Arithmetic Progression MCQ Quiz - Objective Question with Answer for Arithmetic Progression - Download Free PDF
Last updated on Jun 3, 2025
Latest Arithmetic Progression MCQ Objective Questions
Arithmetic Progression Question 1:
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
Answer (Detailed Solution Below)
Arithmetic Progression Question 1 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 - n4 = -1 + 3d - ( -8 + 3d )
m4 - n4 = 7
Hence, option 3 is correct.
Arithmetic Progression Question 2:
The sum of the 9th term and 12th term of the series
Answer (Detailed Solution Below)
Arithmetic Progression Question 2 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
Arithmetic Progression Question 3:
The sum of all two digit natural numbers which are divisible by 3 is
Answer (Detailed Solution Below)
Arithmetic Progression Question 3 Detailed Solution
Given:
Two-digit natural numbers divisible by 3
Formula used:
Use Arithmetic Progression (AP)
Sum = n/2 × (first term + last term)
Where: n = number of terms
Calculation:
Smallest 2-digit number divisible by 3 = 12
Largest 2-digit number divisible by 3 = 99
AP: 12, 15, 18, ..., 99
Common difference (d) = 3
To find number of terms (n):
Last term = a + (n - 1)d
99 = 12 + (n - 1) × 3
⇒ 99 - 12 = (n - 1) × 3
⇒ 87 = (n - 1) × 3
⇒ n - 1 = 29
⇒ n = 30
Sum = 30/2 × (12 + 99)
⇒ 15 × 111 = 1665
∴ The sum is 1665
Arithmetic Progression Question 4:
Sum of first 30 terms of the Arithmetic Sequence 6, 12, 18,...., is 2790. Then, the sum of first 30 terms of the Arithmetic Sequence 13, 19, 25,... is
Answer (Detailed Solution Below)
Arithmetic Progression Question 4 Detailed Solution
Given:
Arithmetic Sequence 13, 19, 25, ...
Formula used:
Sum of first n terms of an Arithmetic Sequence:
Where:
Sn = Sum of n terms
n = Number of terms
a = First term
d = Common difference
Calculation:
The sequence 13, 19, 25, ...:
a = 13, d = 6, n = 30
S30 =
⇒ S30 =
⇒ S30 =
⇒ S30 = 15 × 200
⇒ S30 = 3000
∴ The correct answer is option (1).
Arithmetic Progression Question 5:
If the arithmetic progression whose common difference is non-zero, the sum of first 3n terms is equal to the sum of next n terms. Then, find the ratio of the sum of the first 2 n terms to the sum of next 2 n terms:
Answer (Detailed Solution Below)
Arithmetic Progression Question 5 Detailed Solution
Given:
The nth term of a sequence with first term a and common difference d is:
an = a + (n − 1)d
The sum of n terms of an AP is:
Sn = (n/2) × [2a + (n − 1)d]
Formula used:
Sum of first 3n terms: S3n = (3n/2) × [2a + (3n − 1)d]
Sum of next n terms = Sn = S4n − S3n
Given: S3n = Sn ⇒ 2S3n = S4n
Calculations:
Substitute sum formulas:
2 × (3n/2) × [2a + (3n − 1)d] = (4n/2) × [2a + (4n − 1)d]
Cancel 2:
3n × [2a + (3n − 1)d] = 2n × [2a + (4n − 1)d]
Expand both sides:
3n(2a + 3nd − d) = 2n(2a + 4nd − d)
6a + 9nd − 3d = 4a + 8nd − 2d
Now simplify:
6a − 4a = 8nd − 9nd + (−2d + 3d)
2a = d(1 − n) ......... (i)
Find the ratio of sum of first 2n terms to next 2n terms:
Required Ratio = S2n / (S4n − S2n)
S2n = (2n/2) × [2a + (2n − 1)d] = n × [2a + (2n − 1)d]
S4n = (4n/2) × [2a + (4n − 1)d] = 2n × [2a + (4n − 1)d]
So, S4n − S2n = 2n[2a + (4n − 1)d] − n[2a + (2n − 1)d]
= n[4a + 8nd − 2d − 2a − 2nd + d]
= n[2a + 6nd − d]
Now substitute 2a from equation (i):
2a = (1 − n)d
S2n = n[(1 − n)d + (2n − 1)d] = n × nd
S4n − S2n = n × 5nd
Final Ratio:
S2n / (S4n − S2n) = nd / 5nd = 1 : 5
∴ The correct answer is Option 2: 1 : 5
Top Arithmetic Progression MCQ Objective Questions
What will come in place of the question mark (?) in the following question?
