Important Questions Class 11 Maths Chapter 8 Sequences and Series

Sequences and Series study ordered number patterns, their general terms and the sums formed by adding their terms.
Important Questions Class 11 Maths Chapter 8 help students practise G.P., sum of G.P., geometric mean and A.M.-G.M. relation for 2026-27 exams.

Class 11 Maths Chapter 8 develops the idea of ordered number patterns and their sums. Students first learn how a sequence has a first term, second term and nth term. The chapter then explains series, sigma notation, geometric progression, sum of G.P., geometric mean and the relation between arithmetic mean and geometric mean. This chapter is important because many exam questions are formula-based, but the answers need clean substitution and step-by-step reasoning. Students should practise finding nth terms, sums, inserted geometric means and application questions such as ancestors, bacteria growth, depreciation and compound interest.

Key Takeaways

• Sequence: A sequence is an ordered list of numbers following a rule.
• Series: A series is formed by adding the terms of a sequence.
• G.P.: In a geometric progression, every non-zero term has the same ratio with the previous term.
• A.M.-G.M. relation: For two positive numbers, arithmetic mean is always greater than or equal to geometric mean.

Important Questions Class 11 Maths Chapter 8 Structure 2026-27

Section A: MCQs from Important Questions Class 11 Maths Chapter 8

Section A checks definitions, formulas and quick calculations. Class 11 Maths Sequences and Series important questions in this section usually cover nth term, G.P., common ratio, geometric mean and finite series.

Q1. A sequence is:

1. An unordered collection of numbers
   b. An ordered list of numbers following a rule
   c. A single number only
   d. A diagram only

Answer: b. An ordered list of numbers following a rule

A sequence has terms arranged in a definite order.

Q2. The nth term of a sequence is also called:

1. First term
   b. Last term only
   c. General term
   d. Common difference

Answer: c. General term

The nth term gives the term at the nth position.

Q3. A finite sequence contains:

1. No terms
   b. A fixed number of terms
   c. Infinite terms
   d. Only odd terms

Answer: b. A fixed number of terms

A finite sequence ends after a fixed number of terms.

Q4. A series is formed by:

1. Multiplying all terms of a sequence
   b. Adding the terms of a sequence
   c. Subtracting only alternate terms
   d. Dividing all terms

Answer: b. Adding the terms of a sequence

A series is the indicated sum of sequence terms.

Q5. The Greek letter used in sigma notation is:

1. π
   b. θ
   c. Σ
   d. β

Answer: c. Σ

Sigma notation is used to write sums compactly.

Q6. A sequence is a geometric progression if:

1. Difference between consecutive terms is constant
   b. Ratio of each term to its previous term is constant
   c. Sum of all terms is zero
   d. Terms are always natural numbers

Answer: b. Ratio of each term to its previous term is constant

This constant ratio is called the common ratio.

Q7. The common ratio of the G.P. 2, 4, 8, 16, ... is:

1. 1
   b. 2
   c. 4
   d. 8

Answer: b. 2

Common ratio = 4 / 2 = 2.

Q8. The common ratio of 9, 3, 1, 1/3, ... is:

1. 3
   b. 1/3
   c. 9
   d. -3

Answer: b. 1/3

Common ratio = 3 / 9 = 1/3.

Q9. The nth term of a G.P. with first term a and common ratio r is:

1. an = a + (n - 1)r
   b. an = arn - 1
   c. an = ar^(n - 1)
   d. an = n/2[2a + (n - 1)r]

Answer: c. an = ar^(n - 1)

The nth term of GP class 11 is found using an = ar^(n - 1).

Q10. The 5th term of the G.P. 3, 6, 12, ... is:

1. 24
   b. 36
   c. 48
   d. 96

Answer: c. 48

Here, a = 3 and r = 2.

5th term = 3 × 2^4 = 48.

