Explore Exams
Login Signup
Sign in Open App
All Exams  >   Commerce  >   Online MCQ Tests for Commerce  >   All Questions

All questions of Sequence and Series for Commerce Exam

What is the 50th term of the sequence   √3, 3, 3√3, 9, ......
  • a)
    (√3)49
  • b)
    (√3)50
  • c)
    349
  • d)
    350
Correct answer is option 'B'. Can you explain this answer?

Nandini Iyer answered
an = ar(n-1)
 Given, a = √3, r = 3/√3
​r = √3
a50 = ar(n-1)
= (√3)(√3)(50-1)
= (√3)(√3)49
= (√3)50
Clear all your doubts with EduRev

 Find the missing number. 1, 4, 9, 16, 25, 36, 49, (....)
  • a)
    64
  • b)
    54
  • c)
    56
  • d)
    81
Correct answer is option 'A'. Can you explain this answer?

Anjana Sharma answered
The given number series as follow as 1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, ...
So the next number is 8^2 = 64.

The terms of an A.P. are doubled, then the resulting sequence is
  • a)
    An A.P. with common difference equal to the common difference of the original A.P​
  • b)
    An A.P. with common difference thrice the common difference of the original A.P.
  • c)
    An A.P. with common difference double the common difference of the original A.P.
  • d)
    An A.P. with common difference half the common difference of the original A.P.
Correct answer is option 'C'. Can you explain this answer?

Lavanya Menon answered
The general form of an AP is  a,  a+d, a+2d,.....
where a is the first term and d is the common difference  
If we double the terms,  the new sequence would be
A , a+2d, a+4d,......
We can observe that this sequence is also an AP
First term is a
Common difference is 2d  nth term= 2a+(n-1)2d               
= 2[a+(n-1)d]

The next term of the sequence 1, 2, 4, 7,11,…. Is
  • a)
    18
  • b)
    17
  • c)
    16
  • d)
    15
Correct answer is option 'C'. Can you explain this answer?

Imk Pathsala answered
The given series is: 1,2,4,7,11,...
Difference between second and first term = 2 - 1 = 1
Difference between third and second term = 4 - 2 = 2
Difference between fourth and third term = 7 - 4 = 3
Difference between fifth and fourth term = 11 - 7 = 4
Difference between sixth and fifth term = 16 - 11 = 5

The sum of the first hundred even natural numbers divisible by 5 is
  • a)
    55005
  • b)
    55000
  • c)
    50000
  • d)
    50500
Correct answer is option 'D'. Can you explain this answer?

Preeti Iyer answered
first even no. divisible by 5= 10

d=10

Sn = n/2(2a + (n-1) d)

S100= 100/2 (2(10) + (100-1)10)

S100=50(20 + 990)

S100= 50(1010)

S100 = 50500

The first negative term of the  A.P.62,57,52…. is the
  • a)
    10th term
  • b)
    14th term
  • c)
    12th term
  • d)
    18th term
Correct answer is option 'B'. Can you explain this answer?

Naina Sharma answered
a = 62  d = 57 - 62 = -5
tn = a + (n-1)d
= 62 + (n-1)(-5)
= 62 - 5n + 5
= 67 - 5n
From the options, we take '14'
= 67 - 5(14)
= 67 - 70
= -3 (The first negative term will be at the 14th term)

The three numbers between 1 and 256 such that the sequence is in GP are
  • a)
    4,16,64
  • b)
    2,8,32
  • c)
    8,16,32
  • d)
    8,16,64
Correct answer is option 'A'. Can you explain this answer?

Pooja Shah answered
nth G.M. between a and b is
Gn = arn
Where common ratio is r = (b/a)(1/(n+1))
​So, to insert 3 geometric means between 1 and 256
 r = (b/a)(1/(n+1))
r = (256/(1)(1/(3+1))
r = (256)1/4
r  = (4)(4)1/4
r = 4
Gn = 1 * (4)n
G1 = 1 * (4)1 = 4
G2 = 1 * (4)2 = 16
G3 = 1 * (4)3 = 64
The terms are  4, 16, 64

How many terms of the series
24,20,16,…are required so that their sum is 72?
  • a)
    8 or 6
  • b)
    12
  • c)
    9 or 4
  • d)
    11
Correct answer is option 'C'. Can you explain this answer?

