NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series Exercise 8.2

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2 – Sequences and Series help students understand the concepts of arithmetic progression (AP) and geometric progression (GP). This exercise focuses on solving problems related to the sum of terms in these sequences, particularly focusing on the sum of the first n terms of an arithmetic or geometric series.

Prepared according to the latest CBSE Class 11 Maths syllabus, Exercise 8.2 helps students master the formulas for sum of terms in AP and sum of terms in GP and apply them to solve a variety of problems. It includes both theoretical and application-based problems to test students' understanding of sequences and series.

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series Exercise 8.2

Sequences and Series
Easy
Q.
Write the first five terms of the sequence whose nth terms is:
an = n(n+2)
Sequences and Series
Medium
Q.
Given a G.P. with a = 729 and 7th term 64, determine S7.
Sequences and Series
Medium
Q.
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
Sequences and Series
Medium
Q.
The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.
Sequences and Series
Difficult
Q.
Which term of the following sequences:(a). 2, 22, 4, ... is​ 128?(b). 3, 33, 3, ... is​ 729?(c).  13, 19, 127,...  is  119683?
Sequences and Series
Medium
Q.
For what values of xx​, the numbers27,x,72\frac{2}{7},x,\frac{7}{2}​ are in G.P.?

Sequences and Series
Medium
Q.
Find the sum to indicated number of terms in each of the geometricprogressions given below.i. 0.15, 0.015, 0.0015,… 20 terms.ii. 7,  21,  37,  ...  n terms.iii. 1,− a, a2,− a3, ... n terms (if a≠−1).iv. x3, x5, x7, ... n terms (if x≠± 1).
Sequences and Series
Medium
Q.
Evaluate:  ∑k=111(2+3k).
Sequences and Series
Difficult
Q.
The sum of first three terms of a G.P. is 3910 and their product is 1. Find the common ratio and the terms.
Sequences and Series
Easy
Q.
How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?
Sequences and Series
Medium
Q.
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.
Determine the first term, the common ratio and the sum to n terms of the G.P.
Sequences and Series
Medium
Q.
Find the sum to nn​ terms of the sequence, 8, 88, 888, 8888… .
Sequences and Series
Easy
Q.
Find the 20th and nth terms of the G.P. 52,54,58,...
Sequences and Series
Difficult
Q.
  Find the sum of the products of the corresponding terms of the sequences 2,4,8,16,322,4,8,16,32​ and 128,32,8,2,12128,32,8,2,\frac{1}{2}.​
Sequences and Series
Difficult
Q.
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Sequences and Series
Medium
Q.
Find the value of n so that  an+1+bn+1an+bn may be the geometric mean between a and b.
Sequences and Series
Medium
Q.
 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3+22):(3−22). \begin{array}{l}\mathbf{{The sum of two numbers is 6 times their geometric mean, show that }}\\ \mathbf{{numbers are in the ratio }}\left({3+2}\sqrt{{2}}\right)\mathbf{{:}}({3}-{2}\sqrt{{2}})\mathbf{{.}}\end{array} 
Sequences and Series
Medium
Q.
  If AA​ and GG​ be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A±(A+G)(A−G)A \pm \sqrt{(A + G)(A - G)}​​
Sequences and Series
Difficult
Q.
Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3(S2 – S1).
Sequences and Series
Difficult
Q.
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Sequences and Series
Medium
Q.
If  a+bxa−bx=b+cxb−cx=c+dxc−dx(x≠0), then show that a, b, c and d are in G.P.
Sequences and Series
Medium
Q.
 If a(1b+1c),b(1c+1a),c(1a+1b) are in A.P., prove that a, b and c are in A.P. \mathbf{{If a}}\left(\frac{{1}}{{b}}{+}\frac{{1}}{{c}}\right)\mathbf{{,\hspace{0.17em}b}}\left(\frac{{1}}{{c}}{+}\frac{{1}}{{a}}\right)\mathbf{{,\hspace{0.17em}c}}\left(\frac{{1}}{{a}}{+}\frac{{1}}{{b}}\right)\mathbf{{ are in A.P., prove that a, b and c are in A.P.}} 
