AP, GP, Coordinate Geometry, Reflection & Equation of a Line — Practice

Question 1

The 5th term of an AP is 18 and the 9th term is 30. Find (a) the common difference, (b) the first term, (c) the 15th term.

Accepted Answer

Using aₙ = a + (n−1)d: a₅ = a + 4d = 18 ...(1). a₉ = a + 8d = 30 ...(2). Subtracting (1) from (2): 4d = 12, so d = 3. From (1): a + 4(3) = 18 → a + 12 = 18 → a = 6. (c) a₁₅ = a + 14d = 6 + 14(3) = 6 + 42 = 48. Answers: (a) d = 3, (b) a = 6, (c) a₁₅ = 48. Verify: a₅ = 6+4(3) = 18 ✓. a₉ = 6+8(3) = 30 ✓.

Question 2

Find the coordinates of the point that divides the line segment joining A(−3, 5) and B(9, −7) internally in the ratio 1:2.

Accepted Answer

Using section formula (internal division in ratio m:n = 1:2): x = (m×x₂ + n×x₁)/(m+n) = (1×9 + 2×(−3))/(1+2) = (9 − 6)/3 = 3/3 = 1. y = (m×y₂ + n×y₁)/(m+n) = (1×(−7) + 2×5)/(1+2) = (−7 + 10)/3 = 3/3 = 1. The point is P(1, 1). Verification: AP = √[(1−(−3))² + (1−5)²] = √[16+16] = 4√2. PB = √[(9−1)² + (−7−1)²] = √[64+64] = 8√2. AP:PB = 4√2 : 8√2 = 1:2 ✓.

Question 3

The sum of the first 3 terms of a GP is 21 and the product of these terms is 216. Find the GP.

Accepted Answer

Let the three terms be a/r, a, ar. Sum: a/r + a + ar = 21 ...(1). Product: (a/r) × a × ar = a³ = 216 → a = 6. Substituting a = 6 into (1): 6/r + 6 + 6r = 21 → 6/r + 6r = 15 → Multiply by r: 6 + 6r² = 15r → 6r² − 15r + 6 = 0 → 2r² − 5r + 2 = 0 → (2r − 1)(r − 2) = 0. r = 1/2 or r = 2. If r = 2: terms are 6/2, 6, 6×2 = 3, 6, 12. GP: 3, 6, 12. If r = 1/2: terms are 6/(1/2), 6, 6×(1/2) = 12, 6, 3. GP: 12, 6, 3 (same terms in reverse). Both are valid GPs. Check: 3+6+12 = 21 ✓. 3×6×12 = 216 ✓.

Practice — AP, GP, Coordinate Geometry, Reflection & Equation of a Line

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