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

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 '

a_n = 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 4^th term is as follows,

m₄ - n₄ = -1 + 3d - ( -8 + 3d )

m4 - n4 = 7

Hence, option 3 is correct.

Given:

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

Formula used:

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

General term: T_n = 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

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

Given:

The series is 201, 208, 215, ..., 369

Formula used:

For an arithmetic progression, the n^th term is given by: l = a + (n - 1)d

Where: l = last term, a = first term, n = number of terms and d = common difference

Calculations:

First term (a) = 201 and Common difference (d) = 208 - 201 = 7

Last term (l) = 369

369 = 201 + (n - 1)7

⇒ 369 - 201 = (n - 1)7

⇒ 168 = (n - 1)7

⇒ 168/7 = n - 1

⇒ 24 = n - 1

⇒ n = 24 + 1 = 25

∴ There are 25 terms in the given series.

Given:

Arithmetic Sequence 13, 19, 25, ...

Formula used:

Sum of first n terms of an Arithmetic Sequence:

Where:

S_n = 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).

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

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 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

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: 

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:

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.

Concept 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

Concept 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 1^st and 2^nd term = 1 - 1 = 0

And, difference between 2nd and 3^rd 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 4^th 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,