Sequences and Series
Difficult
Q.
Find the sum of the first n terms of the series: 3+ 7 + 13 + 21 + 31 +…
Sequences and Series
Easy
Q.
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Sequences and Series
Difficult
Q.
The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.
Sequences and Series
Medium
Q.
Write the first five terms of the sequence whose nth terms is:an=nn+1.
Sequences and Series
Difficult
Q.
How many terms of the A.P.  −6,−112,−5,...  are needed to givethe sum −25?
Sequences and Series
Easy
Q.
Write the first five terms of the sequence whose nth terms is:
an = 2n
Sequences and Series
Medium
Q.
Write the first five terms of the sequence whose nth terms is:an=2n−36.
Sequences and Series
Medium
Q.
Write the first five terms of the sequence whose nth terms is an = (– 1)n–1 5n+1.
Sequences and Series
Medium
Q.
Write the first five terms of the sequence whose nth terms isan=nn2+54
Sequences and Series
Medium
Q.
Find the indicated terms in each of the sequences given below whose nth terms are:1. an= 4n–3; a17, a242. an=n22n;  a73. an= (–1)n–1n3; a94. an=n(n−2)n+3; a20
Sequences and Series
Medium
Q.
Write the first five terms of each of the sequences given below andobtain the corresponding series:1. a1=3, an=3an-1+2, for all n>12. a1=–1, an=an-1n, n≥23. a1=a2=2, an=an–1–1, n > 2
Sequences and Series
Difficult
Q.
The Fibonacci sequence is defined by 1=a1=a2 and an=an−1+an−2,  n>2.Find an+1an, for n=1,2,3,4,5.
Sequences and Series
Medium
Q.
Find the sum of odd integers from 1 to 2001.
Sequences and Series
Medium
Q.
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Sequences and Series
Medium
Q.
In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.
Sequences and Series
Difficult
Q.
In​​ an A.P., if pth term is 1p​​ and qth term is 1q, prove that the sum of first pq terms is 12(pq+1), where p≠q.
Sequences and Series
Difficult
Q.
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.
Sequences and Series
Difficult
Q.
If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.
Sequences and Series
Medium
Q.
Find the sum to n terms of the A.P., whose kth term is 5k + 1.
Sequences and Series
Medium
Q.
If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.
Sequences and Series
Medium
Q.
The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.
Sequences and Series
Difficult
Q.
If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Sequences and Series
Difficult
Q.
Sum of the first p, q and r terms of an A.P. are a, b and c,  respectively.Prove that apq−r+bqr−p+crp−q = 0.
Sequences and Series
Difficult
Q.
The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1).
Sequences and Series
Medium
Q.
If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.
Sequences and Series
Difficult
Q.
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Sequences and Series
Medium
Q.
 If an+bnan−1+bn−1is the A.M. between a and b, then find the value of n. \mathbf{{If }}\frac{{\mathbf{{a}}}^{\mathbf{{n}}}{\mathbf{{+b}}}^{\mathbf{{n}}}}{{\mathbf{{a}}}^{\mathbf{{n}}\mathbf{-}\mathbf{{1}}}{\mathbf{{+b}}}^{\mathbf{{n}}\mathbf{-}\mathbf{{1}}}}\mathbf{{\hspace{0.17em}\hspace{0.17em}is the A.M. between a and b, then find the value of n.}} 
Sequences and Series
Medium
Q.
 If S1, S2, S3are the sum of first n natural numbers,their squares and their cubes,respectively, show that9S22=S3(1+8S1). \begin{array}{l}{{If S}}_{{1}}{{, S}}_{{2}}{{, S}}_{{3}}{are the sum of first n natural numbers,\hspace{0.33em}their squares and their cubes,respectively, }\\ {show that\hspace{0.33em}}9{\mathrm{S}}_{2}^{2}={\mathrm{S}}_{3}\left(1+8{\mathrm{S}}_{1}\right).\end{array} 

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series Exercise 8.2

The solutions are explained in a clear, step-by-step manner to help students grasp the methods and techniques to solve the questions effectively.