Q11. If r = 1 in a G.P., the sum of n terms is:

1. 0
   b. a
   c. na
   d. n/a

Answer: c. na

When r = 1, every term is a, so the sum is na.

Q12. If r is not equal to 1, the sum of first n terms of a G.P. is:

1. Sn = a(1 - r^n)/(1 - r)
   b. Sn = n/2[2a + (n - 1)d]
   c. Sn = a + d
   d. Sn = ar^(n - 1)

Answer: a. Sn = a(1 - r^n)/(1 - r)

This form is commonly used when r is less than 1 or convenient for calculation.

Q13. The geometric mean of two positive numbers a and b is:

1. a + b
   b. ab
   c. √ab
   d. (a + b)/2

Answer: c. √ab

Geometric mean class 11 uses the positive square root of the product.

Q14. The geometric mean of 2 and 8 is:

1. 2
   b. 4
   c. 8
   d. 16

Answer: b. 4

G.M. = √(2 × 8) = √16 = 4.

Q15. For two positive numbers, the relation between A.M. and G.M. is:

1. A.M. < G.M.
   b. A.M. = 0 always
   c. A.M. ≥ G.M.
   d. A.M. + G.M. = 1

Answer: c. A.M. ≥ G.M.

AM and GM relation class 11 is A ≥ G for positive numbers.

Q16. Fibonacci sequence begins with:

1. 1, 1, 2, 3, 5, ...
   b. 2, 4, 6, 8, ...
   c. 1, 3, 5, 7, ...
   d. 3, 6, 12, 24, ...

Answer: a. 1, 1, 2, 3, 5, ...

In Fibonacci sequence class 11, each term after the second is the sum of the two previous terms.

Q17. In a G.P., if a = 5 and r = 5, then the nth term is:

1. 5n
   b. 5^n
   c. n^5
   d. 25n

Answer: b. 5^n

an = ar^(n - 1) = 5 × 5^(n - 1) = 5^n.

Q18. If a, G, b are in G.P., then G equals:

1. a + b
   b. a - b
   c. √ab
   d. a/b

Answer: c. √ab

The middle term is the geometric mean.

Q19. Which of the following is a finite series?

1. 1 + 3 + 5 + 7
   b. 1 + 3 + 5 + ... forever
   c. All natural numbers
   d. All prime numbers

Answer: a. 1 + 3 + 5 + 7

It has exactly four terms.

Q20. Assertion: The sequence 2, 4, 8, 16 is a G.P.

Reason: The ratio of each term to the previous term is 2.

1. Both Assertion and Reason are true, and Reason explains Assertion
   b. Both are true, but Reason does not explain Assertion
   c. Assertion is true, Reason is false
   d. Assertion is false, Reason is true

Answer: a. Both Assertion and Reason are true, and Reason explains Assertion

The common ratio is constant.

Section B: Very Short Answer Questions from Class 11 Maths Sequences and Series Important Questions

Section B questions need short formulas and direct substitution. Keep the steps visible because Sequences and Series questions are often awarded marks for method.

Q21. Write the first three terms of the sequence an = 2n + 5.

Put n = 1, 2 and 3.

a1 = 2(1) + 5 = 7.

a2 = 2(2) + 5 = 9.

a3 = 2(3) + 5 = 11.

So, the first three terms are 7, 9 and 11.

Q22. Find the 20th term of the sequence an = (n - 1)(2 - n)(3 + n).

Put n = 20.

a20 = (20 - 1)(2 - 20)(3 + 20)

= 19 × (-18) × 23

= -7866

So, the 20th term is -7866.

Q23. Write the first five terms if a1 = 1 and an = an - 1 + 2 for n ≥ 2.

a1 = 1.

a2 = a1 + 2 = 1 + 2 = 3.

a3 = a2 + 2 = 3 + 2 = 5.

a4 = a3 + 2 = 5 + 2 = 7.

a5 = a4 + 2 = 7 + 2 = 9.

So, the first five terms are 1, 3, 5, 7 and 9.

