Progression MCQ Quiz - Objective Question with Answer for Progression - Download Free PDF
Last updated on Jun 6, 2025
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'?
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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 - n4 = -1 + 3d - ( -8 + 3d )
m4 - n4 = 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?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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.
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 \(99\frac{11}{99}+99\frac{13}{99}+99\frac{15}{99}+\ldots+99\frac{67}{99}\) ?
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:
\(1\over99\)[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.
Answer (Detailed Solution Below)
Progression Question 10 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 = \(\frac{n(a+l)}{2}\)
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?
Answer (Detailed Solution Below)
Progression Question 11 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 12 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
\(S = \dfrac{n}{2}[2a + (n-1)d]\)
\(S = \dfrac{n}{2}( a + l)\)
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, _____?
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.