Here's the complete Q&A text formatted for WordPress copy-paste:

NCERT Solutions for Class 11 Maths – Chapter 8: Sequence and Series (Exercise 8.2)

Q1. Find the 20th and nth term of the G.P. 5/2, 5/4, 5/8, ...

Ans: First term a = 5/2, common ratio r = 1/2.

nth term: aₙ = 5/2ⁿ

20th term: a₂₀ = 5/2²⁰

Q2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Ans: Let first term = a, r = 2.

a₈ = ar⁷ = 192 → a = 3/2

a₁₂ = ar¹¹ = (3/2)(2¹¹) = 3 × 2¹⁰ = 3072

Q3. The 5th, 8th and 11th terms of a G.P. are p, q and s respectively. Show that q² = ps.

Ans: a₅ = ar⁴ = p, a₈ = ar⁷ = q, a₁₁ = ar¹⁰ = s

Dividing a₈ by a₅: r³ = q/p

Dividing a₁₁ by a₈: r³ = s/q

Therefore q/p = s/q → q² = ps (Proved)

Q4. The 4th term of a G.P. is square of its second term, and the first term is −3. Determine its 7th term.

Ans: a = −3, a₄ = ar³ = (−3)r³, a₂ = ar = (−3)r

Condition: (−3)r³ = [(−3)r]² → −3r³ = 9r² → r = −3

a₇ = ar⁶ = (−3)(−3)⁶ = −(3)⁷ = −2187

Q5. Which term of the following sequences:

(a) 2, 2√2, 4, ... is 128?

Ans: a = 2, r = √2

ar^(n−1) = 128 → (2)(√2)^(n−1) = 128 → 2^((n−1)/2 + 1) = 2⁷

Solving: n = 13

(b) √3, 3, 3√3, ... is 729?

Ans: a = √3, r = √3

(√3)(√3)^(n−1) = 729 → 3^(n/2) = 3⁶ → n = 12

(c) 1/3, 1/9, 1/27, ... is 1/19683?

Ans: a = 1/3, r = 1/3

(1/3)ⁿ = (1/3)⁹ → n = 9

Q6. For what values of x, the numbers −2/7, x, −7/2 are in G.P.?

Ans: Common ratio: x/(−2/7) = (−7/2)/x

→ x² = (−2/7)(−7/2) = 1 → x = ±1

Q7. Find the sum up to 20 terms in the G.P. 0.15, 0.015, 0.0015, ...

Ans: a = 0.15, r = 0.1

S₂₀ = 0.15 × [1 − (0.1)²⁰] / (1 − 0.1) = (1/6)[1 − (0.1)²⁰]

Q8. Find the sum of n terms in the G.P. √7, √21, 3√7, ...

Ans: a = √7, r = √3

Sₙ = √7 × [1 − (√3)ⁿ] / (1 − √3)

= −√7(1 + √3)/2 × [(3)^(n/2) − 1]

Q9. Find the sum of n terms in the G.P. 1, −a, a², −a³, ... (if a ≠ −1)

Ans: First term = 1, r = −a

Sₙ = [1 − (−a)ⁿ] / (1 + a)

Q10. Find the sum of n terms in the G.P. x³, x⁵, x⁷, ... (if x ≠ ±1)

Ans: a = x³, r = x²

Sₙ = x³[1 − x²ⁿ] / (1 − x²)

Q11. Evaluate Σ(2 + 3ᵏ) for k = 1 to 11

Ans: Σ(2 + 3ᵏ) = 22 + Σ3ᵏ

Σ3ᵏ (k=1 to 11) = (3/2)(3¹¹ − 1)

= 22 + (3/2)(3¹¹ − 1)

Q12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Ans: Let terms be a/r, a, ar.

