Sequences and Series MCQ Quiz - Objective Question with Answer for Sequences and Series - Download Free PDF

Last updated on Jun 19, 2025

Latest Sequences and Series MCQ Objective Questions

Sequences and Series Question 1:

If the sum of n terms of an A. P. be 3n2 + n and the common difference is 6, then its 1st term is:

  1. 2
  2. 3
  3. 1
  4. 4

Answer (Detailed Solution Below)

Option 4 : 4

Sequences and Series Question 1 Detailed Solution

Given:

The sum of n terms of an A.P. is given by:

Sn = 3n2 + n

The common difference, d = 6

Formula used:

The sum of the first n terms of an A.P. is given by the formula:

Sn = n/2 × (2a + (n - 1)d)

Where, a = first term, d = common difference

Calculations:

We are given Sn = 3n2 + n, and d = 6. We can equate this to the formula for the sum of n terms of an A.P.:

3n2 + n = n/2 × (2a + (n - 1) × 6)

Multiply both sides by 2 to eliminate the fraction:

6n2 + 2n = n × (2a + 6n - 6)

Now, divide both sides by n (n ≠ 0):

6n + 2 = 2a + 6n - 6

Cancel out 6n from both sides:

2 = 2a - 6

2a = 8

a = 4

∴ The first term of the A.P. is 4.

Sequences and Series Question 2:

If p, 1, q are in AP and p, 2, q are in GP, then which of the following statements is/are correct?

I. p, 4, q are in HP.

II. (1/p), (1/4), (1/q) are in AP.

Select the answer using the code given below.

  1. I only
  2. II only
  3. Both I and II
  4. Neither I nor II

Answer (Detailed Solution Below)

Option 3 : Both I and II

Sequences and Series Question 2 Detailed Solution

Concept:

1. Arithmetic Progression (AP): The difference between consecutive terms is constant.

⇒ If p, a, q are in AP, then 2a = p + q.

2. Geometric Progression (GP): The ratio of consecutive terms is constant.

⇒ If p, b, q are in GP, then b2 = pq.

3. Harmonic Progression (HP): Reciprocal of terms are in AP.

⇒ If p, c, q are in HP, then (1/p), (1/c), (1/q) are in AP.

Calculation:

Using AP condition: p, 1, q are in AP.

⇒ 2(1) = p + q

⇒ p + q = 2 ............(1)

Using GP condition: p, 2, q are in GP.

⇒ 22 = p × q

⇒ 4 = pq ............(2)

Now,

2pqp+q=82=4

and 

p+q2pq=14

∴  (1/p), (1/4), (1/q) are in AP and 

 p, 4 and q are in H.P.

Thus, both statements are true.

Hence, the correct answer is Option 3.

Sequences and Series Question 3:

If 5th, 7th and 13th terms of an AP are in GP, then what is the ratio of its first term to its common difference?

  1. -3
  2. -2
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 1 : -3

Sequences and Series Question 3 Detailed Solution

Concept:

In an AP, the nth term is given by:

Tn = a + (n - 1)d

In a GP, terms satisfy the relation:

Tm2 = Tn × Tp

Calculation:

Let the 5th, 7th, and 13th terms of the AP be T5, T7, and T13 respectively.

⇒ T5 = a + 4d

⇒ T7 = a + 6d

⇒ T13 = a + 12d

Since the terms are in GP:

⇒ T72 = T5 × T13

⇒ (a + 6d)2 = (a + 4d) × (a + 12d)

Expanding both sides:

⇒ (a + 6d)2 = a2 + 12ad + 48d2

⇒ a2 + 12ad + 36d2 = a2 + 12ad + 48d2

Cancel out common terms:

⇒ 36d2 = 48d2

⇒ -12d2 = 0

Dividing through by d2 (assuming d ≠ 0):

⇒ a = -3d

Conclusion:

∴ Ratio of the first term to the common difference is:

⇒ a/d = -3

Hence, the correct answer is Option 1.

Sequences and Series Question 4:

The sum of the first 8 terms of a GP is five times the sum of its first 4 terms. If is the common ratio, then what is the number of possible real values of r?

  1. One 
  2. Two 
  3. Three
  4. More than three 

Answer (Detailed Solution Below)

Option 2 : Two 

Sequences and Series Question 4 Detailed Solution

Calculation:

Given,

The sum of the first 8 terms of a GP is 5 times the sum of its first 4 terms.