Q24. Find the common ratio of the G.P. 5, 25, 125, ...

Common ratio = second term / first term.

r = 25 / 5 = 5.

So, the common ratio is 5.

Q25. Find the 10th term of the G.P. 5, 25, 125, ...

Here, a = 5 and r = 5.

nth term of a G.P.:

an = ar^(n - 1)

a10 = 5 × 5^(10 - 1)

= 5 × 5^9

= 5^10

So, the 10th term is 5^10.

Section C: Short Answer Questions from Sequences and Series Class 11 Important Questions

Section C questions usually test G.P. formulas, geometric mean and simple applications. Write the given values first, then apply the formula.

Q26. Which term of the G.P. 2, 8, 32, ... is 131072?

Here, a = 2 and r = 4.

Let the nth term be 131072.

an = ar^(n - 1)

131072 = 2 × 4^(n - 1)

65536 = 4^(n - 1)

Since 65536 = 4^8,

4^(n - 1) = 4^8

n - 1 = 8

n = 9

So, 131072 is the 9th term.

Q27. In a G.P., the 3rd term is 24 and the 6th term is 192. Find the 10th term.

Let the first term be a and common ratio be r.

3rd term = ar^2 = 24.

6th term = ar^5 = 192.

Divide ar^5 by ar^2:

r^3 = 192 / 24 = 8.

r = 2.

Now, ar^2 = 24.

a × 2^2 = 24.

4a = 24.

a = 6.

10th term = ar^9.

a10 = 6 × 2^9

= 6 × 512

= 3072

So, the 10th term is 3072.

Q28. Find the sum of first 5 terms of the G.P. 1, 2/3, 4/9, ...

Here, a = 1 and r = 2/3.

Sum of first n terms:

Sn = a(1 - r^n)/(1 - r)

S5 = 1 × [1 - (2/3)^5] / [1 - 2/3]

= [1 - 32/243] / [1/3]

= [211/243] × 3

= 211/81

So, the sum of first 5 terms is 211/81.

Q29. Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.

Let the sequence be 1, G1, G2, G3, 256.

Here, a = 1 and 256 is the 5th term.

5th term = ar^4.

256 = 1 × r^4.

r^4 = 256.

r = 4.

Now,

G1 = ar = 1 × 4 = 4.

G2 = ar^2 = 1 × 16 = 16.

G3 = ar^3 = 1 × 64 = 64.

So, the numbers are 4, 16 and 64.

Q30. If A.M. and G.M. of two positive numbers are 10 and 8, find the numbers.

Let the numbers be a and b.

A.M. = (a + b) / 2 = 10.

So, a + b = 20.

G.M. = √ab = 8.

So, ab = 64.

Now,

(a - b)^2 = (a + b)^2 - 4ab

= 20^2 - 4(64)

= 400 - 256

= 144

So, a - b = ±12.

Solving a + b = 20 and a - b = 12:

a = 16 and b = 4.

So, the numbers are 16 and 4.

Q31. Prove that A.M. is greater than or equal to G.M. for two positive numbers.

Let A and G be the arithmetic mean and geometric mean of positive numbers a and b.

A = (a + b) / 2.

G = √ab.

Now,

A - G = (a + b) / 2 - √ab

= (a + b - 2√ab) / 2

= (√a - √b)^2 / 2

Since a square is always non-negative,

(√a - √b)^2 / 2 ≥ 0.

So, A - G ≥ 0.

Therefore, A ≥ G.

Section D: Long Answer Questions from Important Questions Class 11 Maths Chapter 8

Section D questions need complete working. Most long-answer questions in this chapter come from G.P., sum of G.P., G.M. and application-based series.

Q32. Find the sum of the sequence 7, 77, 777, 7777, ... to n terms.

Let the sum be Sn.

Sn = 7 + 77 + 777 + 7777 + ... to n terms.

Write each term using 9:

7 = 7/9 × 9.