Product = a³ = 1 → a = 1

Sum: 1/r + 1 + r = 39/10 → 10r² − 29r + 10 = 0 → r = 2/5 or 5/2

Terms: 5/2, 1, 2/5

Q13. How many terms of G.P. 3, 3², 3³, ... are needed to give the sum 120?

Ans: a = 3, r = 3, Sₙ = 120

3(3ⁿ − 1)/2 = 120 → 3ⁿ = 81 = 3⁴ → n = 4

Q14. The sum of the first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio, and the sum to n terms.

Ans: a(1 + r + r²) = 16 and ar³(1 + r + r²) = 128

Dividing: r³ = 8 → r = 2

a(1 + 2 + 4) = 16 → a = 16/7

Sₙ = (16/7)(2ⁿ − 1)

Q15. Given a G.P. with a = 729 and 7th term 64, determine S₇.

Ans: ar⁶ = 64 → 729r⁶ = 64 → r = 2/3

S₇ = 729[1 − (2/3)⁷] / (1 − 2/3) = 3⁷ − 2⁷ = 2187 − 128 = 2059

Q16. Find a G.P. for which sum of the first two terms is −4 and the fifth term is 4 times the third term.

Ans: a₅ = 4a₃ → r² = 4 → r = ±2

For r = 2: a = −4/3 → G.P.: −4/3, −8/3, −16/3, ...

For r = −2: a = 4 → G.P.: 4, −8, 16, −32, ...

Q17. If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively. Prove that x, y, z are in G.P.

Ans: a₄ = ar³ = x, a₁₀ = ar⁹ = y, a₁₆ = ar¹⁵ = z

y/x = r⁶ and z/y = r⁶ → y/x = z/y → x, y, z are in G.P. (Proved)

Q18. Find the sum to n terms of the sequence 8, 88, 888, 8888, ...

Ans: Sₙ = (8/9)(9 + 99 + 999 + ...)

= (8/9)[10(10ⁿ − 1)/9 − n]

= 80(10ⁿ − 1)/81 − 8n/9

Q19. Find the sum of the products of corresponding terms of sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.

Ans: Products: 256, 128, 64, 32, 16 → G.P. with a = 256, r = 1/2

Wait — products are: 2×128=256, 4×32=128, 8×8=64, 16×2=32, 32×(1/2)=16

S₅ = 64[1 − (1/2)⁵] / (1 − 1/2) = 496

Q20. Show that the products of corresponding terms of sequences a, ar, ar², ...arⁿ⁻¹ and A, AR, AR², ...ARⁿ⁻¹ form a G.P. and find the common ratio.

Ans: Product sequence: aA, arAR, ar²AR², ...

Ratio of consecutive terms = rR (constant) → forms a G.P. with common ratio = rR

Q21. Find four numbers forming a G.P. in which the third term is greater than the first by 9, and the second term is greater than the 4th by 18.

Ans: a(r² − 1) = 9 ... (1) and ar(1 − r²) = 18 ... (2)

Dividing: −r = 2 → r = −2, a = 3

G.P.: 3, −6, 12, −24

Q22. If pth, qth and rth terms of a G.P. are a, b and c respectively. Prove that a^(q−r) · b^(r−p) · c^(p−q) = 1.

Ans: Using AR^(p−1) = a, AR^(q−1) = b, AR^(r−1) = c,

a^(q−r) · b^(r−p) · c^(p−q) = A⁰ × R⁰ = 1 (Proved)

Q23. If the first and nth terms of a G.P. are a and b, and P is the product of n terms, prove that P² = (ab)ⁿ.

Ans: P = aⁿ · r^(n(n−1)/2)

P² = a²ⁿ · r^(n(n−1)) = [a · arⁿ⁻¹]ⁿ = (ab)ⁿ (Proved)

Q24. Show that the ratio of sum of first n terms of a G.P. to the sum of terms from (n+1)th to (2n)th term is 1/rⁿ.

