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

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 8 – Sequences and Series help students practice a variety of problems based on arithmetic progressions (AP) and geometric progressions (GP). This exercise is designed to provide students with a deeper understanding of the sum of terms in these progressions, and to apply the concepts of sequences and series in different problem-solving scenarios.

The miscellaneous exercise includes questions related to both finite and infinite sequences, sum of terms of AP and GP, and applications of sequences and series in real-life problems. It combines the knowledge gained from previous exercises to test the students' ability to apply the formulas and techniques to a wider range of problems.

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 step-by-step to help students grasp the concepts clearly and practice solving such problems efficiently.

Q1. If f(x+y) = f(x)·f(y) for all x, y ∈ N, such that f(1) = 3 and Σf(x) = 120 (x=1 to n), find the value of n.

Ans: Given f(x+y) = f(x)·f(y) and f(1) = 3.

Let x = y = 1 → f(2) = f(1)·f(1) = 3×3 = 9

f(3) = 27, f(4) = 81, ...

So f(1), f(2), f(3), ... = 3, 9, 27, ... forms a G.P. with a = 3, r = 3.

Sₙ = 3(3ⁿ − 1)/(3−1) = 120

→ (3/2)(3ⁿ − 1) = 120 → 3ⁿ − 1 = 80 → 3ⁿ = 81 = 3⁴ → n = 4

Q2. The sum of some terms of G.P. is 315 whose first term and common ratio are 5 and 2. Find the last term and the number of terms.

Ans: a = 5, r = 2, Sₙ = 315

5(2ⁿ − 1)/(2−1) = 315 → 2ⁿ − 1 = 63 → 2ⁿ = 64 = 2⁶ → n = 6

6th term = ar⁵ = 5 × 32 = 160

Number of terms = 6, Last term = 160

Q3. The first term of a G.P. is 1. The sum of the third and fifth term is 90. Find the common ratio.

Ans: a = 1, a₃ = r², a₅ = r⁴

r² + r⁴ = 90 → r⁴ + r² − 90 = 0

r² = (−1 + √361)/2 = 9 → r = ±3

Q4. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in order, we get an A.P. Find the numbers.

Ans: Let the three numbers be a, ar, ar².

a(1 + r + r²) = 56 ...(1)

A.P. condition: (ar−7)−(a−1) = (ar²−21)−(ar−7)

→ a(r²−1)² = 8 ...(2)

Dividing (1) by (2): 7(r²−2r+1) = 1+r+r²

→ 6r² − 15r + 6 = 0 → (6r−3)(r−2) = 0 → r = 2 or r = 1/2

For r = 2: a = 8 → Numbers: 8, 16, 32

For r = 1/2: a = 32 → Numbers: 32, 16, 8 (same set)

Q5. A G.P. consists of an even number of terms. If the sum of all terms is 5 times the sum of odd-placed terms, find the common ratio.

Ans: Let G.P. be a, ar, ar², ar³, ...

Sum of all terms = 5 × sum of odd-placed terms

→ T₂ + T₄ + ... = 4(T₁ + T₃ + ...)

→ ar(rⁿ−1)/(r−1) = 4a(rⁿ−1)/(r−1)

→ ar = 4a → r = 4

Q6. If (a+bx)/(a−bx) = (b+cx)/(b−cx) = (c+dx)/(c−dx), (x ≠ 0), show that a, b, c, d are in G.P.

Ans: From (a+bx)/(a−bx) = (b+cx)/(b−cx):

Cross-multiplying → 2b²x = 2acx → b² = ac → b/a = c/b

From (b+cx)/(b−cx) = (c+dx)/(c−dx):

→ 2c²x = 2bdx → c² = bd → c/b = d/c

Therefore b/a = c/b = d/c → a, b, c, d are in G.P. (Proved)

Q7. Let S be the sum, P the product and R the sum of reciprocals of terms in a G.P. Prove that P²Rⁿ = Sⁿ.