Naina Bansal answered
As given above we can clearly see that the given series forms an AP. So we can use the sum formula to calculate it.

= Sn = n / 2 [ 2a + ( n-1 ) d ]

So the values are :

a = 24

d = -4

n = ?

Sn = 72
So substitute in the formula to get the answer:

= 72 = n / 2 [ 2 ( 24 ) + ( n - 1 ) -4 ]

= 72 = n / 2 [ 48 - 4n + 4 ]

= 72 * 2 = n [ 52 - 4n ]

= 144 = 52n - 4n^2

= 4n^2 - 52n + 144 = 0 ------ Dividing by 4 throughout the equation we get,

= n^2 - 13n + 36

Factorizing the above quadratic equation we get,

= n^2 - 9n - 4n + 36 = 0

= n ( n - 9 ) -4 ( n - 9 ) = 0

= ( n - 4 ) ( n - 9 ) = 0

= n = 4 , 9

So the number of terms can be both 4 terms and 9 terms. This is because since the AP is decreasing. 
Hope it helps !!

If a, b, c, d are in H.P., then ab + bc + cd is
  • a)
    3 b d
  • b)
    (a + b) (c + d)
  • c)
    3 a d
  • d)
    3 a c
Correct answer is option 'C'. Can you explain this answer?

Rajesh Gupta answered
 Since a,b,c are in H.P, so b = 2ac/(a+c).
Also, b,c,d are in H.P, so c = 2bd/(b+d).
Therefore, (a+c)(b+d) = 2ac/b × 2bd/c
⇒ab+cb+ad+cd = 4ad
⇒ab+bc+cd = 3ad

 If A1, A2, A3,…., An are n numbers between a and b, such that a, A1, A2, A3,…, An, b are in A.P., then nth term from beginning is:
  • a)
    b
  • b)
    An-1
  • c)
    An
  • d)
    A1
Correct answer is option 'B'. Can you explain this answer?

Krishna Iyer answered
A1,A2,......, An are inserted between a and b then the series will become 
a, A1,A2,A3,......, An,b. Now a becomes the first term, A1 will be second, A2 will become third term
An will become A(n+1)th term 
therefore A(n-1) will become nth term.

The digits of a positive integer having three digits are in AP and sum of their digits is 21. The number obtained by reversing the digits is 396 less than the original number. Find the original number.
  • a)
    876
  • b)
    579
  • c)
    975
  • d)
    678
Correct answer is option 'C'. Can you explain this answer?

Trisha Vashisht answered
Let the digits at ones, tens and hundreds place be (a−d)a and (a+d) respectively. The, the number is
(a+d)×100+a×10+(a−d) = 111a+99d
The number obtained by reversing the digits is
(a−d)×+a×10+(a+d) = 111a−99d
It is given that the sum of the digits is 21.
(a−d)+a+(a+d) = 21                        ...(i)
Also it is given that the number obtained by reversing the digits is 594 less than the original number.
∴111a−99d = 111a+99d−396          ...(ii)
⟹ 3a = 21 and 198d = 396
⟹ a = 7 and d = -2
Original number = (a−d)×+a×10+(a+d)
= 100(9) + 10(7) + 5
= 975

The first 4 terms of the sequence   are
  • a)
    2, 6, 13, 27
  • b)
    2, 5, 11, 23
  • c)
    2, 6, 11, 24
  • d)
    2, 5, 12, 37
Correct answer is option 'B'. Can you explain this answer?

Geetika Shah answered
a1 = 2 
a2 = 2a1 + 1
=> 2(2) + 1 = 5
a3 = 2a2 + 1
=> 2(5) + 1 = 11
a4 = 2a3 + 1
=> 2(11) + 1 = 23
Hence, the required series is : 2,5,11,23………

Sum to 20 terms of the series (1.3)2 + (2.5)2 + (3.7)2 +… is:
  • a)
    168090
  • b)
    198090
  • c)
    178090
  • d)
    188090
Correct answer is option 'D'. Can you explain this answer?

