Arithmetic Progressions — Karnataka (SSLC) Class 10 Mathematics Solutions (Free)

An arithmetic progression (AP) is a sequence in which the difference between any two consecutive terms stays the same throughout — this constant is called the common difference (d). General form: a, a + d, a + 2d, a + 3d, ... Test: 3, 7, 11, 15, ... 7 - 3 = 4, 11 - 7 = 4, 15 - 11 = 4 — the difference is constant, so this IS an AP with a = 3, d = 4. Test: 2, 4, 8, 16, ... 4 - 2 = 2, 8 - 4 = 4, 16 - 8 = 8 — the difference keeps changing (it doubles each time), so this is NOT an AP. It is a geometric progression instead, where the RATIO between terms is constant, not the difference. Key formulas for an AP: - nth term: tₙ = a + (n − 1)d - Sum of n terms: Sₙ = n/2 × [2a + (n − 1)d]

Given AP: 6, 11, 16, 21, ... Step 1: a = 6, d = 11 − 6 = 5 (Check: 16 − 11 = 5, 21 − 16 = 5 ✓) Step 2: Apply tₙ = a + (n − 1)d for n = 12 t₁₂ = 6 + (12 − 1) × 5 t₁₂ = 6 + 55 t₁₂ = 61 The 12th term is 61.

Given: S₂₄ = 1200, d = 4, n = 24. Find a. Step 1: Use Sₙ = n/2 × [2a + (n − 1)d] 1200 = 24/2 × [2a + 23 × 4] 1200 = 12 × [2a + 92] Step 2: Solve for a 100 = 2a + 92 2a = 8 a = 4 Verification: S₂₄ = 12 × [2(4) + 92] = 12 × 100 = 1200 ✓ The first term is a = 4.

Given AP: 6, 11, 16, 21, ... so a = 6, d = 5. Assume 150 is the nth term: 150 = 6 + (n − 1) × 5 144 = 5(n − 1) n − 1 = 28.8 n = 29.8 Since n must be a positive integer and 29.8 is not one, 150 is NOT a term of this AP. The nearest actual terms are: t₂₉ = 6 + 28 × 5 = 146 t₃₀ = 6 + 29 × 5 = 151 150 falls strictly between them, confirming it is not in the sequence.

Step 1: First three-digit multiple of 4 Smallest three-digit number is 100, and 100 ÷ 4 = 25 exactly, so the first term is a = 100. Step 2: Last three-digit multiple of 4 Largest three-digit number is 999; 999 ÷ 4 = 249.75, so the last multiple is 4 × 249 = 996. Step 3: This is an AP: 100, 104, 108, ..., 996 with a = 100, d = 4, l = 996. Using tₙ = a + (n − 1)d: 996 = 100 + (n − 1) × 4 896 = 4(n − 1) n − 1 = 224 n = 225 Step 4: Sum using Sₙ = n/2 × (a + l) S₂₂₅ = 225/2 × (100 + 996) S₂₂₅ = 112.5 × 1096 S₂₂₅ = 123,300 The sum of all three-digit numbers divisible by 4 is 123,300.

Given: t₉ = 53, t₁₈ = 125. Step 1: Write both as equations using tₙ = a + (n − 1)d a + 8d = 53 ... (1) a + 17d = 125 ... (2) Step 2: Subtract (1) from (2) 9d = 72 d = 8 Step 3: Substitute back into (1) a + 8(8) = 53 a + 64 = 53 a = −11 Verification: t₉ = −11 + 8(8) = −11 + 64 = 53 ✓ t₁₈ = −11 + 17(8) = −11 + 136 = 125 ✓ First term a = −11, common difference d = 8.

Given Sₙ = 4n² − n. Step 1: First term S₁ = t₁ = 4(1)² − 1 = 3 Step 2: Second term S₂ = 4(2)² − 2 = 14, and t₂ = S₂ − S₁ = 14 − 3 = 11 Step 3: Common difference d = t₂ − t₁ = 11 − 3 = 8 Step 4: General term, two ways Directly: tₙ = a + (n − 1)d = 3 + 8(n − 1) = 8n − 5 Check via tₙ = Sₙ − Sₙ₋₁: 4n² − n − [4(n−1)² − (n−1)] = 8n − 5 ✓ (matches) The AP is 3, 11, 19, 27, ... with general term tₙ = 8n − 5.

Let the three numbers be (a − d), a, (a + d). Step 1: Sum condition (a − d) + a + (a + d) = 27 3a = 27 a = 9 Step 2: Product condition (9 − d)(9)(9 + d) = 504 9(81 − d²) = 504 81 − d² = 56 d² = 25 d = ±5 Step 3: The three numbers With d = 5: 4, 9, 14. (d = −5 gives the same set in reverse.) Verification: 4 + 9 + 14 = 27 ✓ 4 × 9 × 14 = 504 ✓ The numbers are 4, 9, and 14.

Step 1: First two-digit multiple of 6 Smallest two-digit number is 10; the first multiple of 6 at or above it is 12. Step 2: Last two-digit multiple of 6 Largest two-digit number is 99; the last multiple of 6 at or below it is 96. Step 3: This is an AP: 12, 18, 24, ..., 96 with a = 12, d = 6, l = 96. Using tₙ = a + (n − 1)d: 96 = 12 + (n − 1) × 6 84 = 6(n − 1) n − 1 = 14 n = 15 There are 15 two-digit numbers divisible by 6.