Ans: Let G.P. be a, ar, ar², ...arⁿ⁻¹

S = a(rⁿ−1)/(r−1)

P = aⁿ · r^(n(n−1)/2)

R = (rⁿ−1) / [arⁿ⁻¹(r−1)]

P²Rⁿ = a²ⁿ · r^(n(n−1)) × (rⁿ−1)ⁿ / [aⁿ · r^(n(n−1)) · (r−1)ⁿ]

= aⁿ(rⁿ−1)ⁿ / (r−1)ⁿ = [a(rⁿ−1)/(r−1)]ⁿ = Sⁿ (Proved)

Q8. If a, b, c, d are in G.P., prove that (aⁿ+bⁿ), (bⁿ+cⁿ), (cⁿ+dⁿ) are in G.P.

Ans: Given b² = ac, c² = bd, ad = bc.

To prove: (bⁿ+cⁿ)² = (aⁿ+bⁿ)(cⁿ+dⁿ)

LHS = b²ⁿ + 2bⁿcⁿ + c²ⁿ = (ac)ⁿ + 2bⁿcⁿ + (bd)ⁿ

= aⁿcⁿ + bⁿcⁿ + aⁿdⁿ + bⁿdⁿ = cⁿ(aⁿ+bⁿ) + dⁿ(aⁿ+bⁿ)

= (aⁿ+bⁿ)(cⁿ+dⁿ) = RHS (Proved)

Q9. a and b are roots of x²−3x+p=0 and c, d are roots of x²−12x+q=0, where a, b, c, d form a G.P. Prove that (q+p):(q−p) = 17:15.

Ans: a+b = 3, ab = p; c+d = 12, cd = q

Let a = x, b = xr, c = xr², d = xr³

x(1+r) = 3 and xr²(1+r) = 12

Dividing: r² = 4 → r = ±2

Case I (r=2, x=1): p = ab = x²r = 2, q = cd = x²r⁵ = 32

(q+p)/(q−p) = 34/30 = 17/15

Case II (r=−2, x=−3): p = 18, q = −288

(q+p)/(q−p) = −306/−270 = 17/15

Therefore (q+p):(q−p) = 17:15 (Proved)

Q10. The ratio of A.M. and G.M. of two positive numbers a and b is m:n. Show that a:b = (m+√(m²−n²)) : (m−√(m²−n²)).

Ans: (a+b)/2√(ab) = m/n → (a+b) = 2√(ab)·m/n ...(1)

Using (a−b)² = (a+b)² − 4ab:

(a−b) = 2√(ab)·√(m²−n²)/n ...(2)

Adding (1) and (2): a = (√ab/n)(m + √(m²−n²))

Subtracting: b = (√ab/n)(m − √(m²−n²))

Therefore a:b = (m+√(m²−n²)) : (m−√(m²−n²)) (Proved)

Q11(i). Find the sum of n terms of the series 5 + 55 + 555 + ...

Ans: Sₙ = (5/9)(9 + 99 + 999 + ...)

= (5/9)[(10−1) + (10²−1) + ... to n terms]

= (5/9)[10(10ⁿ−1)/9 − n]

= 50(10ⁿ−1)/81 − 5n/9

Q11(ii). Find the sum of n terms of the series 0.6 + 0.66 + 0.666 + ...

Ans: Sₙ = 6(0.1 + 0.11 + 0.111 + ... to n terms)

= (6/9)(0.9 + 0.99 + 0.999 + ... to n terms)

= (2/3)[n − (1/10)(1−(1/10)ⁿ)/(1−1/10)]

= 2n/3 − (2/27)(1−10⁻ⁿ)

Q12. Find the 20th term of the series 2×4 + 4×6 + 6×8 + ... + n terms.

Ans: nth term aₙ = 2n × (2n+2) = 4n² + 4n

a₂₀ = 4(20)² + 4(20) = 4(400) + 80 = 1600 + 80 = 1680

Q13. A farmer buys a tractor for Rs. 12000. He pays Rs. 6000 cash and pays balance in annual installments of Rs. 500 plus 12% interest on unpaid amount. How much will the tractor cost?