Mira Joshi answered
Given that (1.3)2, (2.5)2 , (3.7)2 .....................20 terms
Here 1,2,3,.................are in A.P.
a = 1, d = 1
tn = a + (n-1) d = 1 + (n -1)1 = n
3,5,7 ............. are in A.P
a = 3, d = 2
tn = a + (n-1) d = 3 + (n -1)2 = 2n + 1
∴ nth term = n(2n + 1)2 = 4n3 + 4n2 + n
Sum of n terms Sn = ∑ ( 4n3 + 4n2 + n ) = 4 { n2(n+1)2}/4 + 4 {n (n+1)(2n+1)} / 6 + n(n+1)/2
= { n2(n+1)2} + 2 {n (n+1)(2n+1)} / 3 + n(n+1)/2
Sum of 20 terms S20 = 400 × 441 + 40 × 41 × 7 + 10 ×21 = 176400 + 11480 + 210
 = 188090.

In an A.P., sum of first p terms is equal to the sum of first q terms. Sum of its first p + q terms is
  • a)
    0
  • b)
    - ( p + q)
  • c)
    p + q
  • d)
    none of these.
Correct answer is option 'A'. Can you explain this answer?

Trisha Vashisht answered
Sp = Sq
⇒  p/2(2a+(p−1)d) = q/2(2a+(q−1)d)
⇒ p(2a+(p−1)d) = q(2a+(q−1)d)
⇒ 2ap + p2d − pd = 2aq + q2d − qd
⇒ 2a(p−q) + (p+q)(p−q)d − d(p−q) = 0
⇒ (p−q)[2a + (p+q)d − d] = 0
⇒ 2a + (p+q)d − d = 0
⇒ 2a + ((p+q) − 1)d = 0
⇒ Sp+q = 0

Which term of the sequence 8 – 6i, 7 – 4i, 6 – 2i, ….is a real number ?
  • a)
    7th
  • b)
    6th
  • c)
    5th
  • d)
    4th
Correct answer is option 'D'. Can you explain this answer?

Preeti Iyer answered
a = 8−6i 
d = 7−4i−8+6i
= −1+2i
an = a+(n−1)d
a+ib = 8−6i+(n−1)(−1+2i)
a+ib = 8−6i+(−1)(n−1)+(n−1)2i
= − 6+2(n−1)=0
= 2(n−1) = 6
n = 4
an = 8−6i+(4−1)(−1+2i)
= 8−6i−3+6i = 5
4th term = 5

The next term of the sequence 1, 5, 14, 30, 55, …… is
  • a)
    95
  • b)
    91
  • c)
    90
  • d)
    80
Correct answer is option 'B'. Can you explain this answer?

Mihir Chaudhary answered
The sequence is obtained by adding consecutive odd numbers, starting with 1.

1 + 0 = 1

1 + 3 = 4

4 + 5 = 9

9 + 7 = 16

16 + 9 = 25

So the next term in the sequence is 25.

Therefore, the complete sequence is:

1, 5, 14, 30, 55, 25

 The A.M. between two numbers is 34 and their G.M. is 16.The numbers are
  • a)
    32 and 8
  • b)
    64 and 4
  • c)
    62 and 6
  • d)
    60 and 18
Correct answer is option 'B'. Can you explain this answer?

Sushil Kumar answered
Let the numbers be x, y 
Then arithmatic mean = (x+y)/2 =34 
→x+y =68 
Also geometric mean =√(xy)=16 
oy xy=16^2=256 
Hence 
x(68−x)=256 
or x^2−68x+256=0 
(x−64)(x−4)=0 
Hence x=64 or x=4 
and y=4 or 64 
Larger number is 64

What is the 10th A.M between 2 and 57 if 10 A.M s are inserted between these numbers?
  • a)
    54
  • b)
    53
  • c)
    52
  • d)
    55
Correct answer is option 'C'. Can you explain this answer?

