Sequences and Series — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Sequences and Series" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Sequences and Series" — 8 important questions with detailed answers for CBSE board…
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NCERT Solutions for Class 11 Mathematics — Sequences and Series. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Sequences and Series — All 8 Questions Solved
Q1. Find the 15th term of the arithmetic progression (AP) 2, 5, 8, 11, ...
Given AP: 2, 5, 8, 11, ...
First term a = 2
Common difference d = 5 - 2 = 3
Step 1: Use the formula for nth term of an AP:
aₙ = a + (n - 1)d
Step 2: Substitute n = 15, a = 2, d = 3:
a₁₅ = 2 + (15 - 1) × 3
a₁₅ = 2 + 14 × 3
a₁₅ = 2 + 42
a₁₅ = 44
Final Answer: The 15th term is 44
Q2. The sum of first n terms of an AP is Sₙ = n²/2 + 3n/2. Find the AP and its 20th term.
Given: Sₙ = n²/2 + 3n/2
Step 1: Find the first term a₁ using S₁:
S₁ = (1)²/2 + 3(1)/2 = 1/2 + 3/2 = 2
So a₁ = 2
Step 2: Find a₂ using S₂:
S₂ = (2)²/2 + 3(2)/2 = 2 + 3 = 5
So a₁ + a₂ = 5
Therefore a₂ = 5 - 2 = 3
Step 3: Find common difference d:
d = a₂ - a₁ = 3 - 2 = 1
Step 4: Verify with a₃:
S₃ = (3)²/2 + 3(3)/2 = 9/2 + 9/2 = 9
a₃ = S₃ - S₂ = 9 - 5 = 4
Common difference d = 4 - 3 = 1 ✓
Step 5: The AP is: 2, 3, 4, 5, ...
For the 20th term:
a₂₀ = 2 + (20 - 1) × 1 = 2 + 19 = 21
Final Answer: AP is 2, 3, 4, 5, ... and a₂₀ = 21
Q3. Find the sum of the first 20 terms of the geometric progression (GP) 3, 6, 12, 24, ...
Given GP: 3, 6, 12, 24, ...
First term a = 3
Common ratio r = 6/3 = 2
Step 1: Use the formula for sum of n terms of a GP (r ≠ 1):
Sₙ = a(rⁿ - 1)/(r - 1)
Step 2: Substitute n = 20, a = 3, r = 2:
S₂₀ = 3(2²⁰ - 1)/(2 - 1)
S₂₀ = 3(2²⁰ - 1)/1
S₂₀ = 3(1048576 - 1)
S₂₀ = 3 × 1048575
S₂₀ = 3145725
Final Answer: Sum of first 20 terms = 3145725
Q4. If the sum of an AP with 10 terms is 250 and the first term is 4, find the common difference and the last term.
Given: n = 10, Sₙ = 250, a = 4
Step 1: Use the sum formula:
Sₙ = n/2[2a + (n - 1)d]
250 = 10/2[2(4) + (10 - 1)d]
250 = 5[8 + 9d]
50 = 8 + 9d
9d = 42
d = 14/3
Step 2: Find the last term (a₁₀):
aₙ = a + (n - 1)d
a₁₀ = 4 + (10 - 1) × 14/3
a₁₀ = 4 + 9 × 14/3
a₁₀ = 4 + 42
a₁₀ = 46
Final Answer: Common difference d = 14/3, Last term = 46
Q5. The ratio of the 5th and 8th terms of a GP is 8:27. If the 6th term is 24, find the first term and common ratio.
Let a be the first term and r be the common ratio.
