CBSE Important Questions Class 11 Maths Chapter 8 – Sequences and Series

Class 11 Maths Chapter 8 Sequences and Series is part of the CBSE Class 11 Maths syllabus. It explains what a sequence and a series are, and then studies two important types: the arithmetic progression (AP) and the geometric progression (GP). Students learn the nth term and the sum formulae for both, the arithmetic mean and geometric mean, and the sum of an infinite GP.

These CBSE important questions for Class 11 Maths Chapter 8 cover finding the nth term of a GP, the sum of n terms, inserting means between two numbers, the sum of an infinite GP, the relation between AM and GM, and the sums of the first n natural numbers, their squares and their cubes. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Write the formulae for the nth term and the sum of n terms of an AP and of a GP.

Arithmetic Progression (AP) Geometric Progression (GP)
nth term aₙ = a + (n − 1)d aₙ = arn−1
Sum of n terms Sₙ = n/2 [2a + (n − 1)d] Sₙ = a(rⁿ − 1)/(r − 1), for r ≠ 1

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

Q2. Find the 12th term of the GP 2, 6, 18, 54, ...

Answer: Here the first term a = 2 and the common ratio r = 6/2 = 3.
Using aₙ = arn−1 with n = 12:
a₁₂ = 2 × 3¹¹
= 2 × 177147
= 354294

Q3. Find the sum of the first 10 terms of the GP 3, 6, 12, 24, ...

Answer: Here a = 3 and r = 6/3 = 2. Since r > 1 we use
Sₙ = a(rⁿ − 1)/(r − 1)
S₁₀ = 3(2¹⁰ − 1)/(2 − 1)
= 3(1024 − 1)/1
= 3 × 1023
= 3069

Q4. Find the sum to infinity of the GP 1, 1/2, 1/4, 1/8, ...

Answer: Here a = 1 and r = 1/2. Since |r| < 1, the infinite sum exists and
S∞ = a / (1 − r)
= 1 / (1 − 1/2)
= 1 / (1/2)
= 2
(The sum of an infinite GP exists only when |r| < 1.)

Q5. Insert three arithmetic means between 3 and 19.

Answer: Inserting 3 means makes an AP of 5 terms, with first term 3 and fifth term 19.
Using a₅ = a + 4d:
19 = 3 + 4d
4d = 16, so d = 4
The terms are 3, 7, 11, 15, 19.
So the three arithmetic means are 7, 11 and 15.

Q6. Insert two geometric means between 2 and 54.

Answer: Inserting 2 means makes a GP of 4 terms, with first term 2 and fourth term 54.
Using a₄ = ar³:
54 = 2r³
r³ = 27, so r = 3
The terms are 2, 6, 18, 54.
So the two geometric means are 6 and 18.

Q7. If the AM and GM of two positive numbers a and b are 10 and 8 respectively, find the numbers.

Answer: AM: (a + b)/2 = 10, so a + b = 20
GM: √(ab) = 8, so ab = 64
Now a and b are the roots of x² − (a + b)x + ab = 0:
x² − 20x + 64 = 0
(x − 16)(x − 4) = 0
So the numbers are 16 and 4.

Q8. Write the formulae for the sum of the first n natural numbers, the sum of their squares and the sum of their cubes. Hence find 1² + 2² + ... + 10².

Answer: • Σn = n(n + 1)/2
• Σn² = n(n + 1)(2n + 1)/6
• Σn³ = [n(n + 1)/2]²
Calculation for n = 10:
Σn² = 10 × 11 × 21 / 6
= 2310/6
= 385

Q9. The sum of the first three terms of a GP is 13/12 and their product is −1. Find the terms.

Answer: Let the three terms be a/r, a and ar.
Product: (a/r)(a)(ar) = a³ = −1, so a = −1.
Sum: a/r + a + ar = 13/12
Putting a = −1: −1/r − 1 − r = 13/12
Multiplying by −12r: 12 + 12r + 12r² = −13r
12r² + 25r + 12 = 0
(3r + 4)(4r + 3) = 0, so r = −4/3 or r = −3/4
Taking r = −4/3, the terms are 3/4, −1 and 4/3.

Q10. Which term of the GP 3, 3√3, 9, ... is 729?

Answer: Here a = 3 and r = 3√3 / 3 = √3.
Let the nth term be 729.
aₙ = arn−1
729 = 3 × (√3)n−1
243 = (√3)n−1
Writing both sides as powers of 3: 3⁵ = 3(n−1)/2
5 = (n − 1)/2
n − 1 = 10
n = 11, so 729 is the 11th term.

Q11. How many terms of the GP 3, 3/2, 3/4, ... are needed to give the sum 3069/512?

Answer: Here a = 3 and r = 1/2. Since r < 1 we use
Sₙ = a(1 − rⁿ)/(1 − r)
3069/512 = 3[1 − (1/2)ⁿ] / (1 − 1/2)
3069/512 = 6[1 − (1/2)ⁿ]
3069/3072 = 1 − (1/2)ⁿ
(1/2)ⁿ = 1 − 3069/3072 = 3/3072 = 1/1024
So 2ⁿ = 1024, which gives n = 10.

Q12. The 5th term of an AP is 11 and the 9th term is 7. Find the 16th term.

Answer: Using aₙ = a + (n − 1)d:
a + 4d = 11 ...(i)
a + 8d = 7 ...(ii)
Subtracting (i) from (ii): 4d = −4, so d = −1
Putting d = −1 in (i): a − 4 = 11, so a = 15
Now the 16th term:
a₁₆ = a + 15d
= 15 + 15(−1)
= 0

Related Study Material on Extramarks

Q.1 The sum of the binomial coefficients in the expansion of (1 + x)n is

Marks:1

Ans

2.

2n-1.

2n.

2n+1.

We have,
(1 + x)n = C0 + C1x + C2x2 + C3x3 ++ Cn-1xn-1 + Cnxn.
Putting x = 1 in this expansion, we get
2n = C0 + C1 + C2 + C3 ++ Cn-1 + Cn.

Q.2 The largest coefficient in the expansion of (1 + x)24 is

Marks:1

Ans

We know that, if n is even then nCr is greatest for r = n/2.

Therefore, 24Cr is greatest for r = 12.

Hence, the greatest coefficient is 24C12.

Q.3 Find the term independent of x in the expansion of

x+1×6.

Marks:2

Ans

Suppose (r+1) the term is independent of x in the expansion of x+1×6.Tr+1=6Crx6r1xrTr+1=6Crx6rrTo be independent of x, we have62r=0r=3T3+1=6C3x631x3=20 Hence, 4thterm is independent of x1 i.e., 20.

Q.4 Find the middle term in the expansion of 

3x13x20.

Marks:1

Ans

Here, n=20, which is an even number. So, 202+1th  term =11th  term is the middle term Hence, the middle term =T11=T10+1=20C103x201013x10=20C10

Q.5 Find a if the 17th and 18th terms of the expansion (2 + a)50 are equal.

Marks:3

Ans

We know Tr+1=n Crxnryr…1 Here x=2andy=a Putting r=16andr=17in1,

we get T16+1=T17=50C1625016a16 and T17+1=T18=50C1725017d17T17=T18

given 50C1625016a16 =50C1725017a1750C162501650C1725017 =a17a16a =5016501623450175017233=1750171650162=1733342a=17234=1

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