The formula for the sum of the first n terms of a GP is:

Sn=a1rn1r

Substitute for S8andS4 into the equation:

a1r81r=5×a1r41r

1r81r4=5

r85r4+4=0

Let x = r4, giving the quadratic equation:

x25x+4=0

Solving for x , we get:

x=4orx=1

Since x = r4, this gives:

r4=4r=±2

The number of possible real values of r  is two: ±2.

Hence, the correct answer is Option 2.

Sequences and Series Question 5:

The sum of the first k terms of a series S is . Which one of the following is correct?

  1. The terms of S form an arithmetic progression with common difference 14.
  2. The terms of S form an arithmetic progression with common difference 6.
  3. The terms of S form a geometric progression with common ratio 10/7.
  4. The terms of S form a geometric progression with common ratio 11/4.

Answer (Detailed Solution Below)

Option 2 : The terms of S form an arithmetic progression with common difference 6.

Sequences and Series Question 5 Detailed Solution

Given:

The sum of the first k terms of a series S is given as Sk = 3k2 + 5k.

Concept:

To find the nth term of a series, we use the formula:

an = Sn - Sn-1

If the nth term forms an arithmetic progression (AP), the common difference (d) is given by:

d = an+1 - an

Calculation:

We are given Sk = 3k2 + 5k.

⇒ an = Sn - Sn-1

Also Sn = 3n2 + 5n and Sn-1 = 3(n-1)2 + 5(n-1)

⇒ an = [3n2 + 5n] - [3(n-1)2 + 5(n-1)]

⇒ an = [3n2 + 5n] - [3(n2 - 2n + 1) + 5n - 5]

⇒ an = [3n2 + 5n] - [3n2 - 6n + 3 + 5n - 5]

⇒ an = 3n2 + 5n - 3n2 + 6n - 3 - 5n + 5

⇒ an = 6n + 2

The series is an arithmetic progression (AP) if the difference between consecutive terms is constant.

⇒ Common difference d = an+1 - an

⇒ an+1 = 6(n+1) + 2 = 6n + 6 + 2 = 6n + 8

⇒ d = (6n + 8) - (6n + 2) = 6

∴ The terms of the series form an arithmetic progression with a common difference of 6.

Hence, the correct answer is Option B.

Top Sequences and Series MCQ Objective Questions

The sum of the series 5 + 9 + 13 + … + 49 is:

  1. 351
  2. 535
  3. 324
  4. 435

Answer (Detailed Solution Below)

Option 3 : 324

Sequences and Series Question 6 Detailed Solution

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Concept:

Arithmetic Progression (AP):

  • The sequence of numbers where the difference of any two consecutive terms is same is called an Arithmetic Progression.
  • If a be the first term, d be the common difference and n be the number of terms of an AP, then the sequence can be written as follows:
    a, a + d, a + 2d, ..., a + (n - 1)d.
  • The sum of n terms of the above series is given by:
    Sn = n2[a+{a+(n1)d}]=(First Term+Last Term2)×n

 

Calculation:

The given series is 5 + 9 + 13 + … + 49 which is an arithmetic progression with first term a = 5 and common difference d = 4.

Let's say that the last term 49 is the nth term.

∴ a + (n - 1)d = 49

⇒ 5 + 4(n - 1) = 49

⇒ 4(n - 1) = 44

⇒ n = 12.

And, the sum of this AP is:

S12(First Term+Last Term2)×12

= (5+492)×12 = 54 × 6 = 324.

Find the value of 111!+12!13!+14!.....

  1. e
  2. 2e
  3. 1e
  4. e2

Answer (Detailed Solution Below)

Option 3 : 1e

Sequences and Series Question 7 Detailed Solution

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Concept:

Expansion of ex:

ex=1+x1!+x22!+x33!+x44!+.....

 

Calculation:

ex=1+x1!++x22!+x33!+x44!+.....

Put x = -1,

e(1)=1+(1)1!++(1)22!+(1)33!+(1)44!+.....

e1=111!+12!13!+14!.....

