Practice Questions on Arithmetic Progression (Basic)

An arithmetic progression (AP), also known as an arithmetic sequence, is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is called the "common difference."

Practice Questions

Question 1: Find the common difference from the given AP: 1, 6, 11, 16, 21.

Solution:

To find common difference we need to subtract consecutive terms

d = a_n - a_n-1

d = 6 - 1 = 5
So, common difference of the given AP is 5.

Question 2: If the 4th term of an AP is 8 with a common difference of 2. Find out the arithmetic progression up to 8 terms.

Solution:

Given that, the fourth term, a4 is 8 and the common difference is 2,

So the fourth term can be written as,
a + (4 - 1) × 2 = 8  [a = first term]
= a + 3 × 2 = 8
= a = 8 - 3 × 2
= a = 8 - 6
= a = 2

So the first term a₁ is 2,
Now, a₂ = a₁ + 2 = 2 + 2 = 4 similarly the series upto 8 term is = 2, 4, 6, 8, 10, 12, 14, 16.

Question 3: Find the 40th term in the 2, 5, 8, 11, 14,... sequence

Solution:

a = 2
d = 5 - 2 = 3
n = 40

a_n = a + ( n - 1 )d
a₄₀ = 2 + ( 40 - 1 )3
= 2 + 117
a₄₀ = 119

Question 4: Find the 9th term in the 1/3, 2/3, 1, 4/3... sequence

Solution:

a = 1/3
d = 2/3 - 1/3 = 1/3 
n = 9

a_n = a + ( n -1 )d

a₉ = 1/3 + ( 9 - 1 )(1/3)
= 1/3 + 8/3
= 9/3 = 3

Question 5: Find the 15th term of the sequence 9, 7, 5, 3, 1,...

Solution:

a = 9
d = 7 - 9 = 5 - 7 = -2
n = 15

a₁₅ = a + ( n - 1 ) × d
= 9 + ( 15 - 1 ) × (-2)
= 9 + ( 14 ) × ( -2 )
= 9 - 28
a₁₅ = -19

Hence, 15th term in the sequence 9,7,5,3,1,... is -19

Question 6: Find the general term of the series 4,7, 10,13......

Solution:

In the given AP, we have
a₁ = 4, d = 3

So, general term of the given AP is
a_n = a₁ + (n - 1)d
a_n = 4 + (n - 1)3
a_n = 4 + 3n - 3
a_n = 3n + 1

General term = 3n + 1.

Question 7: Find the 50th term of the sequence 293, 290, 287, 284, 281,...

Solution:

a = 293
d = 290 - 293 = 287 - 290 = -3
n = 50

a₅₀  = a + (n - 1) × d
= 293 + ( 50 - 1 ) × (-3)
= 293 + ( 49 ) × ( -3 )
= 293 - 147
a₅₀ = 146

Hence, 50th term in the sequence 293, 290, 287, 284, 281,... is 146

Question 8: Find the 100th term of the sequence 100, 50, 0, -50, -100,...

Solution:

a = 100
d = 50 - 100 = 0 - 50 = -50
n = 100

a₁₀₀  = a + ( n - 1 ) × d
= 100 + ( 100 - 1 ) × (-50)
= 100 + ( 99 ) × ( -50 )
= 100 - 4950
a₁₀₀ = -4850

Hence, 100th term in the sequence 100, 50, 0, -50, -100,... is -4850

Question 9: What is the sum of the first 20 odd natural numbers?

Solution.

First odd natural number is 1
Common Difference is 2

So, 20^th term is

a₂₀ = a₁ + (n-1)d

a₂₀ = 1 + (20-1)2
a₂₀ = 1 + 38
a₂₀ = 39
So,

Sum = n/2(a₁ + a₂₀)

Sum = 20/2(1 + 39)
Sum = 10 × 40
Sum = 400

So, sum of first 20 odd numbers = 400

Question 10: Which term of the arithmetic progression 30, 27, 24,.... is 0?

