Progressions — Telangana (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Progressions" — important questions with detailed answers, download PDF for board exam preparation.

Progressions — Telangana (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Progressions" — important questions with detailed answers, download PDF for board exam preparation.

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TL;DR: Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Progressions" — important questions with detailed answers, download PDF for boar…

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Progressions — Textbook Questions Answered (8)

Q1: Find the 10th term of the arithmetic progression 2, 5, 8, 11, ...

Step 1: Identify first term a = 2 Step 2: Find common difference d = 5 - 2 = 3 Step 3: Use formula aₙ = a + (n-1)d Step 4: a₁₀ = 2 + (10-1) × 3 Step 5: a₁₀ = 2 + 9 × 3 = 2 + 27 = 29 Final Answer: The 10th term is 29

Q2: Find the sum of first 20 terms of AP where a = 5 and d = 3.

Step 1: First term a = 5, common difference d = 3, n = 20 Step 2: Use formula Sₙ = n/2 × [2a + (n-1)d] Step 3: S₂₀ = 20/2 × [2(5) + (20-1)(3)] Step 4: S₂₀ = 10 × [10 + 19 × 3] Step 5: S₂₀ = 10 × [10 + 57] = 10 × 67 = 670 Final Answer: Sum of first 20 terms = 670

Q3: Find the 6th term of the geometric progression 3, 6, 12, 24, ...

Step 1: Identify first term a = 3 Step 2: Find common ratio r = 6/3 = 2 Step 3: Use formula aₙ = a × r^(n-1) Step 4: a₆ = 3 × 2^(6-1) = 3 × 2⁵ Step 5: a₆ = 3 × 32 = 96 Final Answer: The 6th term is 96

Q4: Find the sum of the geometric series 2 + 4 + 8 + 16 + ... up to 8 terms.

Step 1: First term a = 2, common ratio r = 2, n = 8 Step 2: Use formula Sₙ = a(rⁿ - 1)/(r - 1) Step 3: S₈ = 2(2⁸ - 1)/(2 - 1) Step 4: S₈ = 2(256 - 1)/1 = 2 × 255 Step 5: S₈ = 510 Final Answer: Sum of 8 terms = 510

Q5: If the 4th term of an AP is 13 and the 9th term is 33, find the AP.

Step 1: Let first term = a, common difference = d Step 2: a₄ = a + 3d = 13 ... (1) Step 3: a₉ = a + 8d = 33 ... (2) Step 4: Subtract (1) from (2): 5d = 20, so d = 4 Step 5: From (1): a + 3(4) = 13, so a = 1 Step 6: AP is 1, 5, 9, 13, 17, ... Final Answer: AP = 1, 5, 9, 13, ... with a = 1 and d = 4

Q6: Between 200 and 400, how many integers are divisible by 3? Find their sum.

Step 1: First multiple of 3 after 200 is 201 Last multiple of 3 before 400 is 399 Step 2: This is an AP: 201, 204, 207, ..., 399 with d = 3 Step 3: Find n: 399 = 201 + (n-1)3 Step 4: 198 = (n-1)3, so n-1 = 66, n = 67 Step 5: Sum Sₙ = n/2 × (first term + last term) = 67/2 × (201 + 399) Step 6: Sₙ = 67/2 × 600 = 67 × 300 = 20,100 Final Answer: 67 numbers; Sum = 20,100

Q7: Find the sum to infinity of the geometric series 1 + 1/2 + 1/4 + 1/8 + ...

Step 1: First term a = 1, common ratio r = 1/2 Step 2: Since |r| < 1, the sum to infinity exists Step 3: Use formula S∞ = a/(1 - r) Step 4: S∞ = 1/(1 - 1/2) = 1/(1/2) = 2 Final Answer: Sum to infinity = 2

Q8: An AP has first term 10, last term 100, and sum 1210. Find the number of terms and common difference.

Step 1: Given a = 10, l = 100, Sₙ = 1210 Step 2: Use Sₙ = n/2 × (a + l) Step 3: 1210 = n/2 × (10 + 100) = n/2 × 110 Step 4: 1210 = 55n, so n = 22 Step 5: Use l = a + (n-1)d Step 6: 100 = 10 + (22-1)d = 10 + 21d Step 7: 90 = 21d, so d = 30/7 ≈ 4.29 Final Answer: Number of terms = 22; Common difference d = 30/7

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