111!+12!13!+14!.....=1e

The third term of a G.P. is 9. The product of its first five terms is

  1. 35
  2. 39
  3. 310
  4. 312

Answer (Detailed Solution Below)

Option 3 : 310

Sequences and Series Question 8 Detailed Solution

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Concept:

Five terms in a geometric progression:

If a G.P. has first term a and common ratio r then the five consecutive terms in the GP are of the form ar2,ar,a,ar,ar2 .

 

Calculation:

Let us consider a general geometric progression with common ratio r.

Assume that the five terms in the GP are ar2,ar,a,ar,ar2.

It is given that third term is 9.

Therefore, a = 9.

Now the product of the five terms is given as follows:

ar2×ar×a×ar×ar2=a5

But we know that a = 9.

Thus, the product is 95=310.

Find sum of 11×2+12×3+13×4+...+1n×(n+1)

  1. n(n + 1)
  2. nn+1
  3. 2nn+1
  4. None of these 

Answer (Detailed Solution Below)

Option 2 : nn+1

Sequences and Series Question 9 Detailed Solution

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Calculation:

11×2+12×3+13×4+...+1n×(n+1)

=211×2+322×3+433×4+...+(n+1)nn×(n+1)

=1112+1213+1314+...+1n1n+1

=11n+1

=n+11n+1

=nn+1

Find the sum to n terms of the A.P., whose nth term is 5n + 1

  1. n2
  2. n2 (7+ 4n)
  3. n2 (7+ 5n)
  4. None of these

Answer (Detailed Solution Below)

Option 3 : n2 (7+ 5n)

Sequences and Series Question 10 Detailed Solution

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Concept:

For AP series, 

Sum of n terms  = n2 (First term + nth term)

Calculations:

We know that, For AP series, 

the sum of n terms  = n2 (First term + nth term)

Given, the nth term of the given series is a= 5n + 1.

Put n = 1, we get

a= 5(1) + 1 = 6.

We know that 

sum of n terms = n2 (First term + nth term)

⇒Sum of n terms = n2 (6 + 5n + 1)

⇒Sum of n terms = n2 (7+ 5n)

If the sum of n numbers in the GP 5, 10, 20, ... is 1275 then n is ?

  1. 6
  2. 7
  3. 8
  4. 9

Answer (Detailed Solution Below)

Option 3 : 8

Sequences and Series Question 11 Detailed Solution

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Concept:

 Let us consider sequence a1, a2, a3 …. an is a G.P.
  • Common ratio = r = a2a1=a3a2==anan1
  • nth  term of the G.P. is an = arn−1
  • Sum of n terms of GP = sn = a(rn1)r1; where r >1
  • Sum of n terms of GP = sn = a(1rn)1r; where r <1
  • Sum of infinite GP = s=a1r ; |r| < 1

Calculation:

Given series is 5, 10, 20, ...

Here, a = 5, r = 2

Sum of n numbers = sn = 1275

To Find: nAs we know that, Sum of n terms of GP = sn = a(rn1)r1; where r >1

∴ sn = 5(2n1)21

1275 = 5 × (2n - 1)

⇒ 255 = (2n - 1)

⇒ 2n = 256

⇒ 2n = 28

∴ n = 8

The third term of a GP is 3. What is the product of the first five terms?

  1. 216
  2. 226
  3. 243
  4. Cannot be determined due to insufficient data

Answer (Detailed Solution Below)

Option 3 : 243

Sequences and Series Question 12 Detailed Solution

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Concepts:

Let us consider sequence a1, a2, a3 …. an is an G.P.

  • Common ratio = r = a2a1=a3a2==anan1
  • nth term of the G.P. is an = arn−1
  • Sum of n terms = s = a(rn1)r1; where r >1
  • Sum of n terms = s = a(1rn)1r; where r <1
  • Sum of infinite GP = s=a1r; |r| < 1

Where a is 1st term and r is common ratio.

Calculation:

Given: The third term of a GP is 3

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

∴ T3 = ar2 = 3

We know that Tn = a rn-1

So, T1 = a, T2 = ar, T3 = ar2, T4 = ar3, T5 = ar4

Now, Product of the first five terms = a × ar × ar2 × ar3 × ar4 = a5r10 = (ar2)5 = 35 = 243

The harmonic mean and the geometric mean of two numbers are 10 and 12 respectively. What is their arithmetic mean?

