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

Concept:  

A.P having Common ratio d = a_n - a_{n -1} 

The first term is a 

a_n  = a + (n - 1)d

Solution:

For the given A.P 

a = 1/p d = ((1 - p)/p) -(1/p) = -1 

4th term for the A.P = a + (4 -1) d = a + 3d = 1/p - 3 

Hence, The correct option is 4

Given:

• The 12th term of the A.P. is 19.75.
• The 8th term of the A.P. is 15.75.
• We need to find the sum of the first 16 terms (n = 16).

Formula used:

• nth term of an A.P.: Tn = a + (n - 1) × d
• Sum of the first n terms of an A.P.: Sn = (n/2) × [2a + (n - 1) × d]

Calculation:

From the given information, we can set up two equations:

12th term: a + (12 - 1)d = 19.75

⇒ a + 11d = 19.75 ---- (1)

8th term: a + (8 - 1)d = 15.75

⇒ a + 7d = 15.75 ---- (2)

Now, subtract equation (2) from equation (1) to find 'd':

(a + 11d) - (a + 7d) = 19.75 - 15.75

a + 11d - a - 7d = 4

4d = 4

d = 4 / 4

d = 1

Substitute the value of 'd' (d=1) into equation (2) to find 'a':

a + 7(1) = 15.75

a + 7 = 15.75

a = 15.75 - 7

a = 8.75

Now we have the first term (a = 8.75) and the common difference (d = 1).

We need to find the sum of the first 16 terms (n = 16).

Use the formula for the sum of n terms (Sn):

Sn = (n/2) × [2a + (n - 1) × d]

S16 = (16/2) × [2 × 8.75 + (16 - 1) × 1]

S16 = 8 × [17.5 + 15 × 1]

S16 = 8 × [17.5 + 15]

S16 = 8 × 32.5

S16 = 260

∴ The sum of the first 16 terms of the A.P. is 260.

Given:

The 10th term of the sequence a, a - b, a - 2b, a - 3b,......is 20. The 20th term is 10.

Calculation:

10^th term = a - 9b = 20    --(1)

20^th 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, x^th term = a - (x - 1)b

⇒ 29 - (x - 1) (1) 

⇒ 29 - x + 1 = 30 - x

∴ The xth term of the sequence is 30 - x.

Given:

Given series: a, a + d, a + 2d, ....., a + 2md

Concept used:

1. A, A + D, A + 2D, ....., N^th term

In case of the aforementioned series,

Sum of N terms = (First Term + Last Term)/2 × N

N^th 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).

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.

Formula used:

Sum of geometric progression (S_n) = {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.

Given:

13 + 23 + …….. + 93 = ?

Formula:

S_n = n/2 [a + l]

T_n = a + (n – 1)d

n = number of term

a = first term

d = common difference

l = last term

Calculation:

a = 13

d = 23 – 13 = 10

T_n = [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

Formula used:

an = a + (n – 1)d

Here, a → first term, n → Total number, d → common difference, an → n^th 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, a_n = 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

Formula used:

S_n = [n x (a + a_n) ] /2

a_n = a + (n-1)d

d = difference

a = initial term

a_n = last term

n = number of terms

S_n = Sum of n terms

Solution:

The series can be written as:

 [99x99+11 + 99x99+13 + ... + 99x99+67]

=  [9812 + 9814 + 9816+ ... + 9868]

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

a = 9812

a_n = 9868

d = 9814 - 9812 = 2

9868= 9812 + (n-1) x 2

n - 1 = 56/2 = 28

n = 29

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

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

Given condition: 

Numbers between 300 and 1000 are divisible by 7.

Concept:

Arithmetic Progression

a_n = 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. 

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:

n^th term of AP 

T_n = 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 n^th 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

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 = 

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,

⇒  = 99

⇒  = 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 = 

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. 

Given: 

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 1^st, 2^nd and 3^rd 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

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

Given:

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

Formula used:

Sum of n^th term of an AP

S_n = (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 80^th terms of an AP is 12,880.

Alternate Method

n^th 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 n^th terms of an AP

⇒ S_n = (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.