Arithmetic Progressions
Punjab Board · Class 10 · Mathematics
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The sum of the first n terms of an AP is given by Sₙ = 3n² + 5n. Find the common difference of the AP.
If the 9th term of an AP is 0, prove which term equals twice the 19th term. Which term is it?
The ratio of the sum of n terms of two APs is (7n + 1) : (4n + 27). Find the ratio of their 11th terms.
How many terms of the AP 63, 60, 57, ... must be taken so that their sum is 693? Choose the correct number of terms.
Sample Questions
In an AP, the sum of first 10 terms is -150 and the sum of its next 10 terms is -550. Find the first term of the AP.
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3
Step 1: S₁₀ = -150 → 5[2a + 9d] = -150 → 2a + 9d = -30 ... (i) Step 2: Sum of next 10 terms = S₂₀ - S₁₀ = -550 → S₂₀ = -700. Step 3: S₂₀ = -700 → 10[2a + 19d] = -700 → 2a + 19d = -70 ... (ii) Step 4: Subtract (i) from (ii): 10d = -40 → d = -4. Substitute in (i): 2a + 9(-4) = -30 → 2a = 6 → a = 3. The first term is 3.
If the pth term of an AP is q and the qth term is p (p ≠ q), what is the (p + q)th term?
The sum of four numbers in AP is 32 and the sum of their squares is 276. Find the largest of the four numbers.
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11
Step 1: Let four numbers in AP be (a-3d), (a-d), (a+d), (a+3d) with common difference 2d. Step 2: Sum = 4a = 32 → a = 8. Step 3: Sum of squares: (a-3d)² + (a-d)² + (a+d)² + (a+3d)² = 276. Step 4: = 4a² + 20d² = 276 → 4(64) + 20d² = 276 → 20d² = 20 → d² = 1 → d = 1. Step 5: Numbers are 5, 7, 9, 11. Largest = 11.
The sum of n terms of an AP is 5n² - 3n. Which term of this AP is 152?
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16th
Step 1: Find general term: aₙ = Sₙ - Sₙ₋₁ = (5n² - 3n) - [5(n-1)² - 3(n-1)]. Step 2: = 5n² - 3n - [5n² - 10n + 5 - 3n + 3] = 5n² - 3n - 5n² + 13n - 8 = 10n - 8. Step 3: Set aₙ = 152: 10n - 8 = 152 → 10n = 160 → n = 16. Step 4: Verify: a₁₆ = 10(16) - 8 = 160 - 8 = 152. ✓ The 16th term is 152.
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