The Arithmetic Progression questions and answers are given here for students to help them better grasp the concept. Arithmetic Progressions (AP) is one of the most important concepts in Maths and it is included in higher education. NCERT standards will be followed for preparing these questions. The problems given here will cover both the basics and more complex topics, for students of all levels. Students can practise the arithmetic progression questions, and can then cross verify their answers with the provided solutions. To learn more about Arithmetic progression, click here.
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Arithmetic Progression Definition: An arithmetic progression (AP) is defined as a sequence in which the differences between every two consecutive terms are the same. For example, the sequence 2, 4, 6, 8, … is an arithmetic progression, as it follows a pattern where each term in the sequence is obtained by adding 2 to its previous term. |
Here, we have provided the Arithmetic Progression questions and answers with complete explanations.
Arithmetic Progression Questions with Solutions
1. Find the common difference for the following AP: 10, 20, 30, 40, 50.
Solution:
Given AP: 10, 20, 30, 40, 50
Common difference:
d = 20 – 10 = 10
d = 30 – 20 = 10
d = 40 – 30 = 10
d = 50 – 40 = 10.
Hence, the common difference for the sequence, 10, 20, 30, 40, 50 is 10.
2. Is a, 2a, 3a, 4a, … an arithmetic progression?
Solution:
Given sequence: a, 2a, 3a, 4a, …
To check whether the given sequence is AP or not, we have to find the common difference.
Hence, d = 2a – a = a
d = 3a – 2a = a
d = 4a – 3a = a
Hence, the common difference is “a”.
Therefore, the sequence a, 2a, 3a, 4a,… is an arithmetic progression.
3. Prove that 7, 11, 15, 19, 23 is an AP.
Solution:
Given sequence: 7, 11, 15, 19, 23.
To prove that the sequence is AP, find the common difference between two consecutive terms.
d = 11 – 7 = 4
d = 15 – 11 = 4
d = 19 – 15 = 4
d = 23 – 19 = 4
Hence, 7, 11, 15, 19, 23 is an AP with a common difference of 4.
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AP Formulas: If “a” is the first term and “d” is the common difference between two consecutive terms in the sequence a, a+d, a+2d, a+3d, +…., the formulas used to calculate the nth term and the sum of first n terms are given as follows: The nth term of an AP: a+(n-1)d Sum of first “n” terms of an AP = (n/2)[2a+(n-1)d] Also, read: Sum of n terms. |
4. The sequence 28, 22, x, y, 4 is an AP. Find the values of x and y.
Solution:
Given AP: 28, 22, x, y, 4
Here, first term, a = 28
Common difference, d = 22 – 28 = -6
Hence, x = 22 – 6 = 16
y = 16 – 6 = 10.
Hence, the values of x and y are 16 and 10, respectively.
5. What is the nth term of an AP 9, 13, 17, 21, 25, …?
Solution:
Given AP: 9, 13, 17, 21, 25, …
Here, a = 9
d = 13 – 9 = 4
The formula to find the nth term of an AP is a + (n-1)d.
The nth term of an AP = 9 + (n-1)4
= 9 + 4n -4
= 4n + 5.
Hence, the nth term of AP 9, 13, 17, 21, 25, … is 4n+5.
6. Find the 5th term of the arithmetic progression 1, 4, 7, ….
Solution:
Given AP: 1, 4, 7, …
a = 1
d = 4 – 1 = 3
n = 5
As we know,
The nth term of AP = a + (n-1)d
Hence, 5th term of AP = 1 + (5-1)3
= 1+(4)3
= 1 + 12
= 13.
Hence, the 5th term of AP is 13.
Also, check: Arithmetic Progression Class 10 Notes.
7. Find the 17th term of AP 4, 9, 14, …
Solution:
Given AP: 4, 9, 14, …
Here, a = 4
d = 9 – 4 = 5
n = 17.
Now, substitute the values in the formula a+(n-1)d,
17th term of AP = 4+(17-1)5
= 4+16(5)
= 4+80 = 84
Hence, the 17th term of AP is 84.
8. If the first, second and last terms of the AP are 5, 9, 101, respectively, find the total number of terms in the AP.
Solution:
Given: First term, a = 5
Common difference, d = 9 – 5 = 4
Last term, an = 101
Now, we have to find the value of “n”.
Hence, an = a+(n-1)d
Substituting the values, we get
5+(n-1)4 = 101
5+4n-4 = 101
4n+1= 101
4n = 100
n=100/4 = 25
Hence, the number of terms in the AP is 25.
9. What is the general term of the series, 4, 7, 10, 13, …?
Solution:
Given sequence is 4, 7, 10, 13, …
Here, a = 4
d = 7-4 = 3
The general term of an AP is:
an = a+(n-1)d
an = 4 +(n-1)3
an = 4 + 3n-3
an = 3n+1
Therefore, the general term of the series 4, 7, 10, 13 is 3n+1.
10. Which term of AP 27, 24, 21, … is 0?
Solution:
Given AP: 27, 24, 21, …
Here, a = 27
d = 24 – 27 = -3.
Also given that an = 0
Now, we have to find the value of n.
Hence, an = a+(n-1)d
0 = 27 +(n-1)(-3)
0 = 27 -3n +3
0 = 30 – 3n
3n = 30
Hence, n = 10
Therefore, the 10th term of AP is 0.
Practice Questions
- The sequence 12b, 8b, 4b is in AP. Find the sum of the first 18 terms.
- The first three terms of a sequence are 8, y, 18. Find the value of y so that the sequence becomes an Arithmetic progression.
- In an Arithmetic progression, the ratio of the 7th term to the 10th term is -1. If the 16th term is -15, find the 3rd term.
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