Progression MCQ Quiz - Objective Question with Answer for Progression - Download Free PDF

Last updated on Jun 6, 2025

Testbook provides Progression MCQ Quiz with logical and easy explanations granting you effortless practice of this highly scoring topic. Detailed solutions for all Progression Objective questions have been provided so that candidates can learn the right strategies and shortcuts to approach a question while solving it in minimum time. These Progression Question Answers will help you practice every nook and corner of your approach to this section.

Latest Progression MCQ Objective Questions

Progression Question 1:

If N1 = 3 + 33 + 333 + .... + 333333 and N2 = 4 + 44 + 444 + .... + 4444444, then what is the sum of digits of 'N1 + N2'?

  1. 33
  2. 31
  3. 35
  4. 29
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 31

Progression Question 1 Detailed Solution

Given:

N1 = 3 + 33 + 333 + .... + 333333 and

N2 = 4 + 44 + 444 + .... + 4444444

Calculations:

N1 = 3 + 33 + 333 + 3333 + 33333 + 333333

⇒ N1 = 3 × (1 + 11 + 111 + 1111 + 11111 + 111111)

Sum of series: 1 + 11 + 111 + 1111 + 11111 + 111111 = 123456

⇒ N1 = 3 × 123456 = 370368

N2 = 4 + 44 + 444 + 4444 + 44444 + 444444 + 4444444

⇒ N2 = 4 × (1 + 11 + 111 + 1111 + 11111 + 111111 + 1111111)

Sum of series: 1 + 11 + 111 + 1111 + 11111 + 111111 + 1111111 = 1234567

⇒ N2 = 4 × 1234567 = 4938268

N1 + N2 = 370368 + 4938268

⇒ N1 + N2 = 5308636

Sum of digits of 5308636 ⇒ 5 + 3 + 0 + 8 + 6 + 3 + 6 = 31

The correct answer is option 2.

Progression Question 2:

The geometric mean of vitiates 32, 4, 8, X, 2 is 8. What is the value of vitiate X?

  1. 2
  2. 4
  3. 8
  4. 16
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : 16

Progression Question 2 Detailed Solution

Concept Used:

The Geometric Mean (G.M) of a series containing n observations is the nth root of the product of the values.

\(\begin{array}{l}G. M = \sqrt[n]{x_{1}× x_{2}× …x_{n}}\end{array}\)

Calculation:

Using the above formula -

⇒ 85 = 32 × 4 × 8 × X × 2

⇒ X =  \(\frac{8^{5}}{32\times 4\times 8\times 2}\)= 16

∴ The correct answer is 16

Progression Question 3:

Two APs have the same common difference. The first term of one of these is –1 and that of the other is – 8. Then the difference between their 4th terms is

  1. -1
  2. -8
  3. 7
  4. -9
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 7

Progression Question 3 Detailed Solution

Given:

Two APs have the same common difference. The first term of one of these is –1 and that of the other is – 8.

Concept:

For any given AP such that its first term is ' a ' and the common difference is ' d '

an = a + (n - 1)d

Solution:

According to the question, the common difference of the two APs is the same,

Let's say the common difference is ' d '.

For the first AP

The first term is -1 and the common difference is ' d '

The fourth term will be,

m4 = -1 + (4 - 1)d = -1 + 3d

For the second AP

The first term is -8 and the common difference is ' d '

The fourth term will be,

n4 = -8 + (4 - 1)d = -8 + 3d

The difference between the 4th term is as follows,

m4 - n= -1 + 3d - ( -8 + 3d )

m4 - n= 7

Hence, option 3 is correct.

Progression Question 4:

The fifth term and the eighth term of a geometric progression are 27 and 729 respectively. What is its 11th term? 

  1. 19683
  2. 59049
  3. 6561
  4. 27729

Answer (Detailed Solution Below)

Option 1 : 19683

Progression Question 4 Detailed Solution

Given:

The fifth term of a geometric progression (GP) = 27

The eighth term of the GP = 729

Formula used:

General term of GP: Tn = a × rn-1

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

Calculation:

Fifth term: T5 = a × r4 = 27

Eighth term: T8 = a × r7 = 729

Dividing both equations:

⇒ (a × r7) / (a × r4) = 729 / 27

⇒ r3 = 27

⇒ r = 3

Substitute r = 3 into T5:

⇒ a × 34 = 27

⇒ a × 81 = 27 

⇒ a = 27 / 81 = 1/3

T11 = (1/3) × 310

⇒ T11 = (1/3) × 59049

⇒ T11 = 19683

∴ The correct answer is option (1).

