F₁OC and F₁A'B': A'B'/OC = (OA'−OF₁)/OF₁ → v/u = (v−f)/f

6. 6

   Cross multiply and rearrange, then divide by uvf: 1/f = 1/v − 1/u

Steps

1. 1

   Draw the lens with object AB (on left) and image A'B' (on right)

2. 2

   Triangles OAB and OA'B' are similar (vertically opposite angles at O, right angles at A and A')

3. 3

   A'B'/AB = OA'/OA

4. 4

   With sign convention: h'/h = v/u

5. 5

   Therefore m = v/u

Steps

1. 1

   Place two thin lenses L₁ (focal length f₁) and L₂ (focal length f₂) in contact. Object O is placed on the principal axis.

2. 2

   Lens L₁ alone forms image I₁. By lens formula: 1/v₁ − 1/u = 1/f₁ …(i)

3. 3

   I₁ acts as virtual object for lens L₂. By lens formula: 1/v − 1/v₁ = 1/f₂ …(ii)

4. 4

   Add equations (i) and (ii): 1/v − 1/u = 1/f₁ + 1/f₂

5. 5

   For the equivalent single lens with focal length f: 1/v − 1/u = 1/f

6. 6

   Therefore 1/f = 1/f₁ + 1/f₂

7. 7

   Since Power P = 1/f (in dioptre, f in metres): P = P₁ + P₂

Steps

1. 1

   In series combination, same current I flows through all resistors; the voltage divides across them

2. 2

   Total voltage: V = V₁ + V₂ + V₃ …(i)

3. 3

   By Ohm's law: V₁ = IR₁, V₂ = IR₂, V₃ = IR₃

4. 4

   Substituting into (i): IRₛ = IR₁ + IR₂ + IR₃

5. 5

   Dividing both sides by I: Rₛ = R₁ + R₂ + R₃

Steps

1. 1

   In parallel combination, same voltage V appears across each resistor; current divides

2. 2

   Total current: I = I₁ + I₂ + I₃ …(i)

3. 3

   By Ohm's law: I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃

4. 4

   Substituting into (i): V/Rₚ = V/R₁ + V/R₂ + V/R₃

5. 5

   Dividing both sides by V: 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃

Steps

1. 1

   Work done to move charge Q through potential difference V: W = VQ

2. 2

   Since Q = It (charge = current × time): W = VIt …(i)

3. 3

   By Ohm's law, V = IR: W = (IR) × It = I²Rt

4. 4

   Assuming all electrical energy converts to heat: H = W = I²Rt

5. 5

   Alternate forms: H = VIt (from i) and H = V²t/R (substituting I = V/R)

Steps

1. 1

   Power is defined as work done per unit time: P = W/t

2. 2

   Work done in time t: W = VIt (from Joule's law derivation)

3. 3

   Therefore: P = VIt/t = VI

4. 4

   Substituting V = IR (Ohm's law): P = (IR) × I = I²R

5. 5

   Substituting I = V/R: P = V × (V/R) = V²/R