RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 10 questions · 24 marks
A sequence is just a list of numbers following a rule, and this chapter organises the two rules you will meet again and again: add a fixed amount each time (arithmetic) or multiply by a fixed ratio each time (geometric). You will learn to write the nth term of each, sum them efficiently without adding term by term, and use the AM-GM inequality to compare averages. These summing tricks quietly resurface in compound interest, physics motion problems, and later calculus.
The 12th term of the AP 4, 9, 14, 19, ... is:
Answer
The 12th term is 59. Here a = 4 and d = 9 - 4 = 5. Using aₙ = a + (n-1)d with n = 12: a₁₂ = 4 + 11×5 = 4 + 55 = 59.
The sum of the first 20 terms of the AP 3, 7, 11, 15, ... is:
Answer
The sum is 820. Here a = 3, d = 4, n = 20. Sₙ = n/2 [2a + (n-1)d] = 10 [6 + 19×4] = 10 [6 + 76] = 10 × 82 = 820.
The sum to infinity of the GP 8, 4, 2, 1, ... is:
Answer
The sum to infinity is 16. Here a = 8 and r = 4/8 = 1/2, and since |r| < 1, S∞ = a/(1-r) = 8/(1 - 1/2) = 8/(1/2) = 16.
If the AM and GM of two positive numbers are 10 and 8 respectively, the two numbers are:
Answer
The numbers are 4 and 16. AM = 10 gives a+b = 20. GM = 8 gives ab = 64. So a and b are roots of t² - 20t + 64 = 0, giving t = [20 ± √(400-256)]/2 = [20 ± 12]/2, so t = 16 or t = 4. Check: 4+16=20 ✓ and √(4×16)=√64=8 ✓.
Assertion (A): For any two distinct positive real numbers a and b, the AM of a and b is always strictly greater than their GM. Reason (R): (√a - √b)² ≥ 0 for all real a, b, with equality only when a = b.
Answer
Both A and R are true and R is the correct explanation of A. Expanding (√a-√b)² = a + b - 2√(ab) ≥ 0 gives a+b ≥ 2√(ab), i.e., AM ≥ GM, with equality only when a = b. Since a ≠ b (distinct), AM > GM strictly, so Assertion A is true. Reason R states the squared-quantity fact that generates this inequality, and it is exactly the algebraic step used to derive AM > GM, so R correctly explains A.
Which term of the AP 5, 11, 17, 23, ... is 209?
Answer
Here a = 5, d = 6. We need aₙ = 209. aₙ = a + (n-1)d gives 209 = 5 + (n-1)×6. So (n-1)×6 = 204, giving n-1 = 34, so n = 35. 209 is the 35th term.
Find the sum of the first 8 terms of the GP 3, 6, 12, 24, ...
Answer
Here a = 3, r = 6/3 = 2, n = 8. Since r > 1, use Sₙ = a(rⁿ-1)/(r-1). S₈ = 3(2⁸-1)/(2-1) = 3(256-1)/1 = 3×255 = 765. So the sum of the first 8 terms is 765.
The sum of the first three terms of an AP is 33, and their product is 1155. Find the three terms.
Answer
Step 1 — Let the three terms be a-d, a, a+d. Step 2 — Sum: (a-d) + a + (a+d) = 3a = 33, so a = 11. Step 3 — Product: (a-d)(a)(a+d) = a(a²-d²) = 1155. Substituting a = 11: 11(121 - d²) = 1155, so 121 - d² = 105, giving d² = 16, so d = ±4. Step 4 — If d = 4, the terms are 7, 11, 15. If d = -4, the terms are 15, 11, 7 — the same set in reverse order. So the three terms are 7, 11 and 15.
Find the sum of the series 1² + 3² + 5² + ... + 19² (sum of squares of the first 10 odd natural numbers).
Answer
Step 1 — The sum of squares of the first n odd numbers equals Σ(2k-1)² for k = 1 to n, which has the closed form n(2n-1)(2n+1)/3. Here the terms go up to 19² = (2×10-1)², so n = 10. Step 2 — Substitute n = 10: sum = 10(2×10-1)(2×10+1)/3 = 10 × 19 × 21 / 3. Step 3 — Compute: 19 × 21 = 399, and 10 × 399 = 3990, so sum = 3990/3 = 1330. Step 4 — Verify by direct method: Σ(1² to 19², all integers) - Σ(even squares up to 18²) = [19×20×39/6] - 4×[9×10×19/6] = 2470 - 1140 = 1330. Matches. So the required sum is 1330.
Read the following and answer the questions that follow: A company's revenue (in lakhs) forms an AP over its first 6 years of operation: year 1 revenue is 20 lakh, and it grows by a fixed amount of 8 lakh every year. (a) Write the revenue in year 6. (b) Find the total revenue over the first 6 years. (c) In which year does the revenue first exceed 60 lakh? (d) Find the average yearly revenue over the first 6 years.
Answer
(a) a = 20, d = 8. a₆ = 20 + 5×8 = 20 + 40 = 60 lakh. (b) S₆ = 6/2 [2×20 + 5×8] = 3[40+40] = 3×80 = 240 lakh. (c) aₙ > 60 means 20 + (n-1)×8 > 60, so (n-1)×8 > 40, n-1 > 5, n > 6. So revenue first exceeds 60 lakh in year 7. (d) Average = S₆/6 = 240/6 = 40 lakh per year.
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