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

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

TL;DR: Free step-by-step Andhra Pradesh (SSC) Class 10 Mathematics solutions for "Arithmetic Progressions" — important questions with detailed answers, downl…

By Syllab.in · Updated Jun 14, 2026

Q1: Define an arithmetic progression and state the conditions for a sequence to be an AP.

Arithmetic Progression (AP): An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant. General form: a, a + d, a + 2d, a + 3d, ... where a = first term d = common difference (constant) Condition for AP: A sequence with terms t₁, t₂, t₃, ... is an AP if and only if: tₙ₊₁ - tₙ = d (constant) for all positive integers n Or equivalently: t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = ... = d Examples of AP: 1) 2, 5, 8, 11, 14, ... (a = 2, d = 3)…

Q2: Find the 15th term of the AP: 4, 7, 10, 13, ...

Given AP: 4, 7, 10, 13, ... Step 1: Identify the parameters First term: a = 4 Common difference: d = 7 - 4 = 3 (Verify: 10 - 7 = 3, 13 - 10 = 3 ✓) We need to find the 15th term, so n = 15. Step 2: Use the formula for nth term tₙ = a + (n - 1)d t₁₅ = 4 + (15 - 1) × 3 t₁₅ = 4 + 14 × 3 t₁₅ = 4 + 42 t₁₅ = 46 Step 3: Verification First few terms: t₁ = 4 t₂ = 7 t₃ = 10 Following the pattern, each term increases by 3. From t₁ to t₁₅, we add 3 fourteen times: 4 + 14(3) = 46 ✓ Therefore, the 15th te…

Q3: The sum of the first 20 terms of an AP is 650 and the common difference is 3. Find the first term.

Given: Sum of first 20 terms: S₂₀ = 650 Common difference: d = 3 Number of terms: n = 20 Find: First term a = ? Step 1: Use the sum formula for AP Sₙ = n/2 × (2a + (n - 1)d) Substitute the known values: 650 = 20/2 × (2a + (20 - 1) × 3) 650 = 10 × (2a + 19 × 3) 650 = 10 × (2a + 57) Step 2: Solve for a 650 = 20a + 570 20a = 650 - 570 20a = 80 a = 4 Step 3: Verification With a = 4 and d = 3: S₂₀ = 20/2 × (2(4) + (20 - 1) × 3) = 10 × (8 + 57) = 10 × 65 = 650 ✓ Therefore, the first t…

Q4: Is 100 a term of the AP: 5, 8, 11, 14, ...? If yes, find which term it is.

Given AP: 5, 8, 11, 14, ... Step 1: Identify the parameters First term: a = 5 Common difference: d = 8 - 5 = 3 Step 2: Assume 100 is the nth term and find n Using tₙ = a + (n - 1)d: 100 = 5 + (n - 1) × 3 100 = 5 + 3n - 3 100 = 2 + 3n 3n = 98 n = 98/3 = 32.67... Since n is not a positive integer, 100 is not a term of this AP. Alternative verification: For a number to be in an AP with a = 5 and d = 3: term = 5 + 3k (where k ≥ 0 is an integer) If 100 is a term: 100 = 5 + 3k 3k = 95 k = 95/3 = …

Q5: Find the sum of all three-digit numbers divisible by 7.

We need to find the sum of all three-digit numbers divisible by 7. Step 1: Find the first three-digit number divisible by 7 Smallest three-digit number: 100 100 ÷ 7 = 14.28... First three-digit number divisible by 7: 7 × 15 = 105 Step 2: Find the last three-digit number divisible by 7 Largest three-digit number: 999 999 ÷ 7 = 142.71... Last three-digit number divisible by 7: 7 × 142 = 994 Step 3: Identify the AP The AP is: 105, 112, 119, ..., 994 First term: a = 105 Common difference: d = 7 L…

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