Ans: Sum of first n terms = a(1 − rⁿ)/(1 − r)

Sum from (n+1)th to (2n)th = arⁿ(1 − rⁿ)/(1 − r)

Ratio = 1/rⁿ (Proved)

Q25. If a, b, c, d are in G.P., show that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².

Ans: Using bc = ad, b² = ac, c² = bd and expanding both sides, LHS = RHS. (Proved)

Q26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Ans: 3, G₁, G₂, 81 → r³ = 27 → r = 3

G₁ = 9, G₂ = 27 → Two numbers are 9 and 27

Q27. Find the value of n so that (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) may be the G.M. between a and b.

Ans: Setting equal to √(ab) and solving:

(a/b)^(2n+1) = 1 → 2n + 1 = 0 → n = −1/2

Q28. The sum of two numbers is 6 times their G.M. Show that numbers are in ratio (3 + 2√2) : (3 − 2√2).

Ans: a + b = 6√(ab) → (a+b)² = 36ab

(a−b)² = 32ab → a−b = 4√2·√(ab)

Solving: a = (3 + 2√2)√(ab), b = (3 − 2√2)√(ab)

a/b = (3 + 2√2)/(3 − 2√2) (Proved)

Q29. If A and G be A.M. and G.M. between two positive numbers, prove that the numbers are A ± √[(A+G)(A−G)].

Ans: a + b = 2A, ab = G²

(a−b)² = 4A² − 4G² = 4(A+G)(A−G)

a−b = 2√[(A+G)(A−G)]

a, b = A ± √[(A+G)(A−G)] (Proved)

Q30. The number of bacteria in a culture doubles every hour. If there were 30 bacteria originally, how many will be present at end of 2nd, 4th and nth hour?

Ans: a = 30, r = 2

End of 2nd hour: a₃ = 30 × 2² = 120

End of 4th hour: a₅ = 30 × 2⁴ = 480

End of nth hour: aₙ₊₁ = 30 × 2ⁿ

Q31. What will Rs. 500 amount to in 10 years at 10% annual compound interest?

Ans: Amount = 500 × (1.1)¹ after year 1, 500 × (1.1)² after year 2, etc.

Amount after 10 years = Rs. 500 × (1.1)¹⁰

Q32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively, obtain the quadratic equation.

Ans: a + b = 16, ab = 25

Quadratic equation: x² − (sum)x + (product) = 0

x² − 16x + 25 = 0

FAQs – Class 11 Maths Chapter 8 Exercise 8.2 Sequences and Series

Q1. What is the focus of Exercise 8.2?
Exercise 8.2 focuses on solving problems related to the sum of the first n terms in arithmetic progressions (AP) and geometric progressions (GP).

Q2. What is the formula for the sum of the first n terms of an arithmetic progression (AP)?
The formula for the sum of the first n terms of an arithmetic progression is:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n - 1) d]

Where:

  • SnS_n is the sum of the first n terms,
  • aa is the first term,
  • dd is the common difference.

Q3. What is the formula for the sum of the first n terms of a geometric progression (GP)?
The formula for the sum of the first n terms of a geometric progression is:

Sn=a(1−rn)1−r,(for r≠1)S_n = \frac{a(1 - r^n)}{1 - r}, \quad \text{(for } r \neq 1 \text{)}

Where:

  • SnS_n is the sum of the first n terms,
  • aa is the first term,
  • rr is the common ratio.

Q4. Why is Exercise 8.2 important for exams?
This exercise is essential for understanding and solving problems related to sequences and series, which are key components of Class 11 and Class 12 exams. Understanding the sum of terms in AP and GP is crucial for solving real-world problems in areas like finance, physics, and engineering.

Q5. How can students prepare effectively for Exercise 8.2?
Students should:

  • Memorize the formulas for sum of terms in AP and GP.
  • Practice solving problems step-by-step to get familiar with the application of these formulas.
  • Work on problems involving word problems to understand the practical applications of sequences and series.