Progression Question 5:

The sum of the 9th term and 12th term of the series \(\frac{1}{7}, \frac{1}{11}, \frac{1}{15}\) ..... is

  1. \(\frac{1}{121}\)
  2. \(\frac{10}{663}\)
  3. \(\frac{10}{221}\)
  4. \(\frac{98}{2365}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{10}{221}\)

Progression Question 5 Detailed Solution

Given:

Series: 1/7, 1/11, 1/15, ...

Formula used:

This is an AP in the denominators: 7, 11, 15, ...

General term: Tn = 1 / [4n + 3]

Calculation:

9th term = 1 / (4×9 + 3) = 1 / 39

12th term = 1 / (4×12 + 3) = 1 / 51

Sum = 1/39 + 1/51

LCM of 39 and 51 = 663

1/39 = 17/663, 1/51 = 13/663

Sum = (17 + 13)/663 = 30/663

Simplify: 30 ÷ 3 = 10, 663 ÷ 3 = 221

∴ The sum is 10/221

Top Progression MCQ Objective Questions

Find the sum of 3 +32 + 33 +...+ 38.

  1. 6561
  2. 6560
  3. 9840
  4. 3280

Answer (Detailed Solution Below)

Option 3 : 9840

Progression Question 6 Detailed Solution

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Formula used:

Sum of geometric progression (Sn) = {a × (rn - 1)}/(r - 1)

Where, a = first term ; r = common ratio ; n = number of term

Calculation:

3 +32 + 33 +...+ 38.

Here, a = 3 ; r = 3 ; n = 8

Sum of the series (S8) = {a × (r8 - 1)}/(r - 1)

⇒ {3 × (38 - 1)}/(3 - 1)

⇒ (3 × 6560)/2 = 3280 × 3 

⇒ 9840

∴ The correct answer is 9840.

What will come in place of the question mark (?) in the following question?

13 + 23 + 33 + ……+ 93 = ?

  1. 477
  2. 565
  3. 675
  4. 776

Answer (Detailed Solution Below)

Option 1 : 477

Progression Question 7 Detailed Solution

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Given:

13 + 23 + …….. + 93 = ?

Formula:

Sn = n/2 [a + l]

Tn = a + (n – 1)d

n = number of term

a = first term

d = common difference

l = last term

Calculation:

a = 13

d = 23 – 13 = 10

Tn = [a + (n – 1)d]

⇒ 93 = 13 + (n – 1) × 10

⇒ (n – 1) × 10 = 93 – 13

⇒ (n – 1) = 80/10

⇒ n = 8 + 1

⇒ n = 9

S9 = 9/2 × [13 + 93]

= 9/2 × 106

= 9 × 53

= 477

How many three digit numbers are divisible by 6?

  1. 196
  2. 149
  3. 150
  4. 151

Answer (Detailed Solution Below)

Option 3 : 150

Progression Question 8 Detailed Solution

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Formula used:

an = a + (n – 1)d

Here, a → first term, n → Total number, d → common difference, an → nth term

Calculation:

First three-digit number divisible by 6, (a) = 102 

Last three-digit number divisible by 6, (an) = 996

Common difference, (d) = 6 (Since the numbers are divisible by 6)

Now, an = a + (n – 1)d

⇒ 996 = 102 + (n – 1) × 6 

⇒ 996 – 102 = (n – 1) × 6

⇒ 894 = (n – 1) × 6

⇒ 149 = (n – 1)

⇒ n = 150

∴ The total three digit number divisible by 6 is 150

What is the value of \(99\frac{11}{99}+99\frac{13}{99}+99\frac{15}{99}+\ldots+99\frac{67}{99}\) ?