Solution:

In the given AP,

a₁ = 30
d = -3
a_n = 0

Now, to find n

a_n = a₁ + (n-1)d

0 = 30 + (n - 1)(-3)
0 = 30 - 3n + 3
3n = 33
n = 11

So, the 11th term is 0

Question 11: Find the first negative term of the arithmetic progression 36, 30, 24,.....?

Solution:

In the given AP, we have

a₁ = 36
d = -6

To find the first negative term,

36 + (n−1)(−6) < 0
36 + 6 - 6n < 0
42 - 6n < 0
-6n < - 42
n > 7

So, the 8th term is the first negative term of the given AP.

Question 12: If the common difference of an A.P. is 4, then a₂₀ - a₁₅ is?

Solution:

So, the given common difference = 4.

Now, we want to calculate a20 - a15,
So,
a₂₀ = a + 19d
a₁₅ = a + 14d

Then,
a₂₀ - a₁₅ = a+ 19d - a - 14d
a₂₀ - a₁₅ = 5d
a₂₀ - a₁₅ = 5 × 4 = 20

So the value of a₂₀ - a₁₅ is 20.

Question 13: Which term of the arithmetic progression 3, 8, 13, ..... is 55 more than its 20th term?

Solution:

In the given AP, we have following values

a = 3
d = 5

Let the unknown term is n term,

Now,
a_n = a₂₀ + 55
a₁ + (n - 1)d = a1 + 19d + 55
3 + (n-1)d = 3 + 19×5 + 55
5n - 5 = 95 + 55
5n = 155
n = 31

So, the 31st term is 55 more than the 20th term.

Question 14: Determine the sum of the first 18 terms, of the arithmetic progression 12b, 8b, 4b,......

Solution:

In the given AP we have,

a₁ = 12b

d = -4b

So, the sum of first 18 terms is

Sum = n/2[2a₁ + (n - 1)d]
Sum = 9[2 × 12b + (18 - 1)(-4b)]
Sum = 9[24b - 68b]
Sum = 9 × (-44b)
Sum = -396b

So sum of first 18 terms is -396b.

Question 15: If the 3rd term of an arithmetic sequence is 12 and the 7th term is 24, what is the first term?

Solution:

The nth term : an = a+ (n-1) d

a₃ = 12 = a + 2d - (1)
a₇ = 24 = a + 6d - (2)

Subtract (1) from (2), (2) - (1) :

a₇ - a3 = 24 - 12 = 12 = a + 6d - a - 2d = 6d - 2d = 4d
12 = 4d
d = 12 / 4 = 3

Put value of d in (1) :
12 = a + 2(3)
a = 12 - 6 = 6

Hence the 1st term of arithmetic sequence is 6.

Practice Questions

Question 1: In an AP, if d = 8, n = 10, and a_n = 80, then Find the value of a.

Question 2: Find the 19th term of the AP: 5, 11, 17, 23, ...?

Question 3: What is the sum of first 30 even natural numbers?

Question 4: Which term of the arithmetic progression 13, 18, 23, ..... is 80 more than its 20th term?

Question 5: If the sum of first p terms of an A.P., is ap² + bp, find its common difference.

Question 6: The first term of an A.P. is p and its common difference is q. Find its 10th term

Question 7: If seven times the 7th term of an A.P. is equal to eleven times the 11th term, then what will be its 18th term?

Question 8: How many terms of the A.P. 18, 16, 14, … … be taken so that their sum is zero?

Question 9: 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

Question 10: Find the First Term Given a Condition:

• The 8th term of an AP is zero, and the 12th term is 20. Find the first term and the common difference.
• If the 5th term of an AP is 15 and the sum of the first 10 terms is 120, find the first term.

Answer key

1. a = 8
2. 19th term of the AP: 113
3. Sum of first 30 even natural numbers: 930
4. 36th term is 80 more than the 20th term.
5. Common difference d = 2a
6. 10th term of the AP: p + 9q
7. 18th term is 34d
8. 19 terms should be taken.
9. Proof: 25th term = 3 × 11th term
10. Middle term: 9
    Middle term: 27