Aryan Khanna answered
2 and 57 have 10 terms between them so including them there would be 12 terms
an = 57, a = 2, n = 12
an = a + (n-1)d
=> 57 = 2 + (12 - 1)d
=> 55 = 11d
d = 5
T10 = a + 10d
=> 2 + 10(5) 
= 52

If the sums of n, 2n and 3n terms of an AP are S1, S2 and S3 respectively, then 
  • a)
  • b)
  • c)
    3
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Pooja Shah answered
Given x 2 - 3x = 0 
Factor x out in the expression on the left. 
x (x - 3) = 0 
For the product x (x - 3) to be equal to zero we nedd to have 
x = 0 or x - 3 = 0 
Solve the above simple equations to obtain the solutions. 
x = 0 
or 
x = 3 

The sum of the series 
  • a)
    49/25
  • b)
    49/24
  • c)
    47/24
  • d)
    49/29
Correct answer is option 'B'. Can you explain this answer?

Knowledge Hub answered
 (1 + 1/52) + (1/2 + 1/52) + (1/22 + 1/54)+......
= (1 + 1/2 + 1/22 +....) + (1/52 + 1/54 + 1/56 +......)
= r = (1/4)/(1/2) , r = (1/54)/(1/52)
⇒ r = 1/2 + 1/25
⇒ S = 1 + [(1/2)/(1-1/2)],    S = [(1/25)/(1-1/25)]
⇒ S = 1 + 1,    S = 1/24
Total sum = 2 + 1/24
⇒ 49/24 

If the 10 times of the 10th term of an AP is equal to 15 times to the 15th term, then the 25th term is:
  • a)
    25 times of 25th term
  • b)
    zero
  • c)
    1
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Nandini Patel answered
According to question ,
10 x 10th term =15 x 15th term 
let a is the first term and d is the common difference .
10 (a+9d)=15 (a+14d)

5a+120d=0

a +24d=0
now ,
25th term =a+(25-1) d=a+24d=0
hence 25th term=0

How many terms of the G.P. 4 + 16 + 64 + … will make the sum 5460?
  • a)
    9
  • b)
    7
  • c)
    8
  • d)
    6
Correct answer is option 'D'. Can you explain this answer?

Nandini Iyer answered
Sum (Sn) = a x (rn -1)/(r-1)
5460 = 4 x (4n -1)/3
16380 = 4n+1 - 4
16384 = 4n+1
47 = 4n+1
7 = n + 1
n = 6

The eleventh term of the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, ….. is
  • a)
    89
  • b)
    66
  • c)
    72
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Nandini Iyer answered
The sequence is the Fibonacci series
1+1 = 0
1+2 = 3
2+3 = 5
3+5 = 8
5+8 = 13
8+13 = 21
13+21 = 34
21+34 = 55
34+55 = 89
The 11th term will be 89.

  • a)
    4
  • b)
    6
  • c)
    8
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Preeti Iyer answered
 6 + 6 + 6.........∞
61/2 * 61/4 * 61/8………..∞
= 61/2 + 1/(2*2) + 1/(2*2*2)   (sum of infinte G.P.= a/(1−r))
= (1/2)/(6(1-½))
= (1/2)/(61/2)
= 6

Which term of the following sequence is 64?
2 , 2√2, 4, .....
  • a)
    12
  • b)
    10
  • c)
    11
  • d)
    8
Correct answer is option 'C'. Can you explain this answer?

Krishna Iyer answered
Given sequence : 2, 2√2, 4….
First term a1 = a = 2 and 2nd term a2 = 2√2, then
Common ratio r = a2/a = (2√2)/2
Let an = 64
∴ ar(n-1) = 64
⇒ 2.(√2)(n-1) = 32
⇒ (2)(n-1)/2 = 32
∴ (2)(n-1)/2 = (2)5
⇒ (n − 1)/2 = 5,
⇒ n = 11

Three geometric means between the numbers 1/4 and 64 are:
  • a)
    1/2, 8, 32
  • b)
    4, 16, 32
  • c)
    1, 8, 32
  • d)
    1, 4, 16
Correct answer is option 'D'. Can you explain this answer?

