One of the most crucial concepts for pupils in Class 10 is Arithmetic Progression (AP). It basically establishes the foundation for comprehending sequences, patterns, and series and is a rather fundamental idea in mathematics. In addition to improving students' problem-solving abilities, completing arithmetic progression problems gets them ready for more complex mathematical ideas in the future.
Here are the Arithmetic Progression problems to help students better comprehend the subject. The questions provided here will cover both fundamental and advanced topics, catering to students of all levels. Let's start with the basics, then go on to a few arithmetic progression answers to questions to help you comprehend.
An AP or Arithmetic Progression is a sequence of nos in which each term after the 1st is obtained by adding a constant difference, known as the common difference (d), to the preceding term. The general form of an AP is:
a, a + d, a + 2d, a + 3d,…
Here:
To find the nth term = Tn = A+(n-1)d
To find the sum of n terms = Sn = n/2 × [2a + (n−1)d]
In Class 10, arithmetic progressions are studied to help you identify and organise patterns in numbers or objects for use in everyday life and in maths. It increases self-assurance and problem-solving abilities. Understanding mathematical progressions is crucial for the following reasons:
By mastering these concepts, students gain a solid foundation for advanced topics while applying them to real-world scenarios.
Here are some arithmetic progression solved questions to guide you:
Question 1: The 7th term of an AP is -39/12 and the 15th term is -103/12. What is the 27th term?
Solution: a + 6d = -39/12 - - -(1)
a + 14d = -103/12 - - -(2)
Subtract equations (1) and (2)
a +14d - (a + 6d) = -103/12 - (-39/12)
8d = (-103+39)/12
8d = (-64)/12
d = [(-64)/12] × (1/8)
d = (-8/12)
d = (-2/3)
Substitute d = (-2/3) in equation (1) or (2) , substituting in equation (1)
a + 6d = (-39/12)
a + 6 (-2/3) = (-39/12)
a + (-4) = (-39/12)
a - 4 = (-39/12)
a = (-39/12) + 4
a = (-39+48) /12
a = 9/12=3/4
The 27 th term = a + 26d
27th term = (3/4) + 26(-2/3)
27th term = (3/4) - (52/3)
27 th term = (9–208) /12
27 th term = 199/12
Answer: The 27th term is 199/12.
Question 2: The first term of an AP is 7 and the common difference is 3. Find the 15th Term of an AP.
Given: a = 7, d = 3
Solution: Using the formula Tn = a + (n−1)d:
T15 = 7 + (15−1)×3
T15 = 7 + 42 = 49
Answer: The 15th term is 49.
Question 3: For what value of n, nth terms of the two Arithmetic Progressions (A.P.'s) 63, 65, 67,... and 3, 10, 17,.... will be equal ?
Solution: The formula for nth term of an AP is aₙ = a + (n - 1) d
Here, aₙ is the nth term, a is the first term, d is the common difference and n is the number of terms.
Let the nth term of the two APs be aₙ and aₙ'
Given that the nth term of the two APs are equal.
In first AP 63, 65, 67, . . ., a = 63 , d = 65 - 63 = 2
and in second AP 3, 10, 17, . . ., a = 3, d = 10 - 3 = 7
Then,
aₙ = aₙ'
63 + (n - 1) 2 = 3 + (n - 1) 7......... equation (1)
By Simplifying equation (1)
7(n - 1) - 2(n - 1) = 63 - 3
7n - 7 - 2n + 2 = 60
5n - 5 = 60
n = 65/5
n = 13
Answer: The 13th term of the two given APs are equal.
Question 4: Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution: aₙ = a + (n - 1)d is the nth term of an AP, where aₙ is the nth term, a is the first term, d is a common difference and n is the number of terms.
Let a be the first term and d the common difference.
According to the question, a₃ = 16 and a₇ - a₅ = 12
a + (3 - 1) d = 16
a + 2d = 16 ... equation(1)
Using a₇ - a₅ = 12
[a + (7 - 1) d] - [a + (5 - 1) d] = 12
[a + 6d] - [a + 4d] = 12
2d = 12
d = 6
By substituting this in equation (1), we obtain
a + 2 × 6 = 16
a + 12 = 16
a = 4
Therefore, A.P. will be 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, ...
Answer: The sequence will be 4, 10, 16, 22, ...
By mastering class 10 arithmetic progression questions, students can approach their exams with greater confidence. Whether solving basic sequences or real-world problems, arithmetic progression remains a vital mathematical tool.
(Session 2025 - 26)