Step 1: Write the given terms:
a₅ = ar⁴
a₈ = ar⁷
Ratio: a₅/a₈ = ar⁴/ar⁷ = 1/r³ = 8/27
Step 2: Solve for r:
1/r³ = 8/27
r³ = 27/8
r = 3/2
Step 3: Use the condition a₆ = 24:
a₆ = ar⁵ = 24
a(3/2)⁵ = 24
a × 243/32 = 24
a = 24 × 32/243
a = 768/243
a = 256/81
Step 4: Verify:
a₅ = (256/81) × (3/2)⁴ = (256/81) × (81/16) = 16
a₈ = (256/81) × (3/2)⁷ = (256/81) × (2187/128) = 54
Ratio = 16:54 = 8:27 ✓
Final Answer: First term a = 256/81, Common ratio r = 3/2
Q6. Find the sum of the infinite GP: 1 + 1/2 + 1/4 + 1/8 + ...
Given infinite GP: 1 + 1/2 + 1/4 + 1/8 + ...
First term a = 1
Common ratio r = 1/2
Step 1: Check if |r| < 1:
|1/2| = 0.5 < 1, so the series converges
Step 2: Use the formula for sum of infinite GP (|r| < 1):
S∞ = a/(1 - r)
Step 3: Substitute values:
S∞ = 1/(1 - 1/2)
S∞ = 1/(1/2)
S∞ = 2
Final Answer: Sum of the infinite GP = 2
Q7. How many terms of the AP 1, 4, 7, 10, ... must be added to get a sum of 715?
Given AP: 1, 4, 7, 10, ...
a = 1, d = 3
Sₙ = 715
Step 1: Use the sum formula:
Sₙ = n/2[2a + (n - 1)d]
715 = n/2[2(1) + (n - 1)(3)]
715 = n/2[2 + 3n - 3]
715 = n/2[3n - 1]
1430 = n(3n - 1)
1430 = 3n² - n
3n² - n - 1430 = 0
Step 2: Use the quadratic formula:
n = [1 ± √(1 + 4(3)(1430))]/(2 × 3)
n = [1 ± √(1 + 17160)]/6
n = [1 ± √17161]/6
n = [1 ± 131]/6
Step 3: Find the positive value:
n = (1 + 131)/6 = 132/6 = 22
or n = (1 - 131)/6 = -130/6 (rejected, negative)
Step 4: Verify:
S₂₂ = 22/2[2(1) + (22 - 1)(3)] = 11[2 + 63] = 11 × 65 = 715 ✓
Final Answer: 22 terms must be added
Q8. Three numbers are in GP. Their sum is 26 and their product is 216. Find the numbers.
Let the three numbers in GP be a/r, a, ar where r is the common ratio.
Step 1: Use the product condition:
(a/r) × a × ar = 216
a³ = 216
a = 6
Step 2: Use the sum condition:
a/r + a + ar = 26
6/r + 6 + 6r = 26
6/r + 6r = 20
6 + 6r² = 20r
6r² - 20r + 6 = 0
3r² - 10r + 3 = 0
Step 3: Solve using the quadratic formula:
r = [10 ± √(100 - 36)]/6
r = [10 ± √64]/6
r = [10 ± 8]/6
r = 3 or r = 1/3
Step 4: Find the three numbers:
When r = 3: a/r = 6/3 = 2, a = 6, ar = 18
When r = 1/3: a/r = 18, a = 6, ar = 2
Both give the same set: 2, 6, 18
Step 5: Verify:
Sum: 2 + 6 + 18 = 26 ✓
Product: 2 × 6 × 18 = 216 ✓
Common ratio: 6/2 = 3, 18/6 = 3 ✓
Final Answer: The three numbers are 2, 6, 18
Sequences and Series — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The 5th term of an Arithmetic Progression is 11 and its common difference is 2. What is the first term?
- A. 3
- B. 5
- C. 7
- D. 9
Answer: A. 3 — Using the formula for the nth term of an AP, a_n = a + (n-1)d. Substitute a_5=11, n=5, d=2: 11 = a + (5-1)2, which simplifies to 11 = a + 8, so a = 3.
MCQ 2. Find the 7th term of the Geometric Progression 2, 6, 18, ...
- A. 486
- B. 972
- C. 1458
- D. 2916
Answer: C. 1458 — For a GP, the nth term is a_n = ar^(n-1). Here, a=2 and r=6/2=3. So, a_7 = 2 * 3^(7-1) = 2 * 3^6 = 2 * 729 = 1458.
