Progression MCQ Quiz - Objective Question with Answer for Progression - Download Free PDF
Last updated on Jun 25, 2025
Latest Progression MCQ Objective Questions
Progression Question 1:
If the 10th term of the sequence a, a - b, a - 2b, a - 3b,......is 20 and the 20th term is 10, then the xth term of the series is
Answer (Detailed Solution Below)
Progression Question 1 Detailed Solution
Given:
The 10th term of the sequence a, a - b, a - 2b, a - 3b,......is 20. The 20th term is 10.
Calculation:
10th term = a - 9b = 20 --(1)
20th term = a - 19b = 10 ---(2)
On subtracting equation (2) from (1):
a - 9b - (a - 19b) = 20 - 10
⇒ a - 9b - a + 19b = 10
⇒ 10b = 10
⇒ b = 1
From equation (1):
a - 9 (1) = 20
⇒ a = 20 + 9 = 29
Now, xth term = a - (x - 1)b
⇒ 29 - (x - 1) (1)
⇒ 29 - x + 1 = 30 - x
∴ The xth term of the sequence is 30 - x.
Progression Question 2:
The arithmetic mean of data with observations a, a + d, a + 2d, ....., a + 2md is
Answer (Detailed Solution Below)
Progression Question 2 Detailed Solution
Given:
Given series: a, a + d, a + 2d, ....., a + 2md
Concept used:
1. A, A + D, A + 2D, ....., Nth term
In case of the aforementioned series,
Sum of N terms = (First Term + Last Term)/2 × N
Nth term = A + (N - 1)D
2. Total = Mean × Number of entities
Calculation:
Given series is an arithmetic progression where a is the first term and d is the common difference.
(2m + 1)th term of the given series = a + (2m + 1 - 1)d = a + 2md
Sum of the (2m + 1) terms of the given series
⇒ [{a + (a + 2md)} ÷ 2] × (2m + 1)
⇒ (a + md)/(2m + 1)
Mean of the (2m + 1) terms of the given series
⇒ (a + md)/(2m + 1) ÷ (2m + 1)
⇒ (a + md)
∴ The arithmetic mean of data with observations a, a + d, a + 2d, ....., a + 2md is (a + md).
Progression Question 3:
The sum of three numbers in an A.P is -6 and their product is 24. Taking the positive value of the common difference as 'd', find the smallest of the three numbers.
Answer (Detailed Solution Below)
Progression Question 3 Detailed Solution
Given:
Sum = - 6
Product = 24
Calculation:
Let, the 3 numbers be (m - d), m, and (m + d)
⇒ m - d + m + m + d = -6
⇒ 3m = - 6
⇒ m = - 2
⇒ (m - d)(m)(m + d) = 24
⇒ (-2 - d)(-2)(-2 + d) = 24
⇒ (4 + 2d)(d - 2) = 24
⇒ 4d - 8 + 2d2 - 4d = 24
⇒ d2 = 16
⇒ d = ± 4
On taking d = 4 (d = +ve)
⇒ m - d = - 2 - 4 = - 6
⇒ m = - 2
⇒ m + d = -2 + 4 = 2
The number is -6, -2, and 2.
∴ The smallest number is -6.
Progression Question 4:
The sum of the first 10 terms of the sequence 6, 21, 36, 51 .... is
Answer (Detailed Solution Below)
Progression Question 4 Detailed Solution
Explanation:
The sum of the first n terms of any A.P. is given by
Sn = n/ 2[ 2a + (n - 1) × d]
Where a is the first term and d is the common difference of the A.P,
- a = 6
- d = 15
- n = 10
S10 = (10/2) × [2(6) + (10 - 1)(15)]
First, calculate inside the brackets:
2(6) = 12
(10 - 1)(15) = 9 × 15 = 135
Add the results:
12 + 135 = 147
Now, multiply by (10/2):
S10 = 5 × 147 = 735
Therefore, the sum of the first 10 terms of the sequence is 735.
Progression Question 5:
There are four numbers forming a GP in which the third term is greater than the first by 9 and the second term is greater than the fourth by 18. What is the first term?
Answer (Detailed Solution Below)
Progression Question 5 Detailed Solution
Given:
The third term is greater than the first by 9.
The second term is greater than the fourth by 18.
Concept:
nth Term of GP whose first term & common ratio are 'a' & 'r', is given by Tn = arn-1
Calculation:
Let a be the first term and r be the common ratio of the G.P.
a1 = a, a2 = ar, a3 = ar2, a4 = ar3
According to the question
a3 = a1 + 9
⇒ ar2 = a + 9 ....(1)
Now, a2 = a4 + 8
⇒ ar = ar3 + 18 ....(2)
From equation (1) and (2),
⇒ a(r2 - 1) = 9 ....(3)
⇒ ar(1 - r2) = 18 ...(4)
Dividing (4) and (3), we get
⇒
⇒ -r = 2
⇒ r = -2
Substituting the value of r in equation (1), we get
⇒ 4a = a + 9
⇒ 3a = 9
⇒ a = 3
So. the first term of G.P. a1 = a = 3
∴ The required value is 3.
Top Progression MCQ Objective Questions
Find the sum of 3 +32 + 33 +...+ 38.
Answer (Detailed Solution Below)
Progression Question 6 Detailed Solution
Download Solution PDFFormula 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 = ?
Answer (Detailed Solution Below)
Progression Question 7 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)
Progression Question 8 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
What is the value of
Answer (Detailed Solution Below)
Progression Question 9 Detailed Solution
Download Solution PDFFormula 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:
=
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
How many numbers between 300 and 1000 are divisible by 7?
Answer (Detailed Solution Below)
Progression Question 10 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)
Progression Question 11 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
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)
Progression Question 12 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.
What will be the 10th term of the arithmetic progression 2, 7, 12, _____?
Answer (Detailed Solution Below)
Progression Question 13 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)
Progression Question 14 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)
Progression Question 15 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.