Arithmetic Progressions — Andhra Pradesh (SSC) Class 10 Mathematics Solutions (Free)

An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference (d). General form: a, a + d, a + 2d, a + 3d, ... Condition: a sequence t₁, t₂, t₃, ... is an AP if and only if: tₙ₊₁ − tₙ = d (the same constant value) for every consecutive pair of terms. Key formulas: - nth term: tₙ = a + (n − 1)d - Sum of n terms: Sₙ = n/2 × [2a + (n − 1)d]

Given AP: 7, 11, 15, 19, ... a = 7, d = 11 − 7 = 4 Using tₙ = a + (n − 1)d for n = 18: t₁₈ = 7 + (18 − 1) × 4 t₁₈ = 7 + 68 t₁₈ = 75 The 18th term is 75.

Given: S₁₅ = 825, d = 6, n = 15. Find a. Using Sₙ = n/2 × [2a + (n − 1)d]: 825 = 15/2 × [2a + 14 × 6] 825 = 7.5 × [2a + 84] 110 = 2a + 84 2a = 26 a = 13 Verification: S₁₅ = 15/2 × [2(13) + 84] = 7.5 × 110 = 825 ✓ The first term is a = 13.

Given AP: 9, 13, 17, 21, ... so a = 9, d = 4. Assume 200 is the nth term: 200 = 9 + (n − 1) × 4 191 = 4(n − 1) n − 1 = 47.75 n = 48.75 Since n must be a positive integer and 48.75 is not one, 200 is NOT a term of this AP. The nearest actual terms are: t₄₈ = 9 + 47 × 4 = 197 t₄₉ = 9 + 48 × 4 = 201 200 falls strictly between these, confirming it never appears in the sequence.

Step 1: Find the first three-digit multiple of 6. Smallest three-digit number is 100; 100 ÷ 6 = 16.67, so the first multiple is 6 × 17 = 102. Step 2: Find the last three-digit multiple of 6. Largest three-digit number is 999; 999 ÷ 6 = 166.5, so the last multiple is 6 × 166 = 996. Step 3: This forms an AP: 102, 108, ..., 996, with a = 102, d = 6, l = 996. Using tₙ = a + (n − 1)d: 996 = 102 + (n − 1) × 6 894 = 6(n − 1) n − 1 = 149 n = 150 Step 4: Sum using Sₙ = n/2 × (a + l) S₁₅₀ = 150/2 × (102 + 996) S₁₅₀ = 75 × 1,098 S₁₅₀ = 82,350 The sum of all three-digit numbers divisible by 6 is 82,350.

Given: t₈ = 50, t₁₆ = 98. Step 1: Set up equations using tₙ = a + (n − 1)d t₈: a + 7d = 50 ... (1) t₁₆: a + 15d = 98 ... (2) Step 2: Subtract (1) from (2) 8d = 48 d = 6 Step 3: Substitute back into (1) a + 7(6) = 50 a + 42 = 50 a = 8 Verification: t₈ = 8 + 7(6) = 8 + 42 = 50 ✓ t₁₆ = 8 + 15(6) = 8 + 90 = 98 ✓ The first term is a = 8, and the common difference is d = 6.

Given: Sₙ = 5n² + 3n. Step 1: First term S₁ = t₁ = 5(1)² + 3(1) = 8 Step 2: Second term S₂ = 5(2)² + 3(2) = 26, and t₂ = S₂ − S₁ = 26 − 8 = 18 Step 3: Common difference d = t₂ − t₁ = 18 − 8 = 10 Step 4: General term tₙ = a + (n − 1)d = 8 + 10(n − 1) = 10n − 2 Verification via tₙ = Sₙ − Sₙ₋₁: [5n² + 3n] − [5(n−1)² + 3(n−1)] = 10n − 2 ✓ (matches) The AP is 8, 18, 28, 38, ... with general term tₙ = 10n − 2.

Let the three numbers in AP be (a − d), a, and (a + d). Step 1: Use the sum condition (a − d) + a + (a + d) = 30 3a = 30 a = 10 So the numbers are (10 − d), 10, and (10 + d). Step 2: Use the product condition (10 − d)(10)(10 + d) = 750 10(100 − d²) = 750 100 − d² = 75 d² = 25 d = ±5 Step 3: Find the three numbers If d = 5: the numbers are (10 − 5), 10, (10 + 5) = 5, 10, 15 Step 4: Verification Sum: 5 + 10 + 15 = 30 ✓ Product: 5 × 10 × 15 = 750 ✓ The three numbers are 5, 10, and 15.

Step 1: Find the first two-digit multiple of 8. Smallest two-digit number is 10; the first multiple of 8 at or above it is 16. Step 2: Find the last two-digit multiple of 8. Largest two-digit number is 99; the last multiple of 8 at or below it is 96. Step 3: This forms an AP: 16, 24, 32, ..., 96 with a = 16, d = 8, l = 96. Using tₙ = a + (n − 1)d: 96 = 16 + (n − 1) × 8 80 = 8(n − 1) n − 1 = 10 n = 11 There are 11 two-digit numbers divisible by 8.