MCQ 3. What is the sum of the first 10 terms of an AP whose first term is 5 and the common difference is 3?
- A. 175
- B. 180
- C. 185
- D. 190
Answer: C. 185 — The sum of n terms of an AP is S_n = n/2[2a + (n-1)d]. For n=10, a=5, d=3: S_10 = 10/2[2*5 + (10-1)*3] = 5[10 + 9*3] = 5[10 + 27] = 5*37 = 185.
MCQ 4. The sum of an infinite Geometric Progression is 12 and its common ratio is 1/3. What is the first term?
- A. 4
- B. 6
- C. 8
- D. 10
Answer: C. 8 — The sum to infinity of a GP is S_∞ = a/(1-r). Given S_∞=12 and r=1/3, we have 12 = a/(1-1/3) = a/(2/3), so a = 12 * (2/3) = 8.
MCQ 5. Calculate the sum of the first 50 natural numbers.
- A. 1225
- B. 1250
- C. 1275
- D. 1300
Answer: C. 1275 — The sum of the first n natural numbers is given by the formula Σn = n(n+1)/2. For n=50, the sum is 50(50+1)/2 = 50*51/2 = 25*51 = 1275.
MCQ 6. If (k-1), (k+1), and (2k+1) are in Arithmetic Progression, find the value of k.
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — If a, b, c are in AP, then 2b = a+c. So, 2(k+1) = (k-1) + (2k+1). This simplifies to 2k+2 = 3k, which gives k = 2.
MCQ 7. If 2, x, 8 are in Geometric Progression, find the positive value of x.
- A. 2
- B. 4
- C. 6
- D. 8
Answer: B. 4 — If a, b, c are in GP, then b^2 = ac. So, x^2 = 2*8 = 16. Taking the positive square root, x = 4.
MCQ 8. If AM and GM of two positive numbers a and b are 10 and 8 respectively, find the numbers.
- A. (4, 16)
- B. (2, 32)
- C. (8, 12)
- D. (6, 14)
Answer: A. (4, 16) — Given AM = (a+b)/2 = 10, so a+b=20. Given GM = sqrt(ab) = 8, so ab=64. The numbers are roots of the quadratic equation x^2 - (a+b)x + ab = 0, which is x^2 - 20x + 64 = 0. Factoring gives (x-4)(x-16)=0, so the numbers are 4 and 16.
MCQ 9. If the terms 1/2, 1/x, 1/8 are in Arithmetic Progression, then 2, x, 8 are in which progression?
- A. AP
- B. GP
- C. HP
- D. None
Answer: C. HP — By the definition of a Harmonic Progression (HP), if the reciprocals of terms are in AP, then the terms themselves are in HP. Since 1/2, 1/x, 1/8 are in AP, their reciprocals 2, x, 8 are in HP.
MCQ 10. What is the Harmonic Mean of 3 and 6?
- A. 4
- B. 4.5
- C. 5
- D. 5.5
Answer: A. 4 — The Harmonic Mean (HM) of two numbers a and b is given by the formula HM = 2ab/(a+b). For a=3 and b=6, HM = (2 * 3 * 6) / (3 + 6) = 36 / 9 = 4.
MCQ 11. Find the sum of the series 1.2 + 2.3 + 3.4 + ... up to n terms.
- A. n(n+1)(n+2)/3
- B. n(n+1)(n+2)/2
- C. n(n+1)(n+2)
- D. n(n+1)(2n+1)/6
Answer: A. n(n+1)(n+2)/3 — The nth term is T_n = n(n+1) = n^2 + n. The sum S_n = Σ(n^2 + n) = Σn^2 + Σn. Using standard formulas, S_n = n(n+1)(2n+1)/6 + n(n+1)/2, which simplifies to n(n+1)(n+2)/3.
MCQ 12. What is the sum of the cubes of the first 10 natural numbers?