13 + 23 + 33 + ……+ 93 = ?
Answer (Detailed Solution Below)
Arithmetic Progression Question 6 Detailed Solution
Download Solution PDFGiven:
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?
Answer (Detailed Solution Below)
Arithmetic Progression Question 7 Detailed Solution
Download Solution PDFFormula 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
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.
Answer (Detailed Solution Below)
Arithmetic Progression Question 8 Detailed Solution
Download Solution PDFGiven:
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 =
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 =
⇒ S = 9790
Now,
The average of 11 observations = n
The sum of all 11 observations = 11n
According to the question,
⇒
⇒
⇒ 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 =
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?
Answer (Detailed Solution Below)
Arithmetic Progression Question 9 Detailed Solution
Download Solution PDFGiven 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 _____.
Answer (Detailed Solution Below)
Arithmetic Progression Question 10 Detailed Solution
Download Solution PDFGiven:
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
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 = 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 = 860
What will be the 10th term of the arithmetic progression 2, 7, 12, _____?
Answer (Detailed Solution Below)
Arithmetic Progression Question 11 Detailed Solution
Download Solution PDFGiven
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.?
Answer (Detailed Solution Below)
Arithmetic Progression Question 12 Detailed Solution
Download Solution PDFGiven:
For a value of k; 2, 3 + k and 6 are to be in A.P
Concept:
According to Arithmetic progression, a2 - a1 = a3 - a2
where a1 ,a2 ,a3 are 1st, 2nd and 3rd term of any A.P.
Calculation:
a1 = 2, a2 = k + 3, a3 = 6 are three consecutive terms of an A.P.
According to Arithmetic progression, a2 - a1 = a3 - a2
(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 ?
Answer (Detailed Solution Below)
Arithmetic Progression Question 13 Detailed Solution
Download Solution PDFGiven:
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.
If a, b, c are in arithmetic progression then:
Answer (Detailed Solution Below)
Arithmetic Progression Question 14 Detailed Solution
Download Solution PDFConcept used:
Let a, b, c… and so on be our series
As we know the common difference = b – a, c – b.
The common difference is the same in arithmetic progression
b – a = c – b
Calculation:
b – a = c – b
⇒ b + b = c + a
⇒ 2b = c + a
⇒ 2b = a + c
∴ a, b, c are in arithmetic progression then 2b = a + c.
Alternate Method
Let number be 1, 2, 3 which are in AP
Only one option satisfied the equation
2(2 ) 1 + 3 =so 2b = a + c is correct option
Which of the following form an AP?
Answer (Detailed Solution Below)
Arithmetic Progression Question 15 Detailed Solution
Download Solution PDFConcept used:
A sequence of numbers is called an arithmetic progression if the difference between any two consecutive terms is always same.
Calculation:
In option 1, we have
Difference between 1st and 2nd term = 1 - 1 = 0
And, difference between 2nd and 3rd term = 2 - 1 = 1
Here, the difference between any two consecutive terms is not equal.
So, option 1 is not in AP.
In option 2, we have
Difference between 1st and 2nd term = 0.33 - 0.3 = 0.03
And, difference between 2nd and 3rd term = 0.333 - 0.33 = 0.003
Here, the difference between any two consecutive terms is not equal.
So, option 2 is not in AP.
Now, in option 3, we have
Difference between 1st and 2nd term = 2 - √2
⇒ 2 - 1.414
⇒ 0.586
And, difference between 2nd and 3rd term = 2√2 - 2
⇒ 2(1.414 - 1)
⇒ 2 × 0.414
⇒ 0.825
Here, the difference between any two consecutive terms is not equal.
So, option 3 is not in AP.
In option 4, we have
Difference between 1st and 2nd term = (3 + √2) - 3
⇒ √2
And, difference between 2nd and 3rd term = (3 + 2√2) - (3 + √2)
⇒ √2
And, difference between 2nd and 3rd term = (3 + 2√2) - (3 + √2)
⇒ √2
And, difference between 3nd and 4th term = (3 + 3√2) - (3 + 2√2)
⇒ √2
∴ Option 4 is in AP.
Mistake Points
In this question go through all the option and check with every term. Don't check for only first three options,