Arithmetic Progressions Exemplar — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Arithmetic Progressions Exemplar" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Arithmetic Progressions Exemplar" — 6 important questions with detailed answers fo…
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Key Questions Covered:
- If first term a = 5 and common difference d = 3, find 10th term.
- Find sum of first 20 terms of AP: 2, 5, 8, 11, ...
- Which term of AP 3, 8, 13, 18, ... is 78?
- Insert 3 arithmetic means between 5 and 21.
- Find sum of all two-digit natural numbers divisible by 4.
- If sum of n terms of AP is 3n^2 + 5n, find the AP.
Solutions Summary:
| Question | Status |
|---|---|
| If first term a = 5 and common difference d = 3, find 10t… | ✓ Solved |
| Find sum of first 20 terms of AP: 2, 5, 8, 11, ... | ✓ Solved |
| Which term of AP 3, 8, 13, 18, ... is 78? | ✓ Solved |
| Insert 3 arithmetic means between 5 and 21. | ✓ Solved |
| Find sum of all two-digit natural numbers divisible by 4. | ✓ Solved |
| If sum of n terms of AP is 3n^2 + 5n, find the AP. | ✓ Solved |
Showing 6 of 6 questions
Q1: If first term a = 5 and common difference d = 3, find 10th term.
an = a + (n-1)d. a10 = 5 + (10-1)(3) = 5 + 27 = 32.
Q2: Find sum of first 20 terms of AP: 2, 5, 8, 11, ...
a = 2, d = 3, n = 20. Sn = n/2 [2a + (n-1)d] = 20/2 [2(2) + 19(3)] = 10[4 + 57] = 10(61) = 610.
Q3: Which term of AP 3, 8, 13, 18, ... is 78?
a = 3, d = 5. an = 78. 3 + (n-1)(5) = 78. (n-1)(5) = 75, so n - 1 = 15, thus n = 16. The 16th term is 78.
Q4: Insert 3 arithmetic means between 5 and 21.
If 3 means are inserted, sequence is 5, A1, A2, A3, 21 (5 terms total). Common difference d = (21 - 5)/(5 - 1) = 16/4 = 4. So means are 9, 13, 17.
Q5: Find sum of all two-digit natural numbers divisible by 4.
Sequence: 12, 16, 20, ..., 96. a = 12, d = 4, l = 96. n = (96 - 12)/4 + 1 = 21 + 1 = 22. Sn = n/2 (a + l) = 22/2 (12 + 96) = 11(108) = 1188.
Q6: If sum of n terms of AP is 3n^2 + 5n, find the AP.
Sn = 3n^2 + 5n. First term a = S1 = 3 + 5 = 8. For n ≥ 2, an = Sn - Sn-1 = 3n^2 + 5n - 3(n-1)^2 - 5(n-1) = 3(2n - 1) + 5 = 6n + 2. So a2 = 14, d = 6. AP: 8, 14, 20, 26, ...
Showing 6 of 6 questions. Visit the full page for complete solutions.
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