- A. 2500
- B. 2750
- C. 3025
- D. 3375
Answer: C. 3025 — The sum of the cubes of the first n natural numbers is given by the formula Σn^3 = [n(n+1)/2]^2. For n=10, the sum is [10(10+1)/2]^2 = [10*11/2]^2 = [55]^2 = 3025.
MCQ 13. The first term of a GP is 3 and the sum of its first three terms is 21. Find the common ratio.
- A. 1
- B. 2
- C. -2
- D. 3
Answer: B. 2 — For a GP, S_3 = a(r^2+r+1). Given a=3 and S_3=21, we have 3(r^2+r+1)=21, so r^2+r+1=7. This simplifies to r^2+r-6=0, which factors as (r+3)(r-2)=0. Thus, r=2 or r=-3. From the options, r=2 is correct.
MCQ 14. The sum of n terms of an AP is given by S_n = 3n^2 + 2n. Find the 5th term of the AP.
- A. 21
- B. 25
- C. 29
- D. 31
Answer: C. 29 — The nth term of an AP can be found using a_n = S_n - S_(n-1). Calculate S_5 = 3(5)^2 + 2(5) = 75 + 10 = 85. Calculate S_4 = 3(4)^2 + 2(4) = 48 + 8 = 56. Thus, a_5 = S_5 - S_4 = 85 - 56 = 29.
MCQ 15. If 5 arithmetic means are inserted between 8 and 26, find the third arithmetic mean.
- A. 14
- B. 17
- C. 20
- D. 23
Answer: B. 17 — With 5 means inserted, there are 7 terms in total (8, A_1, A_2, A_3, A_4, A_5, 26). This forms an AP where a=8, a_7=26. Using a_n = a + (n-1)d, 26 = 8 + (7-1)d, so 18 = 6d, implying d=3. The third AM (A_3) is the 4th term of the AP, so A_3 = a + 3d = 8 + 3(3) = 17.
MCQ 16. Two geometric means are inserted between 1/3 and 9. Find the second geometric mean.
- A. 1
- B. 3
- C. 9
- D. 27
Answer: B. 3 — With 2 means inserted, there are 4 terms in total (1/3, G_1, G_2, 9). This forms a GP where a=1/3, a_4=9. Using a_n = ar^(n-1), 9 = (1/3)r^(4-1), so 9 = (1/3)r^3, which gives r^3=27, hence r=3. The second GM (G_2) is the 3rd term of the GP, so G_2 = ar^2 = (1/3)(3^2) = (1/3)*9 = 3.
MCQ 17. If AM, GM, HM of two distinct positive numbers are A, G, H respectively, which of the following is true?
- A. A < G < H
- B. G < A < H
- C. H < G < A
- D. A = G = H
Answer: C. H < G < A — For any two distinct positive real numbers, the relationship between their Arithmetic Mean (A), Geometric Mean (G), and Harmonic Mean (H) is given by H < G < A.
MCQ 18. Find the last term of an AP with 15 terms, whose first term is 4 and common difference is 5.
- A. 74
- B. 79
- C. 84
- D. 89
Answer: A. 74 — Using the formula for the nth term of an AP, a_n = a + (n-1)d. For n=15, a=4, d=5: a_15 = 4 + (15-1)5 = 4 + 14*5 = 4 + 70 = 74.
MCQ 19. What is the common ratio of the GP: 100, 20, 4, 0.8, ...?
- A. 5
- B. 1/5
- C. -5
- D. -1/5
Answer: B. 1/5 — The common ratio (r) of a GP is found by dividing any term by its preceding term. Here, r = 20/100 = 1/5.
MCQ 20. How many terms of the AP 9, 17, 25, ... must be taken to give a sum of 636?
- A. 10
- B. 11
- C. 12
- D. 13
Answer: C. 12 — The sum of n terms of an AP is S_n = n/2[2a + (n-1)d]. Here, a=9 and d=8. Setting S_n=636 gives 636 = n/2[2*9 + (n-1)8]. This simplifies to 4n^2 + 5n - 636 = 0. Solving the quadratic equation yields n=12 (as n must be positive).
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.