77 = 7/9 × 99.

777 = 7/9 × 999.

So,

Sn = 7/9 [9 + 99 + 999 + ... to n terms]

= 7/9 [(10 - 1) + (10^2 - 1) + (10^3 - 1) + ... + (10^n - 1)]

= 7/9 [(10 + 10^2 + 10^3 + ... + 10^n) - n]

Now,

10 + 10^2 + 10^3 + ... + 10^n = 10(10^n - 1) / 9.

Therefore,

Sn = 7/9 [10(10^n - 1) / 9 - n]

So, the sum is:

Sn = 7[10(10^n - 1) - 9n] / 81.

Q33. How many terms of the G.P. 3, 3/2, 3/4, ... are needed to give the sum 3069/512?

Here, a = 3 and r = 1/2.

Let n terms be needed.

Sum formula:

Sn = a(1 - r^n)/(1 - r)

3069/512 = 3[1 - (1/2)^n] / [1 - 1/2]

3069/512 = 6[1 - (1/2)^n]

Divide by 6:

3069/3072 = 1 - (1/2)^n

So,

(1/2)^n = 1 - 3069/3072

= 3/3072

= 1/1024

Now, 1024 = 2^10.

So,

(1/2)^n = (1/2)^10

Therefore, n = 10.

So, 10 terms are needed.

Q34. The sum of the first three terms of a G.P. is 13/12 and their product is -1. Find the terms.

Let the three terms be a/r, a and ar.

Their product:

(a/r) × a × ar = a^3.

Given product = -1.

So, a^3 = -1.

a = -1.

Now, sum of terms = 13/12.

a/r + a + ar = 13/12.

Substitute a = -1:

-1/r - 1 - r = 13/12.

Multiply by 12r:

-12 - 12r - 12r^2 = 13r.

So,

12r^2 + 25r + 12 = 0.

Factorise:

12r^2 + 16r + 9r + 12 = 0.

4r(3r + 4) + 3(3r + 4) = 0.

(4r + 3)(3r + 4) = 0.

So,

r = -3/4 or r = -4/3.

If r = -3/4, the terms are:

a/r = (-1)/(-3/4) = 4/3.

a = -1.

ar = (-1)(-3/4) = 3/4.

If r = -4/3, the terms are:

a/r = (-1)/(-4/3) = 3/4.

a = -1.

ar = (-1)(-4/3) = 4/3.

So, the terms are 4/3, -1, 3/4 or 3/4, -1, 4/3.

Q35. A person has 2 parents, 4 grandparents, 8 great-grandparents and so on. Find the number of ancestors during 10 generations preceding his own.

The number of ancestors forms a G.P.

2, 4, 8, 16, ...

Here, a = 2, r = 2 and n = 10.

Sum of n terms:

Sn = a(r^n - 1)/(r - 1)

S10 = 2(2^10 - 1)/(2 - 1)

= 2(1024 - 1)

= 2 × 1023

= 2046

So, the number of ancestors in 10 generations is 2046.

Section E: Case Study-Based Questions from Class 11 Maths Chapter 8 Sequences and Series

Section E tests real-life patterns using G.P., series and repeated growth. Identify the first term and common ratio before calculating.

Q36. Case Study: Bacteria growth

The number of bacteria in a culture doubles every hour. Initially, there are 30 bacteria.

Q36(a). What is the first term?

The first term is 30.

So, a = 30.

Q36(b). What is the common ratio?

The number doubles every hour.

So, r = 2.

Q36(c). How many bacteria will be present at the end of the 4th hour?

Use nth term of a G.P.:

an = ar^(n - 1)

For the 4th hour:

a4 = 30 × 2^(4 - 1)

= 30 × 2^3

= 30 × 8

= 240

So, 240 bacteria will be present.

Q36(d). Write the number of bacteria at the nth hour.

an = 30 × 2^(n - 1)

So, the number of bacteria at the nth hour is 30 × 2^(n - 1).