  1. 94220/33
  2. 95120/33
  3. 97120/33
  4. 96220/33

Answer (Detailed Solution Below)

Option 2 : 95120/33

Progression Question 9 Detailed Solution

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Formula used:

Sn = [n x (a + an) ] /2

an = a + (n-1)d

d = difference

a = initial term

an = last term

n = number of terms

Sn = Sum of n terms

Solution:

The series can be written as:

 \(1\over99\)[99x99+11 + 99x99+13 + ... + 99x99+67]

\(1\over99\) [9812 + 9814 + 9816+ ... + 9868]

 Now, our series is, 9812, 9814,...,9868.

a = 9812

an = 9868

d = 9814 - 9812 = 2

9868= 9812 + (n-1) x 2

n - 1 = 56/2 = 28

n = 29

Sn = 29 x (9812 + 9868) / 2 = (29 x 19680)/2 = 570720/2 = 285360

Hence, the sum of the series = 285360/99 = 95120/33

If the sum of all even numbers from 21 to 199 is added to 11 observations whose mean value is n, then the mean value of new set becomes 99. Find the value of n.

  1. 10
  2. 11
  3. 100
  4. 89

Answer (Detailed Solution Below)

Option 1 : 10

Progression Question 10 Detailed Solution

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Given:

The sum of even numbers from 21 to 199 is added to 11 observations whose mean value is n.

The mean of the new set of numbers = 99.

Formula used:

(1) Sum of n numbers in A.P.

S = \(\frac{n(a+l)}{2}\)

Where, 

a, is the value of the first term

l, is the value of the last term

n, is the number of terms

S, is the sum of n numbers in A.P

(2) The value of the last term in A.P.

l = a + (n - 1)d

Where, 

a, is the value of the first term

d, is the common difference between two terms

n, is the number of terms

l, is the value of the last term

Calculation:

Let n be the number of even terms between 21 to 199.

The value of the first even number (between 21 to 199), a = 22

The value of the last even number (between 21 to 199), l = 198

The value of the common difference between two even numbers, d = 2

Now,

⇒ 198 = 22 + (n - 1) × 2

⇒ 198 = 22 + (n - 1)2

⇒ 176 = (n - 1)2

⇒ (n - 1) = 88

⇒ n = 89

Now,

Let S be the sum of all even numbers between 21 to 199.

⇒ S = \(\frac{89(22 + 198)}{2}\)

⇒ S = 9790

Now, 

The average of 11 observations = n

The sum of all 11 observations = 11n

According to the question,

⇒ \(\frac{9790+11n}{89+11}\) = 99

⇒ \(\frac{9790+11n}{100}\) = 99

⇒ 9790 + 11n = 9900

⇒ 11n = 110

⇒ n = 10

∴ The required answer is 10.

Additional InformationFormula is used to find the average of numbers when the first and last term is known.

A = \(\frac{a+l}{2}\)

Where, 

a, is the first term of the Arithmetic Progression

l, is the last term of the Arithmetic Progression

A, is the average of Arithmetic Progression from a to l.

Note: The above formula is only applied for Arithmetic Progression.

If the successive terms have a common difference as a non-zero constant, then that sequence can be termed an Arithmetic sequence. 

How many numbers between 300 and 1000 are divisible by 7?

  1. 101
  2. 301
  3. 994
  4. 100

Answer (Detailed Solution Below)

Option 4 : 100

Progression Question 11 Detailed Solution

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Given condition: 

Numbers between 300 and 1000 are divisible by 7.

Concept:

Arithmetic Progression

an = a + (n - 1)d 

Calculation:

The first number that is divisible by 7 (300 - 1000) = 301 

Likewise: 301, 308, 315, 322...........994 

The above series makes an AP,

Where a = 301, Common Difference/d = 308 - 301 = 7 and Last term (an) = 994

⇒ an = a + (n - 1)d

⇒ 994 = 301 + (n - 1)7 

⇒ (994 - 301)/7 = n - 1 

⇒ 693/7 + 1 = n 

⇒ 99 + 1 = n 

⇒ n = 100 

∴ There are 100 numbers between 300 and 1000 which are divisible by 7. 

The sum of the first 20 terms of an arithmetic progression whose first term is 5 and common difference is 4 is _____.