Gaurav Kumar answered
nth G.M. between a and b is
Gn = arn
Where common ratio is r = (b/a)(1/(n+1))
​So, to insert 3 geometric means between 1/4 and 64
 r = (b/a)(1/(n+1))
r = (64/(¼)(1/(3+1))
r = (256)1/4
r  = (4)(4)1/4
r = 4
Gn = 1 * (4)n
G0 = 1 * (4)0 = 1
G1 = 1 * (4)1 = 4
G2 = 1 * (4)2 = 16
The terms are 1, 4, 16

The sum of the series 13 – 23 + 33 – 43 + ……. + 93 is:
  • a)
    1225
  • b)
    -1525
  • c)
    425
  • d)
    925
Correct answer is option 'C'. Can you explain this answer?

Om Desai answered
13 + 33 + 53 + 73 + 93 − (23 + 43 + 63 + 83)
= Sum of cubes of first 5 odd number -Sum of cubes of first 4 even number
= n2(2n2–1) − 2n2(n+1)2
= 1225 – 800
= 425

The sum of 40 A.M.’s between two number is 120. The sum of 50 A.M.’s between them is equal to
  • a)
    150
  • b)
    130
  • c)
    160
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Mira Sharma answered
Let A1, A2, A3, , A40 be 40 A.M's between two numbers 'a' and 'b'.

Then, 

a, A1, A2, A3, , A40, b is an A.P. with common difference d  = (b - a)/(n + 1) = (b - a)/41
[ where n = 40]

now,

A1, A2, A3, , A40  = 40/2( A1 + A40)

A1, A2, A3, , A40 = 40/2(a + b)
[ a, A1, A2, A3,  , A40, b is an Ap then ,a + b = A1 + A40]

sum of 40A.M = 120(given)

120= 20(a + b)

=> 6 = a + b --(1)

Again ,
consider B1, B2,  B50  be 50 A.M.'s between two numbers a and b.

Then, a, B1, B2,, B50, b will be in A.P. with common difference = ( b - a)/51

now , similarly,

B1, B2, ... , B50 = 50/2(B1 + B2)

= 25(6) -from(1)

= 150

If a, 4, b are in A.P.; a, 2, b are in G.P.; then a, 1, b are in
  • a)
    G.P.
  • b)
    H.P.
  • c)
    A.P.
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Lekshmi Chopra answered
Given: a, 4, b are in A.P. and a, 2, b are in G.P.

To find: a, 1, b are in which progression.

Solution:

Let d be the common difference in the A.P.

Then, 4 - a = b - 4 = d ...(1)

Let r be the common ratio in the G.P.

Then, 2/a = b/2 = r ...(2)

From equation (1), we get:

b - a = 4 - 4a

a - 4a + b = 4

(1 - 4 + 1)r = 4 - a

-3r = 4 - a

r = (a - 4)/3

From equation (2), we get:

b = 2r/a

b = 2(a - 4)/(3a)

Substituting the value of b in equation (1), we get:

(a - 4)/(3a) - a = d

(a - 4) - 3a^2 d = 0

3a^2 - a + 4d = 0

Using the formula for the sum of n terms of an H.P., we get:

a, 1, b are in H.P. with sum 3/(a + 1/b) = 3/(a + 2/(a - 4)/3) = (3a(a - 4))/(4a^2 - 4a - 24)

Simplifying, we get:

3(a - 4)/(4a - 6) = 3(1 - 4/a)/(4 - 6/a)

Therefore, a, 1, b are in H.P.

Hence, the correct option is (b).

Chapter doubts & questions for Sequence and Series - Online MCQ Tests for Commerce 2025 is part of Commerce exam preparation. The chapters have been prepared according to the Commerce exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Commerce 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Sequence and Series - Online MCQ Tests for Commerce in English & Hindi are available as part of Commerce exam. Download more important topics, notes, lectures and mock test series for Commerce Exam by signing up for free.

Top Courses Commerce

© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup