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

Last updated on Apr 16, 2025

Latest Sequences and Series MCQ Objective Questions

Sequences and Series Question 1:

Let x=15+16+15+16+... Which of the following equals x?

  1. 1+5562
  2. 15562
  3. 3+10.2
  4. 3+10.2

Answer (Detailed Solution Below)

Option 4 : 3+10.2

Sequences and Series Question 1 Detailed Solution

x=15+16+x

x=6+x31+5x

5x2+31x=x+6

5x2+30x6=0

x=30±900+12010=3±10.2

The continued fraction has to be positive.

Therefore, We reject the negative value.

Sequences and Series Question 2:

The given series n=1(1)n1n5n1

  1. Convergent

  2. Divergent
  3. Oscillatory
  4. None of these

Answer (Detailed Solution Below)

Option 3 : Oscillatory

Sequences and Series Question 2 Detailed Solution

Concept:

A series in which the terms are alternatively positive or negative is called an alternating series.

Liebnitz’s series:

An alternating series u1 – u2 + u3 – u4 + … converges if

(i) Each term is numerically less than its preceding term,

(ii) limnun=0

If limnun0, the given series is oscillatory.

Calculation:

Given series is n=1(1)n1n5n1

Now un=n5n1

⇒ unun1=n5n1n15n8=2n1(5n1)(5n8) < 0

So each term is numerically less than its preceding term.

Now limit,

limnun=limnn5n1=12

⇒ limnun0

Series is oscillatory

Sequences and Series Question 3:

Test for convergence Σ5n25n+17n37n2+2

  1. Convergent 
  2. Divergent 
  3. Neither Convergent nor Divergent 
  4. None of these

Answer (Detailed Solution Below)

Option 1 : Convergent 

Sequences and Series Question 3 Detailed Solution

Given:

Σ5n25n+17n37n2+2

Concept used:

Limit Comparision test:

if an and bn are two positive series such that Ltnanbn=c 

where c > 0 and finite then, either Both series converges or diverges together

P - Series test: 

1npis convergent for p > 1 and divergent for p ≤  1

Calculations:

nth term of the given series = un = Σ5n25n+17n37n2+2

Let vn=1n2

Ltnunvn=Ltn[n55n+1n2n3(77n+2n3)×n21]

=Ltn[55n+1n2(77n+2n3)]=570

∴ By comparison test, Σun and Σvn both converge or diverge.

But Σvn is convergent. [p series test  - p = 2 > 1]

 ∴ Σun is convergent.

Sequences and Series Question 4:

Test for convergence Σn=1(5n+36n+1)1/2

  1. Convergent 
  2. Divergent 
  3. Neither Convergent nor Divergent
  4. None of these

Answer (Detailed Solution Below)

Option 1 : Convergent 

Sequences and Series Question 4 Detailed Solution

Given:

Σn=1(5n+36n+1)1/2

Concept used:

Limit Comparision test:

if an and bn are two positive series such that Ltnanbn=c  where c > 0 and finite then, either Both series converge or diverge together

P - Series test: 

1npis convergent for p > 1 and divergent for p ≤  1

Calculations:
 
un=[5n(1+35n)6n(1+16n)]1/2
 
Take vn=5n6n
 

unvn=(1+35n1+16n)1/2

Ltnunvn=10 ;

∴ By comparison test, Σun and Σvn behave the same way.

But Σvn = Σn=1(56)n/2=56+56+(56)3/2+.... which is a geometric series with common ratio 56 which is less than 1.

∴ Σvn is convergent.

Hence Σun is convergent.

Sequences and Series Question 5:

Let α = 1+ 42 + 82 + 13+ 19+ 262 + ........... upto 10 terms and β=n=110n4. If 4α - β = 55k + 40, then k is equal to ___________.

Answer (Detailed Solution Below) 353

Sequences and Series Question 5 Detailed Solution

Calculation

α = 12 + 42 + 82 …. 

⇒ tn = an2 + bn + c 

⇒ 1 = a + b + c 

⇒ 4 = 4a + 2b + c 

⇒ 8 = 9a + 3b + c 

On solving we get, a=12, b=32,c=1

⇒ α=n=110(n22+3n21)2

⇒ 4α=n=110(n2+3n2)2,β=n=110n4

⇒ 4αβ=n=110(6n3+5n212n+4)=55(353)+40

On comparing

  k = 353

Top Sequences and Series MCQ Objective Questions

The sum of the series 1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ........... will be:

  1. 1a4
  2. 1
  3. 1a2
  4. -1

Answer (Detailed Solution Below)

Option 1 : 1a4

Sequences and Series Question 6 Detailed Solution

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Concept:

a + ar + ar2 + ar3 +….. 

Sum of the above infinite geometric series:

=a1r

Analysis:

Given:

1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ......

let x = (a2 + 1)

The series now becomes

S = 1 + 2x + 3x2 + 4x3 + ......  ----(1)

By multiplying x on both sides we get

xS = x + 2x2 + 3x3 + 4x4 + ...... ----(2)

Subtracting (1) and (2), we get

S(1 - x) = 1 + x + x2 + x3 + ..... ---(3)

The right hand side of (3) forms infinite geometric series with a = 1, r = x

∴ S(1 - x) = 11x

S=1(1x)2

putting the value of x, we get

S=1(1a21)2

S=1a4

Identify the next number in the sequence.

1, 2, 4, 7, 11, _____

  1. 14
  2. 16
  3. 12
  4. 10

Answer (Detailed Solution Below)

Option 2 : 16

Sequences and Series Question 7 Detailed Solution

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The pattern followed here is –

F1 Shubanshi Ravi 03.11.21 D1

Hence 16 will complete the series.

The sequence \(\left\) is

  1. Convergent
  2. Divergent to ∞
  3. Divergent to -∞
  4. None of the above

Answer (Detailed Solution Below)

Option 3 : Divergent to -∞

Sequences and Series Question 8 Detailed Solution

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Concept:

The Nth term test

If limn(n=0an)=L, where L is any tangible number other than zero. Then, (n=0an) diverges.

This is also called the Divergence test.

Calculation:

We have, \(\left\)

n=1log(1n)

limn[n=1log(1n)]

limnlog(1n)

So, as n → ∞, 1n → 0

limnlog(1n)=0

Thus, our series diverges to -∞ by the nth term test.

Hence, The sequence \(\left\) is divergent to -∞.

If (1+ x + x2)nr=02n axr, then a1 − 2a2 + 3a3 − …. −2n a2n is

  1. (n + 1)2n
  2. n
  3. −n
  4. n(2n)

Answer (Detailed Solution Below)

Option 3 : −n

Sequences and Series Question 9 Detailed Solution

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Concept Used:-

If the summation from 0 to n such that r=0n ar xis given, then the expanded form of it can be written as,

 ⇒ r=0n ar x= a0+a1 x+a2 x2+a3 x3+a4 x4+..........+anxn

Explanation:-

Given,

(1+ x + x2)n = r=02n ar xr

On the expanding right-hand side, we get,

(1+x+x2)n=a0+a1x+a2x2+a3x3+a4x4++a2nx2n

Differentiate it with respect to x,

n(1+x+x2)n1(1+2x)=a1+2a2x+3a3x2+4a4x3++2nanx2x1Now put x = -1 in the above equation,

n(11+1)n1(12)=a12a2+3a34a4+5a52na2nn1(1)n1(1)=a12a2+3a34a4+5a52na2nn=a12a2+3a34a4+5a52na2n

So, the value of a1 − 2a2 + 3a3 − …. −2n a2n is -n.

Hence, the correct option is 3.

If ai > 0 for i = 1, 2, 3,..,n and a1, a2, a3, ...an = 1 then the greatest value of (1 + a1)(1 + a2)... (1 + an) is:

  1. 22n
  2. 2n
  3. 1
  4. 2n2

Answer (Detailed Solution Below)

Option 2 : 2n

Sequences and Series Question 10 Detailed Solution

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Let the given expansion be f(n)

f(n) = (1 + a1)(1 + a2)... (1 + an)

Also given, a1 = a= a3 = ... = an = 1

Consider for n = 2

f(2) = (1 + a1)(1 + a2)

f(2) = (1 + 1)(1 + 1) = 22

Consider for n = 5

f(5) = (1 + a1)(1 + a2)(1 + a3)(1 + a4)(1 + a5)

f(5) = (1 + 1)(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 25

Similary for n times, it is given as

f(n) = (1 + 1)(1 + 1) ...... (1 + 1) = 2n

(1 + a1)(1 + a2)... (1 + an) = 2n

The true statement for the series (n=1xn1n(3n)) is:

  1. convergent if x > 3
  2. convergent if x > 1
  3. convergent if x > 0
  4. convergent if x < 3

Answer (Detailed Solution Below)

Option 4 : convergent if x < 3

Sequences and Series Question 11 Detailed Solution

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Concept Used:

Ratio Test:

Let L=limn|an+1||an|

  • If L < 1, then an converges absolutely.
  • If L > 1, or the limit goes to ∞ , then an diverges.
  • If L = 1 or if L does not exist, then the test fails, and we know nothing.

Calculation:

(n=1xn1n(3n)) = (n=0an)

an=xn1n(3n) and an+1=xn(n+1)3n+1

⇒ L=limnxn(n+1)3n+1xn1n(3n)

⇒ L=limnx3nn+1

⇒ L=limnx3(11n+1)

⇒ L=x3

For Convergent, x3<1

⇒ x < 3

Consider the following series:

Σn=1ndcn

For which of the following combinations of c, d values does this series converge?

  1. c = 1, d  = -1
  2. c = 2, d = 1
  3. c = 0.5, d = -10
  4. c = 1, d = -2

Answer (Detailed Solution Below)

Option :

Sequences and Series Question 12 Detailed Solution

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It is a MSQ question.

Correct answers would be option(2) and option (4)

Concept:

To check whether the given expression converges or not we will apply Ration test and P test.

Ratio Test:

Let an is the given expression.

Let L=limn|an+1an|

if L<1, then the series Converges

if L>1, then the series Diverges

if L<1, then the test is Inconclusive.

P series test:

 n=11np=11p+12p+13p+..... where p> 0 by definition

If p > 1 , then the series converges. If 0 < p <= 1 , then the series diverges.

Calculation:

Option (1): Substitute c= 1, d= -1

Perform  test:

n=1ndcn=n=1n11n=n=11n   

As, p=1 , hence series will diverge.

Hence, option (1) is incorrect.

Option (2):

Substitute c=2, d= 1.

Perform Ration test

limn|an+1an|=limnn+12n+1×2nn=12

​​​​As, L < 1 , the series converges.

Hence, option (2) is correct 

Option (3): 

​Substitute c= 0.5, d= -10

Perform Ratio test;

an=n10(0.5)nlimn|an+1an|=limn(n+1)10(0.5)n+1×0.5nn10=10.5=2

As L>1, the series will diverge.

Hence, option (3) is incorrect.

Option (4):

​Substitute c= 1, d= -2

Perform P test:

  • an=n2(1)n=1n2
  • Here, p > 1 , therefore the series converges.
  • Hence, option (4) is correct.

The series n=15n+13n1 is

  1. Convergent
  2. Divergent
  3. Oscillatory
  4. None of these

Answer (Detailed Solution Below)

Option 2 : Divergent

Sequences and Series Question 13 Detailed Solution

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Concept:

The comparison test in the limit form,

Consider two positive term series n=1an and n=1bn such that limnanbn=L (finite), then both the series either converge or diverge.

Calculation:

Given series is

n=15n+13n1

⇒ an=5n+13n1=(52)n21+15n113n

Lets take bn=(52)n2 

By comparison test in limit form,

limn(anbn)=limn1+15n113n=1(finite)

Therefore, both the series either converge or diverge.

The series ∑bn is a geometric progression with r = 2.5, since r > 1, series is divergent.

As bn is divergent, an is also divergent.

Sequences and Series Question 14:

The sum of the series 1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ........... will be:

  1. 1a4
  2. 1
  3. 1a2
  4. -1

Answer (Detailed Solution Below)

Option 1 : 1a4

Sequences and Series Question 14 Detailed Solution

Concept:

a + ar + ar2 + ar3 +….. 

Sum of the above infinite geometric series:

=a1r

Analysis:

Given:

1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ......

let x = (a2 + 1)

The series now becomes

S = 1 + 2x + 3x2 + 4x3 + ......  ----(1)

By multiplying x on both sides we get

xS = x + 2x2 + 3x3 + 4x4 + ...... ----(2)

Subtracting (1) and (2), we get

S(1 - x) = 1 + x + x2 + x3 + ..... ---(3)

The right hand side of (3) forms infinite geometric series with a = 1, r = x

∴ S(1 - x) = 11x

S=1(1x)2

putting the value of x, we get

S=1(1a21)2

S=1a4

Sequences and Series Question 15:

Consider two series n=1an and n=1bn where an=1nn and bn=1n!, then

  1. The series n=1an is convergent and n=1bn is divergent.
  2. The series n=1an is divergent and n=1bn is convergent.
  3. Both the series n=1an and n=1bn are divergent.
  4. Both the series n=1an and n=1bn are convergent.

Answer (Detailed Solution Below)

Option 4 : Both the series n=1an and n=1bn are convergent.

Sequences and Series Question 15 Detailed Solution

Concept:

A p-series is a specific type of infinite series. It's a series of the form as shown below,

n=11np=11p+12p+13p+...

With p-series,

If p > 1, the series will convergeIf 0 < p ≤ 1, the series will diverge.

The Ratio test:

Theorem: Let ∑an be a series and If limnak+1ak=L, and

if L < 1 then the series converges, but if L > 1 the series diverges.

If L = 1, then the test is inconclusive, then we have to go for another test.

Explanation:

Consider the first series:

 n=1ann=11nnn=11n32

This takes the form of p series n=11np

Here p = 3/2

p > 1 ⇒ Convergent

Consider the second series n=1bnn=11n!

By ratio test, limnbn+1bn=lim1n+1=0

Here L = 0 (< 1) ⇒ Convergent

Therefore, both the series are convergent.

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