Q37. Case Study: Depreciation of a machine

A machine costs Rs 15625. Its value depreciates by 20 per cent every year.

Q37(a). What percentage of value remains each year?

If depreciation is 20 per cent, then 80 per cent value remains.

So, the common ratio is 80/100 = 4/5.

Q37(b). What is the value after 1 year?

Value after 1 year = 15625 × 4/5.

= 12500

So, the value after 1 year is Rs 12500.

Q37(c). What is the value at the end of 5 years?

Value after 5 years = 15625 × (4/5)^5.

Now,

15625 = 5^6.

(4/5)^5 = 4^5 / 5^5.

Value = 5^6 × 4^5 / 5^5

= 5 × 4^5

= 5 × 1024

= 5120

So, the value at the end of 5 years is Rs 5120.

Q38. Case Study: Chain letter

A person writes a letter to 4 friends. Each friend sends the letter to 4 more people. The chain continues in the same way.

Q38(a). How many letters are mailed in the first set?

The first set has 4 letters.

So, a = 4.

Q38(b). What is the common ratio?

Each person sends the letter to 4 people.

So, r = 4.

Q38(c). How many letters are mailed in the 8th set?

Use nth term of a G.P.:

an = ar^(n - 1)

a8 = 4 × 4^(8 - 1)

= 4^8

= 65536

So, 65536 letters are mailed in the 8th set.

Q38(d). If mailing one letter costs 50 paise, find the postage cost for the 8th set.

Cost per letter = 50 paise = Rs 0.50.

Total cost = 65536 × 0.50

= Rs 32768

So, the postage cost for the 8th set is Rs 32768.

Formula-Based Revision for Important Questions Class 11 Maths Chapter 8

Important questions class 11 maths chapter 8 should be revised through definitions, formulas and example patterns. Keep formulas simple and copy-friendly.

Sequence Class 11

A sequence is an ordered list of numbers.

The nth term is the term at the nth position.

Example:

For even numbers 2, 4, 6, 8, ...

an = 2n.

Series Class 11

A series is the sum of the terms of a sequence.

Example:

1 + 3 + 5 + 7 is a finite series.

Sigma notation:

a1 + a2 + a3 + ... + an = Σak, where k goes from 1 to n.

Geometric Progression Class 11

A sequence is a G.P. if the ratio of each term to the previous term is constant.

General G.P.:

a, ar, ar^2, ar^3, ...

Here, a is the first term and r is the common ratio.

Nth Term of GP Class 11

nth term of a G.P.:

an = ar^(n - 1)

Here:

a = first term
r = common ratio
n = term number

Sum of GP Class 11

If r = 1:

Sn = na.

If r is not equal to 1:

Sn = a(1 - r^n)/(1 - r)

or

Sn = a(r^n - 1)/(r - 1)

Use the form that makes calculation easier.

Geometric Mean Class 11

For two positive numbers a and b:

G.M. = √ab.

If a, G, b are in G.P., then G = √ab.

AM and GM Relation Class 11

For two positive numbers:

A.M. = (a + b) / 2

G.M. = √ab

A.M. ≥ G.M.

Equality holds when a = b.

Chapter-Wise Revision for Sequences and Series Class 11 Important Questions

Sequences and Series class 11 important questions should be revised in five parts: sequences, series, G.P., geometric mean and applications.

Start with sequence class 11. Practise writing terms from nth term formulas and recurrence relations.

Next, revise series class 11. Understand sigma notation and finite series before moving to formula-based sums.

Then revise geometric progression class 11. Focus on common ratio, nth term of GP class 11 and sum of GP class 11.

After that, revise geometric mean class 11 and AM and GM relation class 11. These are useful in proof-based questions.

Finally, practise application sums. Bacteria growth, depreciation, ancestors, chain letters and compound interest are common G.P. applications.

Important Questions Class 11 Maths Chapter-Wise