  1. 830
  2. 850
  3. 820
  4. 860

Answer (Detailed Solution Below)

Option 4 : 860

Progression Question 12 Detailed Solution

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Given:

First term 'a' = 5, common difference 'd' = 4

Number of terms 'n' = 20

Concept:

Arithmetic progression:

  • Arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
  • The fixed number is called common difference 'd'.
  • It can be positive, negative or zero.


Formula used:

nth term of AP 

Tn = a + (n - 1)d

The Sum of n terms of AP is given by

\(S = \dfrac{n}{2}[2a + (n-1)d]\)

\(S = \dfrac{n}{2}( a + l)\)

Where, 

a = first term of AP, d = common difference, l = last term 

Calculation:

We know that sum of n terms of AP is given by

\(S = \dfrac{n}{2}[2a + (n-1)d]\)

\(⇒ S = \dfrac{20}{2}[2× 5 + (20-1)× 4]\)

⇒ S = 10(10 + 76)

⇒ S = 860

Hence, the sum of 20 terms given AP will be 860.


We know that nth term of AP is given by

Tn = a + (n - 1)d

If l is the 20th term (last term) of AP, then

l = 5 + (20 - 1) × 4 = 81

So the sum of AP

\(S = \dfrac{n}{2}( a + l)\)

\(⇒ S = \dfrac{20}{2}(5 + 81)\)

⇒ S = 860

What will be the 10th term of the arithmetic progression 2, 7, 12, _____?

  1. 245
  2. 243
  3. 297
  4. 47

Answer (Detailed Solution Below)

Option 4 : 47

Progression Question 13 Detailed Solution

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Given

2, 7, 12, ____________

Concept used

Tn = a + (n - 1)d

Where a = first term, n = number of terms and d = difference

Calculation

in the given series

a = 2

d = 7 - 2 = 5

T10 = 2 + (10 - 1) 5

T10 = 2 + 45

T10 = 47

Tenth term = 47

For which value of k; the series 2, 3 + k and 6 are in A.P.?

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

Answer (Detailed Solution Below)

Option 3 : 1

Progression Question 14 Detailed Solution

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Given: 

For a value of k; 2, 3 + k and 6 are to be in A.P 

Concept:

According to Arithmetic progression, a2 - a= a3 - a

where a1 ,a2 ,aare 1st, 2nd and 3rd term of any A.P.

Calculation:

a1 = 2, a= k + 3, a3 = 6 are three consecutive terms of an A.P.

According to Arithmetic progression, a2 - a= a3 - a

(k + 3) – 2 = 6 – (k + 3)

⇒ k + 3 - 8 + k + 3 = 0

⇒ 2k = 2

After solving, we get k = 1

What will be the sum of 3 + 7 + 11 + 15 + 19 + ... upto 80 terms ?

  1. 12880
  2. 12400
  3. 25760
  4. 24800

Answer (Detailed Solution Below)

Option 1 : 12880

Progression Question 15 Detailed Solution

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Given:

An AP is given
3 + 7 + 11 + 15 + 19 + ... upto 80 terms 

Formula used:

Sum of nth term of an AP

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

where, 

'n' is Number of terms, 'a' is First term, 'd' is Common difference

Calculations:

According to the question, we have

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

where, a = 3, n = 80, d = 7 - 3 = 4

Put these values in (1), we get

⇒ S80 = (80/2){2 × 3 + (80 - 1) × 4}

⇒ S80 = 40(6 + 79 × 4)

⇒ S80 = 40 × 322

⇒ S80 = 12,880

∴ The sum of 80th terms of an AP is 12,880.

Alternate Method

nth term = a + (n - 1)d

Here n = 80, a = 3 and d = 4

⇒ 80th term = 3 + (80 - 1)4

⇒ 80th term = 3 + 316

⇒ 80th term = 319

Now, the sum of nth terms of an AP

⇒ Sn = (n/2) × (1st term + Last term)

⇒ S80 = (80/2) × (3 + 319)

⇒ S80 = 40 × 322

⇒ S80 = 12,880

∴ The sum of 80th terms of an AP is 12,880.

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