  1. 253
  2. √120
  3. 11
  4. 14.4

Answer (Detailed Solution Below)

Option 4 : 14.4

Sequences and Series Question 13 Detailed Solution

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Given :

The harmonic mean and the geometric mean of two numbers are 10 and 12 respectively

Concept used :

(Geometric mean)2 = Harmonic mean × Arithmetic mean 

Calculations :

according to the formula given above 

(12)2 = 10 × Arithmetic mean 

⇒ Arithmetic mean = 144/10 

⇒ 14.4 

∴ Option 4 will be the correct answer.

Alternate Method 

Concept:

A.M. between a and b = a+b2

G.M. between a and b = ab

H.M. between a and b = 2aba+b

Calculation:

Given, G.M. = 12, H.M. = 10

GM=ab

122=ab

ab = 144 ........(1)

HM=2aba+b

10=2aba+b

2ab = 10a + 10b

ab = 5 (a + b)........(2)

From equation (1) and (2)

144 = 5 (a + b)

1445=a+b

14410=a+b2

a+b2=14.4

But, we know that,

A.M. between a and b = a+b2

Therefore, AM = 14.4

4th term of a G. P is 8 and 10th term is 27. Then its 6th term is?

  1. 12
  2. 14
  3. 16
  4. 18

Answer (Detailed Solution Below)

Option 1 : 12

Sequences and Series Question 14 Detailed Solution

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Concept:

Let us consider sequence a1, a2, a3 …. an is a G.P.

  • Common ratio = r = a2a1=a3a2==anan1
  • nth  term of the G.P. is an = arn−1
  • Sum of n terms = s = a(rn1)r1; where r >1
  • Sum of n terms = s = a(rn1)r1; where r <1

Calculation:

Given:

4th term of a G. P is 8 and 10th term is 27

nth  term of the G.P. is Tn = a rn-1

∴ T4 = a. r3 = 8      ----(1)

T10 = a r9 = 27      ----(2)

Equation (2) ÷ (1), we get 

r6=278

(r2)3=(32)3

∴ r2=32

T6 = a r5

= a r3.r2

=8×32

= 12

What is the sum of all the common terms between the given series S1 and S2 ?

S1 = 2, 9, 16, .........., 632

S2 = 7, 11, 15, .........., 743

  1. 6974
  2. 6750
  3. 7140
  4. 6860

Answer (Detailed Solution Below)

Option 1 : 6974

Sequences and Series Question 15 Detailed Solution

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GIVEN:

Two series given i.e. S1 and S2

FORMULA USED:

an = a + ( n - 1 ) d 

Sn = n/2 [2a + (n - 1) d ] 

Where,

a= nth term in the sequence , n= number of terms , a = first term in sequence, d = common difference , Sn = Sum 

CALCULATION:

Here, given series S1  and S2 are in A.P.

So, series will move by adding a fixed common difference ( second term - first term) in consecutive terms 

S1 = 2 , 9 , 16 , 23, 30 , 37 , 44 , 51 ,........ 632      [As  here d = 7 .So, add 7 in previous term to get next term]

S2  = 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51 , ......... 743  [ Here d = 4 ] 

Now, let us take a third series S3 which is common series                    [It will contain common numbers of both series only]

So, from S1 and S2 series, we have 1st common term = 23 , 2nd common term = 51 , d = ( 51 - 23) = 28 

So, add 28 in second term to get third term and so on

S3 = 23 , 51 , ..................   632              [ As, 632 is less than 743 so common between them should be less than 632]

Now, we have a = 23 , d = 28 

⇒ an = a + ( n - 1) d     632 

⇒ 23 + ( n - 1) × 28  632 

⇒ ( n -  1) × 28    ( 632 - 23 ) 

⇒ ( n - 1) × 28   609 

⇒ n -1  609 /28 

⇒ n - 1  21.75 

⇒ n  22 .75

 As, n should be equal to or less than 22.75 . So, take n = 22 

Now, as we know that 

Sn = n/2 [ 2a + ( n - 1) d ] 

⇒ Sn = 22/2 [ 2 × 23 + ( 22 - 1) 28]  = 11 [46 + 21 × 28 ] 

⇒ 11 [ 46 + 588 ] = 11 × 634  = 6974

Hence, Sum of all the common terms of the series are 6974 .

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