Ans: Unpaid amount = Rs. 6000

Interest each year: 12% of 6000, 5500, 5000, ..., 500

Total interest = 12% of (500 + 1000 + ... + 6000)

Series is A.P. with a = 500, d = 500, last term = 6000 → n = 12

Sum = (12/2)[2(500) + 11(500)] = 6 × 6500 = Rs. 39000

Total interest = 12% of 39000 = Rs. 4680

Total cost = Rs. 12000 + Rs. 4680 = Rs. 16680

Q14. Shamshad Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and pays balance in annual installment of Rs. 1000 plus 10% interest on unpaid amount. How much will the scooter cost?

Ans: Unpaid amount = Rs. 18000

Interest each year: 10% of 18000, 17000, ..., 1000

Total interest = 10% of (1000 + 2000 + ... + 18000)

A.P. with a = 1000, d = 1000, n = 18

Sum = (18/2)[2(1000) + 17(1000)] = 9 × 19000 = Rs. 171000

Total interest = 10% of 171000 = Rs. 17100

Total cost = Rs. 22000 + Rs. 17100 = Rs. 39100

Q15. A person writes letters to 4 friends, each asked to mail to 4 more. It costs 50 paise per letter. Find amount spent when 8th set of letters is mailed.

Ans: Letters mailed: 4, 4², 4³, ..., 4⁸ → G.P. with a = 4, r = 4, n = 8

S₈ = 4(4⁸−1)/(4−1) = 4(65535)/3 = 87380

Cost = 87380 × 50/100 = Rs. 43690

Q16. A man deposits Rs. 10000 at 5% simple interest annually. Find amount in 15th year and total amount after 20 years.

Ans: Annual interest = 5% of 10000 = Rs. 500

Amount in 15th year = Rs. 10000 + 14 × Rs. 500 = Rs. 10000 + Rs. 7000 = Rs. 17000

Amount after 20 years = Rs. 10000 + 20 × Rs. 500 = Rs. 10000 + Rs. 10000 = Rs. 20000

Q17. A machine costs Rs. 15625 and depreciates 20% each year. Find value at end of 5 years.

Ans: Each year value = 80% = 4/5 of previous year.

Value after 5 years = 15625 × (4/5)⁵ = 15625 × 1024/3125 = 5 × 1024 = Rs. 5120

Q18. 150 workers were to finish a job. 4 workers dropped out from 2nd day onwards each day. It took 8 more days. Find total days to complete the work.

Ans: Let original number of days = x.

150x = 150 + 146 + 142 + ... (x+8 terms)

A.P. with a = 150, d = −4, n = (x+8)

150x = (x+8)/2 × [2(150) + (x+7)(−4)]

→ 150x = (x+8)(150 − 2x − 14) = (x+8)(136−2x)

→ 75x = (x+8)(68−x)

→ x² + 15x − 544 = 0 → (x−17)(x+32) = 0

→ x = 17 (rejecting negative)

Total days = 17 + 8 = 25

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

Q1. What is the focus of the Miscellaneous Exercise in Chapter 8?
The Miscellaneous Exercise in Chapter 8 focuses on solving various types of problems related to arithmetic and geometric progressions, including the sum of terms and applications of these progressions in real-world scenarios.

Q2. What types of problems are included in this exercise?
The exercise includes problems based on:

  • Sum of the first n terms of AP and GP.
  • Problems involving finite and infinite geometric progressions.
  • Applications of sequences and series in practical scenarios.
  • Problems that involve finding the nth term and sum of terms in a progression.

Q3. Why is this exercise important for exams?
This exercise is important because it helps students practice applying the concepts of sequences and series to different types of questions. The problems are designed to improve understanding and problem-solving skills, which are crucial for the exams.

Q4. How can students prepare effectively for this exercise?
Students should:

  • Review the formulas for the sum of terms in AP and GP.
  • Practice solving a variety of problems to get familiar with different types of questions.
  • Understand how to apply sequences and series to real-life situations.
  • Solve previous years’ exam questions related to sequences and series for additional practice.

Q5. How does this exercise relate to real-life applications?
Sequences and series are used in many fields like finance, physics, engineering, and computer science. For example, the concept of geometric progression is applied in calculating compound interest, while arithmetic progressions are used to model situations like salary increments and loan repayments.